[PHP-DB] Array problem...
I am migrating a web app from an old server running PHP 4.2.3 to
a new server running 4.3.8. The problem I am running into at the moment
is building an array and passing it to another page. This code works
fine on the old server, but the array is not coming across to the new
page on the new server. Help...
mysql_select_db($database, $Prod);
$shell_list = getEnumOptions('accounts','shell');
$shell_tmp = select size=\1\ name=\shell[]\\n;
$shell_tmp .= optionDefault Shell/option\n;
$shell_tmp .= option-/option\n;
foreach ($shell_list as $item) {
$shell_tmp .= option$item/option\n;
}
$shell_tmp .= /select\n;
$query_groups = SELECT name FROM grps;
$groups_tmp = mysql_query($query_groups, $Prod) or die(mysql_error());
$grp_list = select size=\1\ name=\grp[]\\n;
$grp_list .= optionPrimary Group/option\n;
$grp_list .= option---/option\n;
while($item2 = mysql_fetch_row($groups_tmp)) {
$grp_list .= option$item2[0]/option\n;
}
$grp_list .= /select\n;
SNIP
?php
while (isset($_POST['system'][$cntr])) {
echo td width=\20%\div align=\center\font
class=\style3\.$_POST['system'][$cntr]./font/div/td;
echo td width=\20%\div
align=\center\.$shell_tmp./div/td;
echo td width=\20%\div
align=\center\.$grp_list./div/td;
echo td width=\20%\div align=\center\input
type=\text\ name=\other[]\ value=\\ size=\15\/div/td/tr;
$tmp = $sbcuid.-.$_POST['system'][$cntr];
$cntr++;
array_push( $accnts, $tmp );
}
?
/table
p align=centerfont class=style3strongIf the systems listed
above are correct, please
proceed. Otherwise, please hit the Back button on your browser and
review your
system selections again./strong/font/pp
align=centernbsp;/p
?php
session_register(accnts, shell, grp, other, cntr);
?
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Problem with foreach looping...
I am attempting to use a simple foreach loop on a query result,
and I am only getting the first element of the array back. Here is the
code...
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results = mysql_query($exclude_query, $Prod);
$exclude = mysql_fetch_array($exclude_results, MYSQL_NUM);
foreach ($exclude as $row) {
echo $row\n;
}
I don't understand how I am just getting a single item back from
this. The query actually returns all 13 entries in the exclusion table.
I have tested the query via the mysql client. I also get all 13
elements back if I loop via the following.
while ($test = mysql_fetch_array($results, MYSQL_NUM)) {
echo $test[0]\n;
}
Is my PHP somehow broke, or am I just missing something here?
Thanks.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] MySQL Subquery problems...
I am working with one of our Oracle DBAs to have him try and help me get a couple of queries working. The problem is that in looking at the documentation for MySQL it seems that what we are attempting should work. Unfortunately, it doesn't. Here is what we are trying... mysql SELECT CPU_Hostname FROM AllMid_Data - WHERE CPU_Hostname NOT IN - (SELECT hostname FROM exclusion); ERROR 1064: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT hostname FROM exclusion)' at line 3 This is an error returned from the mysql client. Any help would be most appreciated. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] MySQL Subquery problems...
Yup... Thanks, I just discovered that myself. Thankfully the docs have instructions on how to write a subquery for earlier versions that don't support subqueries. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -Original Message- From: Bastien Koert [mailto:[EMAIL PROTECTED] Sent: Wednesday, March 30, 2005 9:35 AM To: NIPP, SCOTT V (SBCSI); php-db@lists.php.net Subject: RE: [PHP-DB] MySQL Subquery problems... only mysql 4.1+ supports subqueries..if you version is lower it won;t work Bastien From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: php-db@lists.php.net Subject: [PHP-DB] MySQL Subquery problems... Date: Wed, 30 Mar 2005 09:25:36 -0600 I am working with one of our Oracle DBAs to have him try and help me get a couple of queries working. The problem is that in looking at the documentation for MySQL it seems that what we are attempting should work. Unfortunately, it doesn't. Here is what we are trying... mysql SELECT CPU_Hostname FROM AllMid_Data - WHERE CPU_Hostname NOT IN - (SELECT hostname FROM exclusion); ERROR 1064: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT hostname FROM exclusion)' at line 3 This is an error returned from the mysql client. Any help would be most appreciated. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Further MySQL query problems...
I am having trouble in the following code. Basically, I want to
query two separate database tables and then generate an array from these
two queries. The first query will return a list of ALL systems. The
second query will return a list of systems that should be excluded. The
problem is that this is getting me nowhere. Here is the pertinent
code...
$query = SELECT CPU_Hostname FROM AllMid_Data
WHERE CPU_IN_SVC = \Y\
AND CPU_DNS = \sbcld.sbc.com\
ORDER BY CPU_Hostname ASC;
$results = mysql_query($query, $Prod) or die(mysql_error());
$system = mysql_fetch_array($results, MYSQL_NUM);
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results = mysql_query($exclude_query, $Prod);
$exclude = mysql_fetch_array($exclude_results, MYSQL_NUM);
$temp = array_diff($system, $exclude);
?php do {
$sys = $blah[0];
echo tr; ?
?php echo $sys; ?
?php echo /tr; } ?
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
I have cut out a lot of extraneous stuff that I don't think is
pertinent. Please let me know if you see something causing my problems.
The first query table has about 900 entries and the query returns about
90 hostnames. The exclude table query has only 13 hostnames and
obviously returns all. I am looking for a way to display the
approximately 77 entries that are NOT in the exclude table. Thanks
again.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] RE: Further MySQL query problems...
Oops... Correction to the last line of the script.
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
Should be...
?php } while ($blah = array_shift($temp)); ?
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
-Original Message-
From: NIPP, SCOTT V (SBCSI)
Sent: Monday, March 28, 2005 10:47 AM
To: 'php-db@lists.php.net'
Subject: Further MySQL query problems...
I am having trouble in the following code. Basically, I want to
query two separate database tables and then generate an array from
these two queries. The first query will return a list of ALL systems.
The second query will return a list of systems that should be
excluded. The problem is that this is getting me nowhere. Here is
the pertinent code...
$query = SELECT CPU_Hostname FROM AllMid_Data
WHERE CPU_IN_SVC = \Y\
AND CPU_DNS = \sbcld.sbc.com\
ORDER BY CPU_Hostname ASC;
$results = mysql_query($query, $Prod) or die(mysql_error());
$system = mysql_fetch_array($results, MYSQL_NUM);
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results = mysql_query($exclude_query, $Prod);
$exclude = mysql_fetch_array($exclude_results, MYSQL_NUM);
$temp = array_diff($system, $exclude);
?php do {
$sys = $blah[0];
echo tr; ?
?php echo $sys; ?
?php echo /tr; } ?
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
I have cut out a lot of extraneous stuff that I don't think is
pertinent. Please let me know if you see something causing my
problems. The first query table has about 900 entries and the query
returns about 90 hostnames. The exclude table query has only 13
hostnames and obviously returns all. I am looking for a way to
display the approximately 77 entries that are NOT in the exclude
table. Thanks again.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] RE: Further MySQL query problems...
If I could figure out how to structure the query, I would be
happy to do so. The problem is that I am working with 3 different
tables for this. Additionally, I ran into a brick wall when trying to
structure the query to remove/eliminate the entries in the exclusion
table. The exclusion table has 13 entries currently, and the last thing
I tried with the query was outputting 13 entries of every system. Here
is the query that was yielding this...
#$query = SELECT AllMid_Data.CPU_Hostname FROM AllMid_Data, exclusion
# WHERE AllMid_Data.CPU_IN_SVC = \Y\
# AND AllMid_Data.CPU_DNS = \sbcld.sbc.com\
# AND AllMid_Data.CPU_Hostname != exclusion.hostname
# ORDER BY CPU_Hostname ASC;
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
-Original Message-
From: Bastien Koert [mailto:[EMAIL PROTECTED]
Sent: Monday, March 28, 2005 11:13 AM
To: NIPP, SCOTT V (SBCSI); php-db@lists.php.net
Subject: RE: [PHP-DB] RE: Further MySQL query problems...
why wouldn't you just run a query to return only the elements that you
want?
Bastien
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] RE: Further MySQL query problems...
Date: Mon, 28 Mar 2005 10:49:24 -0600
Oops... Correction to the last line of the script.
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
Should be...
?php } while ($blah = array_shift($temp)); ?
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
-Original Message-
From: NIPP, SCOTT V (SBCSI)
Sent: Monday, March 28, 2005 10:47 AM
To: 'php-db@lists.php.net'
Subject:Further MySQL query problems...
I am having trouble in the following code. Basically, I want to
query two separate database tables and then generate an array from
these two queries. The first query will return a list of ALL
systems.
The second query will return a list of systems that should be
excluded. The problem is that this is getting me nowhere. Here is
the pertinent code...
$query = SELECT CPU_Hostname FROM AllMid_Data
WHERE CPU_IN_SVC = \Y\
AND CPU_DNS = \sbcld.sbc.com\
ORDER BY CPU_Hostname ASC;
$results = mysql_query($query, $Prod) or die(mysql_error());
$system = mysql_fetch_array($results, MYSQL_NUM);
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results = mysql_query($exclude_query, $Prod);
$exclude = mysql_fetch_array($exclude_results, MYSQL_NUM);
$temp = array_diff($system, $exclude);
?php do {
$sys = $blah[0];
echo tr; ?
?php echo $sys; ?
?php echo /tr; } ?
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
I have cut out a lot of extraneous stuff that I don't think is
pertinent. Please let me know if you see something causing my
problems. The first query table has about 900 entries and the query
returns about 90 hostnames. The exclude table query has only 13
hostnames and obviously returns all. I am looking for a way to
display the approximately 77 entries that are NOT in the exclude
table. Thanks again.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] RE: Further MySQL query problems...
OK. Maybe I am going about this in the wrong manner. Here is
the SQL statement that I am starting out with.
$query = SELECT Hostname FROM Data
WHERE CPU_IN_SVC = \Y\
AND CPU_DNS = \blah.com\
ORDER BY Hostname ASC;
This query provides me a complete list of all the systems in our
organization. The following query provides me with a list of systems in
an exclusion table that should not be displayed.
$exclude_query = SELECT hostname FROM exclusion;
This one is obviously quite simple and straight-forward. The
next query is one that will generate a list of systems on which the user
already has an account.
$query2= SELECT Data.Hostname FROM Data, acct_db
WHERE Data.CPU_IN_SVC = \Y\
AND Data.CPU_DNS = \blah.com\
AND acct_db.key1 LIKE '$var%'
AND acct_db.key1 LIKE CONCAT(\%\, Data.Hostname)
ORDER BY Hostname ASC;
This query is very much like the first, but it also looks into
the acct_db table to match userids and hostnames. This is currently a
separate query because this data is used to display information in a
separate section of the web page.
$query3= SELECT Data.Hostname FROM Data, accounts
WHERE Data.CPU_IN_SVC = \Y\
AND Data.CPU_DNS = \sbcld.sbc.com\
AND accounts.uid = '$var'
AND accounts.system = Data.Hostname
AND accounts.ctime IS NULL
ORDER BY Hostname ASC;
This query is also similar to the first , but it looks into the
accounts table to determine if the user has already submitted an account
request for the server, AND that this account has not yet been created
(ctime IS NULL). Like the query above, this is a separate query since
it is used to populate a different section of the web page.
I do not know if it is possible to somehow combine all 4 of
these queries into a single complex query. This would be preferred in
my mind as it would let the server do more work in fewer queries. What
I would like to do is have a single query that would return the
hostnames with some kind of key that could then be used to determine
what criteria the query matched to specifically. I am guessing that
this would be a numeric value that would be a secondary array element
associated with the hostname. I would then write my display code to
check the value of this element to determine how/where to display the
data.
Thanks in advance for any help in this. I have no idea if this
is possible, but I am guessing that it is.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
-Original Message-
From: Bastien Koert [mailto:[EMAIL PROTECTED]
Sent: Monday, March 28, 2005 11:13 AM
To: NIPP, SCOTT V (SBCSI); php-db@lists.php.net
Subject: RE: [PHP-DB] RE: Further MySQL query problems...
why wouldn't you just run a query to return only the elements that you
want?
Bastien
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] RE: Further MySQL query problems...
Date: Mon, 28 Mar 2005 10:49:24 -0600
Oops... Correction to the last line of the script.
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
Should be...
?php } while ($blah = array_shift($temp)); ?
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
-Original Message-
From: NIPP, SCOTT V (SBCSI)
Sent: Monday, March 28, 2005 10:47 AM
To: 'php-db@lists.php.net'
Subject:Further MySQL query problems...
I am having trouble in the following code. Basically, I want to
query two separate database tables and then generate an array from
these two queries. The first query will return a list of ALL
systems.
The second query will return a list of systems that should be
excluded. The problem is that this is getting me nowhere. Here is
the pertinent code...
$query = SELECT CPU_Hostname FROM AllMid_Data
WHERE CPU_IN_SVC = \Y\
AND CPU_DNS = \sbcld.sbc.com\
ORDER BY CPU_Hostname ASC;
$results = mysql_query($query, $Prod) or die(mysql_error());
$system = mysql_fetch_array($results, MYSQL_NUM);
$exclude_query = SELECT hostname FROM exclusion;
$exclude_results = mysql_query($exclude_query, $Prod);
$exclude = mysql_fetch_array($exclude_results, MYSQL_NUM);
$temp = array_diff($system, $exclude);
?php do {
$sys = $blah[0];
echo tr; ?
?php echo $sys; ?
?php echo /tr; } ?
?php } while ($blah = mysql_fetch_array($exclude_results)); ?
I have cut out a lot of extraneous stuff that I don't think is
pertinent. Please let me know if you see something causing my
problems. The first query table has about 900 entries and the query
returns about 90
RE: [PHP-DB] MySQL query problems...
OK. I think I have resolved this issue. The problem I am
having now is that I have a simple table that is a list of systems to
exclude from the display. I am having trouble figuring out how to
structure my query to actually exclude the systems in this table. Here
is what I tried, and it gives me 13 instances of every entry, 13 is the
number of hostnames in the exclusion table.
$query = SELECT AllMid_Data.CPU_Hostname FROM AllMid_Data, exclusion
WHERE AllMid_Data.CPU_IN_SVC = \Y\
AND AllMid_Data.CPU_DNS = \sbcld.sbc.com\
AND AllMid_Data.CPU_Hostname != exclusion.hostname
ORDER BY CPU_Hostname ASC;
$results = mysql_query($query, $Prod) or die(mysql_error());
$system = mysql_fetch_assoc($results);
Additionally, what I am actually working towards is that if a
system shows up in either of the other queries it gets excluded from
this one. I have not tried anything to make that happen yet. I will
tackle that challenge once I can resolve this part. Thanks again.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
-Original Message-
From: J. Connolly [mailto:[EMAIL PROTECTED]
Sent: Thursday, March 24, 2005 7:36 AM
Cc: php-db@lists.php.net
Subject: Re: [PHP-DB] MySQL query problems...
yeah...errora are important. I do not escape my quotation marks in a
query. I use single quotes within my insert and select statements.
If it says No valis MySQL resource then you are returning 0 records
which may be caused by escaping the quotations.
I follow this pattern
$sql =INSERT INTO table VALUES ('some value','another value','etc');
NIPP, SCOTT V (SBCSI) wrote:
I am getting errors for the following queries. This query seems
to work fine in phpMyAdmin.
$query2 = SELECT AllMid_Data.CPU_Hostname FROM AllMid_Data, accounts
WHERE AllMid_Data.CPU_IN_SVC = \Y\
AND AllMid_Data.CPU_DNS = \sbcld.sbc.com\
AND accounts.sbcuid = $sbcuid
AND accounts.system = AllMid_Data.CPU_Hostname
AND accounts.ctime IS NULL
ORDER BY CPU_Hostname ASC;
$results2 = mysql_query($query2, $Prod) or die(mysql_error());
$system2 = mysql_fetch_assoc($results2);
$query3 = SELECT AllMid_Data.CPU_Hostname FROM AllMid_Data, acct_db
WHERE AllMid_Data.CPU_IN_SVC = \Y\
AND AllMid_Data.CPU_DNS = \sbcld.sbc.com\
AND acct_db.key1 LIKE \$sbcuid%\
AND acct_db.key1 LIKE \%\.AllMid_Data.CPU_Hostname
ORDER BY CPU_Hostname ASC;
$results3 = mysql_query($query3, $Prod) or die(mysql_error());
$system3 = mysql_fetch_assoc($results3);
I am assuming that I have an issue with query3. Thanks in
advance.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] MySQL query problems...
I am getting errors for the following queries. This query seems to work fine in phpMyAdmin. $query2 = SELECT AllMid_Data.CPU_Hostname FROM AllMid_Data, accounts WHERE AllMid_Data.CPU_IN_SVC = \Y\ AND AllMid_Data.CPU_DNS = \sbcld.sbc.com\ AND accounts.sbcuid = $sbcuid AND accounts.system = AllMid_Data.CPU_Hostname AND accounts.ctime IS NULL ORDER BY CPU_Hostname ASC; $results2 = mysql_query($query2, $Prod) or die(mysql_error()); $system2 = mysql_fetch_assoc($results2); $query3 = SELECT AllMid_Data.CPU_Hostname FROM AllMid_Data, acct_db WHERE AllMid_Data.CPU_IN_SVC = \Y\ AND AllMid_Data.CPU_DNS = \sbcld.sbc.com\ AND acct_db.key1 LIKE \$sbcuid%\ AND acct_db.key1 LIKE \%\.AllMid_Data.CPU_Hostname ORDER BY CPU_Hostname ASC; $results3 = mysql_query($query3, $Prod) or die(mysql_error()); $system3 = mysql_fetch_assoc($results3); I am assuming that I have an issue with query3. Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Working with multiple tables...
I am working on revising a web request app that I have already developed. The app basically provides a list of system names with checkboxes for a user to select to request accounts on the systems. One feature I am intending to add is that the app should check a separate database of user accounts to see if the user already has an account on the system. I think I could get it to work by making a query of the userid for each system, but this seems quite inefficient. I would rather somehow have a single query that gets a list of system names and also checks the user account database at the same time to determine if the user already has an account. Then I would like to display ALL of the systems, but grey out the ones that the user already has accounts on. Can someone help me figure out the best way to go about this? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Working with multiple tables...
Within the same server, even the same database, just a different table. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -Original Message- From: Micah Stevens [mailto:[EMAIL PROTECTED] Sent: Monday, March 21, 2005 1:47 PM To: php-db@lists.php.net Subject: Re: [PHP-DB] Working with multiple tables... Seperate on another MySQL server, or seperate within the same MySQL server? On Monday 21 March 2005 11:44 am, NIPP, SCOTT V (SBCSI) wrote: I am working on revising a web request app that I have already developed. The app basically provides a list of system names with checkboxes for a user to select to request accounts on the systems. One feature I am intending to add is that the app should check a separate database of user accounts to see if the user already has an account on the system. I think I could get it to work by making a query of the userid for each system, but this seems quite inefficient. I would rather somehow have a single query that gets a list of system names and also checks the user account database at the same time to determine if the user already has an account. Then I would like to display ALL of the systems, but grey out the ones that the user already has accounts on. Can someone help me figure out the best way to go about this? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] MySQL error...
No... That is just some strange error generated by the cut and paste. The IS NOT NU LL that is... The other issue with 'tablename'... I simply typed 'tablename' in the e-mail because I didn't remember the exact tablename when I was typing the e-mail. It does actually provide the name of the table that the script is working with. Thanks for the feedback. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -Original Message- From: Norland, Martin [mailto:[EMAIL PROTECTED] Sent: Friday, December 10, 2004 2:34 PM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] MySQL error... -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED] I have been using a PHP page to update a database table for a long time now. Unfortunately, I have noticed that frequently when I perform an update I get back an error saying Table 'tablename' doesn't exist. This is becoming more and more annoying. [snip] $query = SELECT `id_sys`, atime, gid, shell FROM accounts WHERE atime IS NOT NU LL AND ctime IS NULL ORDER BY rtime ASC; $result = mysql_query($query, $Prod) or die(mysql_error()); ? snip Strange behavior, can't say I've ever heard of/seen it - however... IS NOT NU LL is not, strictly speaking, valid sql. If that's not just a weird line wrapping issue - that may point to part of your problem. If the error you're getting back is literally Table 'tablename' doesn't exist - then you're looking in the wrong code, and somewhere you have code that says tablename instead of $tablename. You may want to grep for tablename to try to track that down. Cheers, - Martin Norland, Database / Web Developer, International Outreach x3257 The opinion(s) contained within this email do not necessarily represent those of St. Jude Children's Research Hospital. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] MySQL error...
I have been using a PHP page to update a database table for a
long time now. Unfortunately, I have noticed that frequently when I
perform an update I get back an error saying Table 'tablename' doesn't
exist. This is becoming more and more annoying. The table obviously
exists as the page that I hit the update button on is populated from
this same page. Additionally, if I backout and reload the page most of
the time it will work. Any help would be greatly appreciated.
Mysql 3.23.49
PHP 4.2.3
?php require_once('useraccounts.lib.php');
session_start();
if (!isset($_SESSION['valid_user']))
{
echo You must be logged in to use this application. br;
$return_url = $_SERVER['PHP_SELF'];
session_register(return_url);
echo Please a href=\../sa_login.php\ login/a now. br;
exit();
} else {
$sbcuid = $valid_user;
}
while(isset($entry[0])) {
$tmp = $entry[0];
$update = UPDATE accounts SET ctime=NOW() WHERE id_sys='.$tmp.';
$results = mysql_query($update, $Prod) or die(mysql_error());
array_shift($entry);
}
mysql_select_db($database, $Prod);
$query = SELECT `id_sys`, atime, gid, shell FROM accounts WHERE atime
IS NOT NU
LL AND ctime IS NULL ORDER BY rtime ASC;
$result = mysql_query($query, $Prod) or die(mysql_error());
?
html
head
titleSBCLD User Account Request System/title
meta http-equiv=Content-Type content=text/html; charset=iso-8859-1
/head
body bgcolor=#CC
p align=centerfont color=#0033CC
size=6strongAdministration/strong
/font/p
p align=centerfont size=4Here are the details for the accounts
approved
and pending creation:/font/p
p align=center
form method=post action=account_details.php
table width=90% bgcolor=#CC cellspacing=0
tr bordercolor=#CC
td width=20% height=23div align=centerfont
color=#CC size=
3Requesting
User/font/div/td
td width=10% valign=topdiv align=centerfont
color=#CCSyste
m/font/div/td
td width=10% valign=topdiv align=centerfont
color=#CCPrima
ry Group/font/div/td
td width=10% valign=topdiv align=centerfont
color=#CCD
efault Shell/font/div/td
td width=30% valign=topdiv align=centerfont
color=#CCReque
st Time/font/div/td
td width=10% valign=topdiv align=centerfont
color=#CCC
ompleted/font/div/td
/tr
?php
do {
$entry = $list['id_sys'];
$id = split('-', $list['id_sys']);
$sbcuid = $id[0];
$sys = $id[1];
if (isset($list['id_sys'])) {
echo trtd width=\20%\div
align=\center\.$sbcuid./div/t
d;
echo td width=\10%\div
align=\center\.$sys./div/td;
echo td width=\15%\div
align=\center\.$list['gid']./div/
td;
echo td width=\15%\div
align=\center\.$list['shell']./div
/td;
echo td width=\30%\div
align=\center\.$list['atime']./div
/td;
echo td width=\10%\div align=\center\input
name=\entry[]\
type=\checkbox\ value=$entry/div/td/tr;
}
} while ($list = mysql_fetch_assoc($result));
?
/table
p align=centerPlacing a check in the completed box will update the
databas
e entry for
this request with a completed time and remove this entry from this
page upon
clicking the Update button below./p
div align=center
input type=submit value=Update
/div
/form
/body
/html
Thanks in advance. This is really beginning to bug the crap out
of me.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] Time Field
Why would you not simply use a 'datetime' field type? This gets you exactly what you are looking for unless I am misreading what you are doing here... Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -Original Message- From: Bomgardner, Mark A [mailto:[EMAIL PROTECTED] Sent: Thursday, October 28, 2004 10:10 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Time Field When creating a time field in MySQL, its purpose is for elapsed time, not time of day. Is it better to use this field or to combine it with the date field when looking at inserting the start date and time of an event? Mark A. Bomgardner Technology Specialist KLETC -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Auto-increment questions...
I am working on a database application for Unix user accounts. I want to be able to have a system that will provide me the next available numeric user ID. I have created a 2 column index table that simply includes UID and user name. This field is autoincrement and the primary key for the table. Currently this field only contains a very small sample of users. Basically, I have numeric IDs 3000, 3001, 3008, and 3028. My problem is that I am not sure how to query this table to get the next available numeric ID which is 3002. If I do an insert into this table without specifying a value for the auto_increment column it comes up with 3029. Is there a way to query and/or insert such that it comes up with the next unused numeric value? Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Auto-increment questions...
OK. The problem is I don't want the next highest number. There are gaps in the UID sequence. I need to find the next UNUSED number in the sequence which is rarely the highest number. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -Original Message- From: Bastien Koert [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 12:45 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Subject: RE: [PHP-DB] Auto-increment questions... If its an autoincrement, the next highest number will be assigned by the db. There is no need to query the db to find it. Simply insert the record and leave the id field out of the insert statement. bastien From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: [PHP-DB] Auto-increment questions... Date: Tue, 5 Oct 2004 11:58:41 -0500 I am working on a database application for Unix user accounts. I want to be able to have a system that will provide me the next available numeric user ID. I have created a 2 column index table that simply includes UID and user name. This field is autoincrement and the primary key for the table. Currently this field only contains a very small sample of users. Basically, I have numeric IDs 3000, 3001, 3008, and 3028. My problem is that I am not sure how to query this table to get the next available numeric ID which is 3002. If I do an insert into this table without specifying a value for the auto_increment column it comes up with 3029. Is there a way to query and/or insert such that it comes up with the next unused numeric value? Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php _ Scan and help eliminate destructive viruses from your inbound and outbound e-mail and attachments. http://join.msn.com/?pgmarket=en-capage=byoa/premxAPID=1994DI=1034SU =http://hotmail.com/encaHL=Market_MSNIS_Taglines Start enjoying all the benefits of MSN(r) Premium right now and get the first two months FREE*. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] LAST_INSERT_ID?????
OK. I know that a BUNCH of people have had this question, and I have read numerous Google threads on this, but nothing really seems to point out the resolution... I have a table with the Primary Key as an auto_increment field. Once I insert a new row of data into the table I want to direct the user to a Submitted page. I am trying to pull the entry number to display for the user as a ticket number. I would assume that I should use the LAST_INSERT_ID function for this. Unfortunately, I am getting either failures or all zeroes for this output. I can't even get this function to work in a SQL query window. Please help!!! I would attach code to this, but I am not sure what to attach. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Array sorting question...
I am trying to populate an array from a MySQL database. I think I
have the code correct to populate the array, but I need to somehow sort the
array eliminating duplicate values. I think I can loop through the array
doing a comparison and building a new array by discarding the value if it is
a duplicate, but I was not sure if there was a more efficient way of doing
this. I am already looping through the database query results, and I am
just thinking about efficiency here.
Here is a code sample with a little more description of what I am
attempting...
?php
do {
$entry = $list['id_sys'];
$id = split('-', $list['id_sys'], 1);
$sbcuid = $id[0];
$users[] = '$sbcuid';
} while ($list = mysql_fetch_assoc($result));
?
This query pulls a value from the database that is a
username-hostname value. What I am working on is building a page that has a
unique list of usernames that link to a second page of user account
information. There may be twenty user account on different hosts for the
same username. I only want to have the username link created once on this
page. I hope that makes since to everyone. Thanks in advance for any help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Parse Errors...
I keep receiving a parse error every time I try and view the page I am working on. I am developing this very simple application in DreamWeaver MX 2004. Here are lines 3-16: ?php mysql_select_db($database_Prod, $Prod); $query_Host = SELECT Name FROM systems; $Host = mysql_query($query_Host, $Prod) or die(mysql_error()); $row_Host = mysql_fetch_assoc($Host); $totalRows_Host = mysql_num_rows($Host); mysql_select_db($database_ProdUsers, $ProdUsers); $query_UserMod = SELECT shell, home_dir FROM modification; $UserMod = mysql_query($query_UserMod, $ProdUsers) or die(mysql_error()); $row_UserMod = mysql_fetch_assoc($UserMod); $totalRows_UserMod = mysql_num_rows($UserMod); ? I am simply attempting to pull data from two separate tables in two separate databases. Every time I try to view my work, I am getting a parse error on the 'mysql_select_db($database_ProdUsers, $ProdUsers);' regardless of whether this is the first or second mysql_select_db. I imagine that I am simply doing something obviously wrong here. Please help. Thanks. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Logic Help...
Thanks for all the help. I was finally able to get this working
properly by viewing the variable data as suggested. I then ran into another
minor issue with session variables keeping their old values on a second pass
through the application, but I resolved this by unregistering these
variables before sending the user back through the application. One strange
thing I noticed was that you can register multiple variables with a single
session_register, but you have to unregister only a single variable at a
time. Anyone know anything more about this? Am I doing something wrong, or
is it this way by design? Thanks.
-Original Message-
From: CPT John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Wednesday, August 27, 2003 4:10 PM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Logic Help...
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
Thanks for the feedback. This seems to have helped, but something
is still not working with the logic. Here is the updated code:
?php
while($cntr != 0) {
$cntr--;
if (($shell[$cntr] == ) || ($grp[$cntr] == )) {
echo p align=\center\font color=\#99\
size=\4\Sorry, but your request was not filled out completely. Please
resubmit this request ensuring that you have selected a Primary Group and
Default Shell./font/p;
echo p align=\center\font color=\#99\ size=\4\Please
follow this a
href=\http://ldsa.sbcld.sbc.com/DW/NewUser/newuser.php\;link/a to
return
to the beginning of the request process./font/p;
exit;
}
}
?
Now, the test conditions never seem to enter into this portion of
code. I am unable to figure out why at this point. Thanks again for the
help.
Okay... let's do some basic debugging now. What does print_r($shell) and
print_r($grp) display? Are there any entries that are equal to an empty
string?
---John Holmes...
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RE: [PHP-DB] Logic Help...
Thanks for the feedback. This seems to have helped, but something
is still not working with the logic. Here is the updated code:
?php
while($cntr != 0) {
$cntr--;
if (($shell[$cntr] == ) || ($grp[$cntr] == )) {
echo p align=\center\font color=\#99\
size=\4\Sorry, but your request was not filled out completely. Please
resubmit this request ensuring that you have selected a Primary Group and
Default Shell./font/p;
echo p align=\center\font color=\#99\ size=\4\Please
follow this a
href=\http://ldsa.sbcld.sbc.com/DW/NewUser/newuser.php\;link/a to return
to the beginning of the request process./font/p;
exit;
}
}
?
Now, the test conditions never seem to enter into this portion of
code. I am unable to figure out why at this point. Thanks again for the
help.
-Original Message-
From: CPT John W. Holmes [mailto:[EMAIL PROTECTED]
Sent: Tuesday, August 26, 2003 9:15 AM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Logic Help...
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
This isn't specifically a DB related question, but... The following
code snippet is not functioning for me. I keep receiving a parse error on
the if line. Basically I want to stop execution of this page if either
of
the conditions in the if statement exist. Thanks in advance.
# while($cntr != 0) {
#if (($shell[$cntr] eq ) || ($grp[$cntr] eq Primary Gr)) {
Umm... you don't use 'eq' in PHP, you use == (that's two equal signs).
---John Holmes...
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[PHP-DB] Logic Help...
This isn't specifically a DB related question, but... The following
code snippet is not functioning for me. I keep receiving a parse error on
the if line. Basically I want to stop execution of this page if either of
the conditions in the if statement exist. Thanks in advance.
# while($cntr != 0) {
#if (($shell[$cntr] eq ) || ($grp[$cntr] eq Primary Gr)) {
# echo p align=\center\font color=\#99\
size=\4\Sorry, but your request was not filled out completely. Please
resubmit this request ensuring that you have selected a Primary Group and
Default Shell./font/p;
# echo p align=\center\font color=\#99\ size=\4\Please
follow this a
href=\http://ldsa.sbcld.sbc.com/sa_bulletin/sa_board.htm\;link/a to
return to the beginning of the request process./font/p;
# exit;
# }
# }
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] PHP/MySQL Default Field Value...
I have a script that is going into semi-production and during this
process a new problem was uncovered. I have the table a couple of table
fields setup as Not NULL with a Default value. The problem is that if
the user fails to select anything I get garbage in one of the fields and
nothing in the other. I think this might be due to the PHP variable
grabbing something, but I am not sure what. Here is the pertinent code.
{ Code snippet }
$a = 0;
$accnts = array();
$shells = array();
$grps = array();
$others = array();
mysql_select_db($database, $Prod);
$shell_list = getEnumOptions('accounts','shell');
$shell_tmp = select size=\1\ name=\shell[]\\n;
$shell_tmp .= optionDefault Shell/option\n;
foreach ($shell_list as $item) {
$shell_tmp .= option$item/option\n;
}
$shell_tmp .= /select\n;
$query_groups = SELECT name FROM grps;
$groups_tmp = mysql_query($query_groups, $Prod) or die(mysql_error());
$grp_list = select size=\1\ name=\grp[]\\n;
$grp_list .= optionPrimary Group/option\n;
$grp_list .= option---/option\n;
while($name = mysql_fetch_row($groups_tmp)) {
$grp_list .= option$name[0]/option\n;
}
$grp_list .= /select\n;
{ Code snippet }
form name=form1 method=post action=account_final.php
?php
while (isset($_POST['system'][$a])) {
echo td width=\20%\div align=\center\font
color=\99\.$_POST['system'][$a]./font/div/td;
echo td width=\20%\div
align=\center\.$shell_tmp./div/td;
echo td width=\20%\div
align=\center\.$grp_list./div/td;
echo td width=\20%\div align=\center\input type=\text\
name=\other[]\ value=\\ size=\15\/div/td/tr;
$tmp = $sbcuid.-.$_POST['system'][$a];
$a++;
array_push( $accnts, $tmp );
}
?
Thanks in advance for any help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] RE: Session help...
Thanks for the responses. I ended up resolving this by fixing the
form destination. I had hard coded a location in there, and did not realize
this. Once I replaced this with the PHP variable everything magically
started to work. My bad on that. Thanks again for the feedback though.
-Original Message-
From: Ow Mun Heng [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 29, 2003 11:13 PM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: RE: Session help...
As mentioned previously in this list,
$_SESSION and session_register, session_is_register, is not compatible with
each other.
Anyway, I used your idea as inspiration for my own code.. My code works..
Try to change From:
if (!session_is_registered(valid_user)) {
To:
if ($_SESSION['valid_user']){
//your code
}
Cheers,
Mun Heng, Ow
H/M Engineering
Western Digital M'sia
DID : 03-7870 5168
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 29, 2003 9:19 PM
To: [EMAIL PROTECTED]
Subject: Session help...
Sorry for the slightly off-topic post... I have a couple pages that
use a login page. If the user tries to bring up one of these pages without
being logged in, then the pages redirects him to the login page. After
successful login, the user is directed back to the page they were attempting
to originally load. For some reason, this is not working and I have been
looking at this off and on for two days now without much luck. The first
section of code is the snippet that directs users to the login page, and the
second snippet is the code from the login page that sends users back.
First code snippet...
session_start();
if ($sbcuid $passwd) {
mysql_select_db($database, $Prod);
$query = select * from contacts_sa
. where sbcuid='$sbcuid' and passwd='$passwd';
$result = mysql_query($query, $Prod) or die(mysql_error());
# $data = mysql_fetch_assoc($result);
$test = mysql_num_rows($result);
if (mysql_num_rows($result) 0 )
{
$valid_user = $sbcuid;
$_SESSION['valid_user'] = $sbcuid;
}
}
if (!session_is_registered(valid_user)) {
$return_url = $_SERVER['PHP_SELF'];
$_SESSION['return_url'] = $return_url;
header('Location: http://ldsa.sbcld.sbc.com/DW/sa_login.php');
exit();
} else {
$sbcuid = $valid_user;
}
End of first snippet...
Second code snippet...
session_start();
if ($sbcuid $passwd) {
if (isset($_SESSION['return_url'])) {
$link = $_SESSION['return_url'];
} else {
$link = 'oncall_log.php';
}
mysql_select_db($database, $Prod);
$query = select * from contacts_sa
. where sbcuid='$sbcuid' and passwd='$passwd';
$result = mysql_query($query, $Prod) or die(mysql_error());
# $data = mysql_fetch_assoc($result);
$test = mysql_num_rows($result);
if (mysql_num_rows($result) 0 ) {
$valid_user = $sbcuid;
# session_register(valid_user);
$_SESSION['valid_user'] = $sbcuid;
header(Location: http://.$_SERVER['HTTP_HOST'].$link);
}
}
End of second snippet...
Thanks in advance for any help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Session help...
Sorry for the slightly off-topic post... I have a couple pages that
use a login page. If the user tries to bring up one of these pages without
being logged in, then the pages redirects him to the login page. After
successful login, the user is directed back to the page they were attempting
to originally load. For some reason, this is not working and I have been
looking at this off and on for two days now without much luck. The first
section of code is the snippet that directs users to the login page, and the
second snippet is the code from the login page that sends users back.
First code snippet...
session_start();
if ($sbcuid $passwd) {
mysql_select_db($database, $Prod);
$query = select * from contacts_sa
. where sbcuid='$sbcuid' and passwd='$passwd';
$result = mysql_query($query, $Prod) or die(mysql_error());
# $data = mysql_fetch_assoc($result);
$test = mysql_num_rows($result);
if (mysql_num_rows($result) 0 )
{
$valid_user = $sbcuid;
$_SESSION['valid_user'] = $sbcuid;
}
}
if (!session_is_registered(valid_user)) {
$return_url = $_SERVER['PHP_SELF'];
$_SESSION['return_url'] = $return_url;
header('Location: http://ldsa.sbcld.sbc.com/DW/sa_login.php');
exit();
} else {
$sbcuid = $valid_user;
}
End of first snippet...
Second code snippet...
session_start();
if ($sbcuid $passwd) {
if (isset($_SESSION['return_url'])) {
$link = $_SESSION['return_url'];
} else {
$link = 'oncall_log.php';
}
mysql_select_db($database, $Prod);
$query = select * from contacts_sa
. where sbcuid='$sbcuid' and passwd='$passwd';
$result = mysql_query($query, $Prod) or die(mysql_error());
# $data = mysql_fetch_assoc($result);
$test = mysql_num_rows($result);
if (mysql_num_rows($result) 0 ) {
$valid_user = $sbcuid;
# session_register(valid_user);
$_SESSION['valid_user'] = $sbcuid;
header(Location: http://.$_SERVER['HTTP_HOST'].$link);
}
}
End of second snippet...
Thanks in advance for any help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] MySQL Date insert
I actually think I know part of this answer... I don't think you
need to define a variable to insert into your date field. Inserting an
empty value into this filed will populate the field in the database with the
current time, which it appears from your code is what you are trying to do
here. Try changing the field name and give that a shot. I think I am
actually right here though.
Scott
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]
Sent: Monday, July 28, 2003 1:19 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] MySQL Date insert
Don't know about your date format, but you have named your table column DATE
which is a reserved keyword in MySQL. Try naming that something different
and see if you still get the error.
-Original Message-
From: Andrew D. Luebke [mailto:[EMAIL PROTECTED]
Sent: Monday, July 28, 2003 2:17 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] MySQL Date insert
Hello,
I have the following PHP code:
$date = date(Y-m-d H:i:s);
$result = mysql_query(INSERT INTO Boats (Make,
Model, Serial,
Stock, Extension, Cust_Name, Store, Date) VALUES
('$make', '$model', '$serial', '$stock',
'$extension',
'$name', '$store', '$date')
or die(Invalid query: . mysql_error() .
/body/html);
The problem is with the Date column. It is of type DATETIME.
I get an
error on the second line of the insert statement. I assume
I'm not putting
formatting the date-time variable correctly but I've tried
everything I can
think of, so any help is very much appreciated. Thanks.
Andrew.
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[PHP-DB] 4.2.3 session login problem...
I have an application that is currently working fine under 4.1.2, and I am in the process of upgrading to 4.2.3. The problem is that after upgrading PHP the application login no longer works. Additionally, this same application works fine on another machine running 4.2.3. The only difference in the machines is that the working one uses PHP as a static module and the broken one uses PHP as a loadable module. Anyone run into this problem before? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 2:14 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 Are you sure $td has a value? If it's blank, you'd get an error like that. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
OK. I guess the next question would be what is the best option for performance? This portion of the app currently doesn't have any session functionality built in. I would be setting up the session for this sole purpose. I haven't worked with cookies yet. I suppose the URL method might work, I have just never like it because I think it looks unclean, just personal preference there. As to the $td question... This is a variable that holds the userid of the Technical Director who is the account approver. This portion of the app is simply an approval mechanism to an account request system that I am developing. The TD receives an e-mail with a link back to the approval page to approve the requested accounts. This approval then fires off an e-mail to the SA responsible for user accounts to inform him to create the newly approved account. Well, thanks again for the feedback. Hopefully I can get this going pretty quick. We are currently planning on rolling this app out to our user community on Monday. This is the last issue I have that I am aware of. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:06 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
Not sure exactly how I would implement this. Currently the reload URL is being provided via the $PHP_SELF variable. I like using this, but suppose I could change this if truly necessary. Also, the $td is not currently being passed into the function that generates the two buttons and hence defines the reload URL. I would rather not pass any additional variables into this function as I just feel this is unclean. Thanks again for the idea, and any additional advice or suggestions are most appreciated. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:01 AM To: NIPP, SCOTT V (SBCSI) Subject: RE: [PHP-DB] Datetime help in an INSERT... You could alway place it as a GET echo a href=\yourscript.com/yourpage.php?td=$td\; Gary Every Sr. UNIX Administrator Ingram Entertainment (615) 287-4876 Pay It Forward mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] http://accessingram.com http://accessingram.com -Original Message- From: NIPP, SCOTT V (SBCSI) [ mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] ] Sent: Thursday, June 05, 2003 8:57 AM To: 'CPT John W. Holmes'; [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. -Original Message- From: CPT John W. Holmes [ mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] ] Sent: Wednesday, June 04, 2003 2:14 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 Are you sure $td has a value? If it's blank, you'd get an error like that. ---John Holmes... -- PHP Database Mailing List ( http://www.php.net/ http://www.php.net/ ) To unsubscribe, visit: http://www.php.net/unsub.php http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
I just responded to a separate but similar suggestion. The form elements, including the buttons, are all generated in a separate function. I am currently passing about 5 variables into this function and I am a little hesitant to pass any others. I am not sure about complexity, performance, etc. in passing so many variables into a function. Please let me know if anyone has any ideas on this. I doubt that this is happening, but I would really like for the code to be as easily maintainable as possible. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:48 AM To: NIPP, SCOTT V (SBCSI) Subject: RE: [PHP-DB] Datetime help in an INSERT... a button is being pressed? so it is processing a form? if its in a form just pass along the value of $td via hidden input data... input type=hidden name=td value=? echo $td; ? jay merritt [EMAIL PROTECTED] [EMAIL PROTECTED] OK. I guess the next question would be what is the best option for performance? This portion of the app currently doesn't have any session functionality built in. I would be setting up the session for this sole purpose. I haven't worked with cookies yet. I suppose the URL method might work, I have just never like it because I think it looks unclean, just personal preference there. As to the $td question... This is a variable that holds the userid of the Technical Director who is the account approver. This portion of the app is simply an approval mechanism to an account request system that I am developing. The TD receives an e-mail with a link back to the approval page to approve the requested accounts. This approval then fires off an e-mail to the SA responsible for user accounts to inform him to create the newly approved account. Well, thanks again for the feedback. Hopefully I can get this going pretty quick. We are currently planning on rolling this app out to our user community on Monday. This is the last issue I have that I am aware of. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:06 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
Thought I was on track, but stumped again... I am now getting a completely different error that I cannot figure out how it is related. I have passed the $td variable into the function and then get it back as a POST variable from a hidden field within the function. Here is the error I am now seeing; Unknown column 'sn4265' in 'field list' This one is really throwing me at the moment as the only place the value sn4265 is stored is in the $sbcuid variable. This $sbcuid variable is not a part of the INSERT that I am performing which I think is generating this error. I am guessing that you guys will need more info to help out with this now. Just let me know what you want. Thanks again. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 10:07 AM To: NIPP, SCOTT V (SBCSI) Subject: Re: [PHP-DB] Datetime help in an INSERT... Just pass the variable to your function. If you're using a form, then make it a hidden element. ---John Holmes... - Original Message - From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: 'CPT John W. Holmes' [EMAIL PROTECTED] Sent: Thursday, June 05, 2003 10:53 AM Subject: RE: [PHP-DB] Datetime help in an INSERT... Do you think URL propagation is a better option than passing another variable into the function and then passing it back out through a hidden form element? Here was another snippet from another suggestion on this. a button is being pressed? so it is processing a form? if its in a form just pass along the value of $td via hidden input data... input type=hidden name=td value=? echo $td; ? jay merritt [EMAIL PROTECTED] [EMAIL PROTECTED] -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:53 AM To: NIPP, SCOTT V (SBCSI) Subject: Re: [PHP-DB] Datetime help in an INSERT... Propagating it in the URL is probably the best method. ---John Holmes... - Original Message - From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: 'CPT John W. Holmes' [EMAIL PROTECTED]; [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Thursday, June 05, 2003 10:44 AM Subject: RE: [PHP-DB] Datetime help in an INSERT... OK. I guess the next question would be what is the best option for performance? This portion of the app currently doesn't have any session functionality built in. I would be setting up the session for this sole purpose. I haven't worked with cookies yet. I suppose the URL method might work, I have just never like it because I think it looks unclean, just personal preference there. As to the $td question... This is a variable that holds the userid of the Technical Director who is the account approver. This portion of the app is simply an approval mechanism to an account request system that I am developing. The TD receives an e-mail with a link back to the approval page to approve the requested accounts. This approval then fires off an e-mail to the SA responsible for user accounts to inform him to create the newly approved account. Well, thanks again for the feedback. Hopefully I can get this going pretty quick. We are currently planning on rolling this app out to our user community on Monday. This is the last issue I have that I am aware of. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:06 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] PHP Upgrade question...
I know there is another mailing list for installation questions, but I am hoping for some help here. I get enough mail from this list alone without having to join another. Anyway, while attempting to compile version 4.3.2 I am running into the following error during the configure... checking whether byte ordering is bigendian... configure: error: can not run tes t program while cross compiling At this point the configure simply dies. I have no idea on how to proceed from this. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Datetime help in an INSERT...
I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 1:45 PM To: NIPP, SCOTT V (SBCSI) Subject: Re: [PHP-DB] Datetime help in an INSERT... what kind of error are you getting? I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
No, the field is a Datetime field rather than a Timestamp field. -Original Message- From: Rankin, Randy [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 1:59 PM To: '[EMAIL PROTECTED]' Subject: RE: [PHP-DB] Datetime help in an INSERT... Correction ( I typed this in a hurry ) ... If you are using MySql and the date field is of type timestamp, you should _not_ have to insert anything using the query. It will update automatically when the record is inserted and/or updated. ie; $denylog = INSERT INTO deny (account, td ) VALUES ($tmp, $td ); Randy -Original Message- From: Rankin, Randy [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 1:56 PM To: '[EMAIL PROTECTED]' Subject: RE: [PHP-DB] Datetime help in an INSERT... If you are using MySql with the date field is of type timestamp, you should have to insert anything using the query. It will update automatically when the record is inserted and/or updated. ie; $denylog = INSERT INTO deny (account, td ) VALUES ($tmp, $td ); Randy -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 1:47 PM To: '[EMAIL PROTECTED]' Cc: '[EMAIL PROTECTED]' Subject: RE: [PHP-DB] Datetime help in an INSERT... The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 1:45 PM To: NIPP, SCOTT V (SBCSI) Subject: Re: [PHP-DB] Datetime help in an INSERT... what kind of error are you getting? I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
I think this is the real problem. Basically, the $td variable is defined intially, but upon clicking a button the page reloads itself and I believe that at this time it is losing the $td variable. I am not quite sure how to prevent this from occurring. Thanks again for all the help so far. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 2:14 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 Are you sure $td has a value? If it's blank, you'd get an error like that. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Design question...
Anyone have an links to websites that offer good tutorials on web design? I am most interested in learning about templates and how to design these for use with PHP pages. I am using DreamWeaver MX as a development environment and would really like some help with DreamWeaver templates if anyone knows of any. Any good book recommendations would be most appreciated too. Thanks, and sorry for the slightly off topic post. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: Empty display loop...
OK... I moved the form tag up before the table tag and added a tr tag to the beginning of the first line of display in the loop. No difference in the output. The first line of the output loop is still a completely empty line. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Tuesday, March 25, 2003 6:30 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Re: Empty display loop... Scott V Nipp [EMAIL PROTECTED] wrote: I am working on a page that displays a line of information with a checkbox at the end. I began by getting all of the information I wanted displayed working properly and now I am adding the checkbox. Everything is going find with the exception that all of a sudden with the checkbox code in I am getting an empty line and checkbox at the beginning of the loop. I cannot figure out where this is coming from. Please let me know where I have gone wrong. Snipping a lot so you can see the problem: table width=90% bgcolor=#CC cellspacing=0 form method=post action=account_details.php /table /form Your nesting is bad. Put the form tag before the table. And you don't have a tr in your loop. Regards... Michael -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Empty display loop...
I am working on a page that displays a line of information with a
checkbox at the end. I began by getting all of the information I wanted
displayed working properly and now I am adding the checkbox. Everything is
going find with the exception that all of a sudden with the checkbox code
in I am getting an empty line and checkbox at the beginning of the loop. I
cannot figure out where this is coming from. Please let me know where I
have gone wrong.
table width=90% bgcolor=#CC cellspacing=0
tr
td width=20% height=23div align=centerRequesting User
ID/div/td
td width=10% valign=topdiv align=centerSystem Name/div/td
td width=10% valign=topdiv align=centerPrimary
Group/div/td
td width=10% valign=topdiv align=centerDefault
Shell/div/td
td width=30% valign=topdiv align=centerRequest Time/div/td
td width=10% valign=topdiv
align=centerCompleted/div/td
/tr
form method=post action=account_details.php
?php
do {
$entry = $list['id_sys'];
$id = split('-', $list['id_sys']);
$sbcuid = $id[0];
$sys = $id[1];
echo td width=\20%\div
align=\center\.$sbcuid./div/td;
echo td width=\10%\div align=\center\.$sys./div/td;
echo td width=\15%\div
align=\center\.$list['gid']./div/td;
echo td width=\15%\div
align=\center\.$list['shell']./div/td;
echo td width=\30%\div
align=\center\.$list['atime']./div/td;
echo td width=\10%\div align=\center\input
name=\entry[]\ type=\checkbox\ value=$entry/div/td/tr;
} while ($list = mysql_fetch_assoc($result));
?
/table
div align=center
input type=submit value=Details
/div
/form
This is the code for all of the section dealing with the data
display. The rest of the page is just some extra text and formatting, as
well as the actual query code. Thanks again.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] compare php and perl
In response to your bullet points... 1. Built in mysql_ functions in php (does perl have this?) Perl does not have built in functions for MySQL or any other database. The Perl functions are all added through modules. This is somewhat annoying as compared to PHP in that if you want to keep things current, for the sake of keeping current, you have to update not only Perl, but the DBI module and the DBD driver module also. This is not to say that you need to keep all of these individual components current. 2. Php may be faster. This is true, more so in some situations than in others. If you are dealing with Apache, you can compile Perl directly into the web server as a module. This is very efficient. However, my understanding is that getting Perl compiled into Apache is very tricky. 3. Embedded in html - no separate cgi directory needed. Yes. The PHP code can be directly imbedded into your HTML pages, but this is not always the best way to go. Often you will want to extract some information like database user IDs and passwords and store these in an alternate directory location for security reasons. You can also include other files with commonly used functions and these may or may not reside in your regular website structure. 4. Perl is more mature - more support - more depth to language. Yes. Perl is more mature in that it has most definitely been around much longer. In fact, PHP was derived from the authors frustrations with using certain aspects of Perl. That said, remember that Perl is NOT a web scripting language, it is a multipurpose scripting language that has been extended to handle web scripting. PHP on the other hand is a dedicated web scripting language. The key thing to remember is that Perl and PHP are both tools. Some people may prefer one tool over another for their own personal reasons. Pick the tool that best suits your purpose, and the one that you feel most comfortable with. If you have a LOT of Perl CGI experience, I don't know that PHP would really offer you a great reason to go out and learn another language (thought PHP is quite similar). I would recommend PHP if you are new to web scripting though, simply because this is exactly what PHP has been developed for from the ground up. Good luck, and I hope this helps. I am sure that some of the other experts, of which I am not one, will chime in to add their opinions and correct any of my errors. One other big advantage of PHP is this mailing list. The people in this list are quite knowledgeable and extremely helpful. I honestly have not found a group this supportive and responsive for my Perl scripting. (This last may sound like sucking up, but it's the honest truth.) -Original Message- From: Mignon Hunter [mailto:[EMAIL PROTECTED] Sent: Thursday, March 20, 2003 12:43 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] compare php and perl Can anyone out there compare these two using with mysql and web apps - What I can gather so far from my limited experience with both and a little googling: 1. Built in mysql_ functions in php (does perl have this?) 2. Php may be faster. 3. Embedded in html - no separate cgi directory needed. 4. Perl is more mature - more support - more depth to language. Thx -- Mignon Hunter Web Developer Toshiba International 713.466.0277 x 3461 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Login and link back...
I am curious about what you guys may have along the lines of best practices for forwarding from a URL to a login, and then jumping back to the original URL automatically. I have several separate applications that all need to utilize the same login mechanism. I want the user to be able to enter the URL for the application and if they are not logged in it redirects them to a login screen. I already have the sessions junk setup and understand all of that portion. I am mainly interested in how people are handling the return to a URL after successful login. I have done some research on this, and discovered the $HTTP_REFERER variable however the PHP site discourages using this. I have also thought of adding code to each page to export an origin variable to be passed to the login page such that it can be used to return the user. I thought of this method, but I am not real clear on how to manage this. Does anyone have any suggestion on implementing this, or another alternative that I have not touched on yet? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Easy array question...
I create an array on one page and then process it on the next page as I have registered the array as a session variable. I need to basically process one of these arrays twice, so I want to make a copy of the array. The question is how do I do this? I have tried the following: $array2 = $array1; This doesn't seem to do the trick for me. Is this correct and I am just missing something elsewhere, or is there another way to copy an array? Thanks. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Easy array question...
I don't think that is the case because I am able to access the original array in this same page without having to use the session variable reference. Won't hurt to give it a try though. Thanks for the feedback. -Original Message- From: Hutchins, Richard [mailto:[EMAIL PROTECTED] Sent: Wednesday, March 12, 2003 10:40 AM To: '[EMAIL PROTECTED]' Subject: RE: [PHP-DB] Easy array question... Are you using: $array2 = $_SESSION[array1]; Not sure that will do the trick, but it was my first inclination. Rich -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED] Sent: Wednesday, March 12, 2003 11:37 AM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] Easy array question... I create an array on one page and then process it on the next page as I have registered the array as a session variable. I need to basically process one of these arrays twice, so I want to make a copy of the array. The question is how do I do this? I have tried the following: $array2 = $array1; This doesn't seem to do the trick for me. Is this correct and I am just missing something elsewhere, or is there another way to copy an array? Thanks. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Easy array question...
Here is basically what I have and what I am getting...
The section of code to copy and then process the copied array:
$other2 = array();
array_push($other2, $other);
foreach ($other2 as $test) {
if ($test == ) {
echo NOTHING.br;
} else {
echo $test.br;
}
}
Basically, what I want is if the array element is nothing, then
output as a test NOTHING. If the array element has some value, then output
as a test that value. Currently, from this code the output I am actually
getting is:
Array
Thanks again.
-Original Message-
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:58 AM
To: [EMAIL PROTECTED]; NIPP, SCOTT V (SBCSI); 'Hutchins, Richard';
[EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...
Sorry, I just reread your post
$array_copy = array();
array_push($array_copy, $original_array);
echo $array_copy[0];//or whatever to test
--- Jonathan
-Original Message-
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:56 AM
To: 'NIPP, SCOTT V (SBCSI)'; 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...
$array_copy = array();
array_push($array_copy, $_SESSION['original_array']);
echo $array_copy[0];//or whatever to test
--- Jonathan
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:43 AM
To: 'Hutchins, Richard'; '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] Easy array question...
I don't think that is the case because I am able to access the
original array in this same page without having to use the session
variable
reference. Won't hurt to give it a try though. Thanks for the
feedback.
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:40 AM
To: '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] Easy array question...
Are you using:
$array2 = $_SESSION[array1];
Not sure that will do the trick, but it was my first inclination.
Rich
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 11:37 AM
To: '[EMAIL PROTECTED]'
Subject: [PHP-DB] Easy array question...
I create an array on one page and then process it on
the next page
as I have registered the array as a session variable. I need
to basically
process one of these arrays twice, so I want to make a copy
of the array.
The question is how do I do this? I have tried the following:
$array2 = $array1;
This doesn't seem to do the trick for me. Is this
correct and I am
just missing something elsewhere, or is there another way to
copy an array?
Thanks.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
--
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RE: [PHP-DB] Easy array question...
GREAT. This does the trick for me. Thanks a lot. I am now getting
the output I want, so I can now feel confident about inputting this into the
actual database.
-Original Message-
From: Katie Evans-Young [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 11:21 AM
To: NIPP, SCOTT V (SBCSI)
Subject: RE: [PHP-DB] Easy array question...
It's actually putting the whole of the array $other as the first element of
the array $other2. You're going to have to step through the array to copy
it.
foreach($other as $element) {
$other2[] = $element;
} //end foreach
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 11:10 AM
To: '[EMAIL PROTECTED]'; 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...
Here is basically what I have and what I am getting...
The section of code to copy and then process the copied array:
$other2 = array();
array_push($other2, $other);
foreach ($other2 as $test) {
if ($test == ) {
echo NOTHING.br;
} else {
echo $test.br;
}
}
Basically, what I want is if the array element is nothing, then
output as a test NOTHING. If the array element has some value, then output
as a test that value. Currently, from this code the output I am actually
getting is:
Array
Thanks again.
-Original Message-
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:58 AM
To: [EMAIL PROTECTED]; NIPP, SCOTT V (SBCSI); 'Hutchins, Richard';
[EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...
Sorry, I just reread your post
$array_copy = array();
array_push($array_copy, $original_array);
echo $array_copy[0];//or whatever to test
--- Jonathan
-Original Message-
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:56 AM
To: 'NIPP, SCOTT V (SBCSI)'; 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Easy array question...
$array_copy = array();
array_push($array_copy, $_SESSION['original_array']);
echo $array_copy[0];//or whatever to test
--- Jonathan
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:43 AM
To: 'Hutchins, Richard'; '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] Easy array question...
I don't think that is the case because I am able to access the
original array in this same page without having to use the session
variable
reference. Won't hurt to give it a try though. Thanks for the
feedback.
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 10:40 AM
To: '[EMAIL PROTECTED]'
Subject: RE: [PHP-DB] Easy array question...
Are you using:
$array2 = $_SESSION[array1];
Not sure that will do the trick, but it was my first inclination.
Rich
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 12, 2003 11:37 AM
To: '[EMAIL PROTECTED]'
Subject: [PHP-DB] Easy array question...
I create an array on one page and then process it on
the next page
as I have registered the array as a session variable. I need
to basically
process one of these arrays twice, so I want to make a copy
of the array.
The question is how do I do this? I have tried the following:
$array2 = $array1;
This doesn't seem to do the trick for me. Is this
correct and I am
just missing something elsewhere, or is there another way to
copy an array?
Thanks.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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[PHP-DB] Stumped again...
OK... I have once again thought of something cool to do, and can't
quite figure out how to make it happen. I am developing a user account
request system for work. Things are coming along nicely. The user logs
into the system, kind of, selects the systems that he wants to have accounts
added on, and then the next page is where the problems begin. Once the user
selects the systems, the next page lists the systems selected along with a
dropdown for Default Shell and another dropdown for Primary Group and a
small text box for Other Group. The dropdowns are populating fine, and
everything is displaying on the screen as expected. The problem I am having
is that the selections don't seem to work. I have this page feed a final
page that also displays the selections as a result test. I am seeing the
system names, but no shell or group information.
Here are some pertinent portions of the code with comments:
Above code snipped
$shell_list = getEnumOptions('accounts','shell');
$shell_tmp = select size=\1\ name=\shell\\n;
$shell_tmp .= optionDefault Shell/option\n;
foreach ($shell_list as $item) {
$shell_tmp .= option$item/option\n;
}
$shell_tmp .= /select\n;
mysql_select_db($database, $Prod);
$query_groups = SELECT name FROM grps;
$groups_tmp = mysql_query($query_groups, $Prod) or die(mysql_error());
$grp_list = select size=\1\ name=\grp\\n;
$grp_list .= optionPrimary Group/option\n;
$grp_list .= option---/option\n;
while($name = mysql_fetch_row($groups_tmp)) {
$grp_list .= option$name[0]/option\n;
}
$grp_list .= /select\n;
Code snipped
This section of code creates the Default Shell dropdown list and
the Primary Group dropdown list. One of these lists is derived from the
ENUM values of the field, the other is derived from the values of a seperate
table.
Code snipped
form name=form1 method=post action=account_final.php
?php
while (isset($_POST['system'][$a])) {
echo td width=\20%\div
align=\center\.$_POST['system'][$a]./div/td;
echo td width=\20%\div
align=\center\.$shell_tmp./div/td;
echo td width=\20%\div
align=\center\.$grp_list./div/td;
echo td width=\20%\div align=\center\input type=\text\
name=\other\ value=\\ size=\15\/div/td/tr;
$tmp = $sbcuid.-.$_POST['system'][$a];
$a++;
array_push( $accnts, $tmp );
array_push( $shells, $shell );
array_push( $grps, $grp );
array_push( $others, $other );
}
?
Code snipped
This section of code actually displays one line for each system
previously selected with additional input items for Default Shell,
Primary Group, and Other Group. This code also creates arrays of the
systems, shells, groups, and others which are passed onto the next page.
This system array is used to display the results as well as populate the
database, and the system array data is the only data actually being
displayed. I do not think that the data for the other arrays is actually
getting populated.
Code snipped
?php
session_register(accnts, shells, grps, others);
?
div align=center
input type=submit name=Submit value=Submit
/div
/form
Code snipped
This section of code simply registers the arrays for use on the next
page.
Thanks in advance for any help. Please let me know if you need any
other information to help figure out what is wrong with this.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] RE: Stumped again...
I don't really understand this suggestion. Could someone possibly
expound on this further? Maybe an example? Thanks.
-Original Message-
From: Ben Walling [mailto:[EMAIL PROTECTED]
Sent: Tuesday, March 11, 2003 12:19 PM
To: NIPP, SCOTT V (SBCSI)
Subject: Re: Stumped again...
You need the value parameter for the option tags:
OPTION VALUE=1First
OPTION VALUE=2Second
Scott V Nipp [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
OK... I have once again thought of something cool to do, and can't
quite figure out how to make it happen. I am developing a user account
request system for work. Things are coming along nicely. The user logs
into the system, kind of, selects the systems that he wants to have
accounts
added on, and then the next page is where the problems begin. Once the
user
selects the systems, the next page lists the systems selected along with a
dropdown for Default Shell and another dropdown for Primary Group and
a
small text box for Other Group. The dropdowns are populating fine, and
everything is displaying on the screen as expected. The problem I am
having
is that the selections don't seem to work. I have this page feed a final
page that also displays the selections as a result test. I am seeing the
system names, but no shell or group information.
Here are some pertinent portions of the code with comments:
Above code snipped
$shell_list = getEnumOptions('accounts','shell');
$shell_tmp = select size=\1\ name=\shell\\n;
$shell_tmp .= optionDefault Shell/option\n;
foreach ($shell_list as $item) {
$shell_tmp .= option$item/option\n;
}
$shell_tmp .= /select\n;
mysql_select_db($database, $Prod);
$query_groups = SELECT name FROM grps;
$groups_tmp = mysql_query($query_groups, $Prod) or die(mysql_error());
$grp_list = select size=\1\ name=\grp\\n;
$grp_list .= optionPrimary Group/option\n;
$grp_list .= option---/option\n;
while($name = mysql_fetch_row($groups_tmp)) {
$grp_list .= option$name[0]/option\n;
}
$grp_list .= /select\n;
Code snipped
This section of code creates the Default Shell dropdown list and
the Primary Group dropdown list. One of these lists is derived from the
ENUM values of the field, the other is derived from the values of a
seperate
table.
Code snipped
form name=form1 method=post action=account_final.php
?php
while (isset($_POST['system'][$a])) {
echo td width=\20%\div
align=\center\.$_POST['system'][$a]./div/td;
echo td width=\20%\div
align=\center\.$shell_tmp./div/td;
echo td width=\20%\div
align=\center\.$grp_list./div/td;
echo td width=\20%\div align=\center\input type=\text\
name=\other\ value=\\ size=\15\/div/td/tr;
$tmp = $sbcuid.-.$_POST['system'][$a];
$a++;
array_push( $accnts, $tmp );
array_push( $shells, $shell );
array_push( $grps, $grp );
array_push( $others, $other );
}
?
Code snipped
This section of code actually displays one line for each system
previously selected with additional input items for Default Shell,
Primary Group, and Other Group. This code also creates arrays of the
systems, shells, groups, and others which are passed onto the next page.
This system array is used to display the results as well as populate the
database, and the system array data is the only data actually being
displayed. I do not think that the data for the other arrays is actually
getting populated.
Code snipped
?php
session_register(accnts, shells, grps, others);
?
div align=center
input type=submit name=Submit value=Submit
/div
/form
Code snipped
This section of code simply registers the arrays for use on the next
page.
Thanks in advance for any help. Please let me know if you need any
other information to help figure out what is wrong with this.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] RE: Stumped again...
OK... I think I understand what you are getting at here, but I
don't think this actually addresses the problem that I am having. I have,
in other places, the direct value successfully passing from one page to
another. In this case, what I am lacking is the selected data actually
getting populated into the array in the first place.
-Original Message-
From: Ben Walling [mailto:[EMAIL PROTECTED]
Sent: Tuesday, March 11, 2003 1:48 PM
To: NIPP, SCOTT V (SBCSI)
Cc: [EMAIL PROTECTED]
Subject: RE: Stumped again...
Your code:
while($name = mysql_fetch_row($groups_tmp)) {
$grp_list .= option$name[0]/option\n;
}
should be changed to:
while($name = mysql_fetch_row($groups_tmp)) {
$grp_list .= option value=.$name[0].$name[0]/option\n;
}
The value property of the option tag is actually what gets submitted to the
next page. The text that immediately follows option value=x is what is
displayed to the user.
So, for this (HTML) code:
SELECT NAME=Test
OPTION VALUE=1Thing One
OPTION VALUE=2Thing Two
/SELECT
The user would see a list with:
Thing One
Thing Two
The next page would get either 1 or 2 as the value for Test.
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]
Sent: Tuesday, March 11, 2003 2:43 PM
To: Ben Walling; '[EMAIL PROTECTED]'
Subject: RE: Stumped again...
I don't really understand this suggestion. Could someone possibly
expound on this further? Maybe an example? Thanks.
-Original Message-
From: Ben Walling [mailto:[EMAIL PROTECTED]
Sent: Tuesday, March 11, 2003 12:19 PM
To: NIPP, SCOTT V (SBCSI)
Subject: Re: Stumped again...
You need the value parameter for the option tags:
OPTION VALUE=1First
OPTION VALUE=2Second
Scott V Nipp [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
OK... I have once again thought of something cool to do, and can't
quite figure out how to make it happen. I am developing a user account
request system for work. Things are coming along nicely. The user logs
into the system, kind of, selects the systems that he wants to have
accounts
added on, and then the next page is where the problems begin. Once the
user
selects the systems, the next page lists the systems selected along with a
dropdown for Default Shell and another dropdown for Primary Group and
a
small text box for Other Group. The dropdowns are populating fine, and
everything is displaying on the screen as expected. The problem I am
having
is that the selections don't seem to work. I have this page feed a final
page that also displays the selections as a result test. I am seeing the
system names, but no shell or group information.
Here are some pertinent portions of the code with comments:
Above code snipped
$shell_list = getEnumOptions('accounts','shell');
$shell_tmp = select size=\1\ name=\shell\\n;
$shell_tmp .= optionDefault Shell/option\n;
foreach ($shell_list as $item) {
$shell_tmp .= option$item/option\n;
}
$shell_tmp .= /select\n;
mysql_select_db($database, $Prod);
$query_groups = SELECT name FROM grps;
$groups_tmp = mysql_query($query_groups, $Prod) or die(mysql_error());
$grp_list = select size=\1\ name=\grp\\n;
$grp_list .= optionPrimary Group/option\n;
$grp_list .= option---/option\n;
while($name = mysql_fetch_row($groups_tmp)) {
$grp_list .= option$name[0]/option\n;
}
$grp_list .= /select\n;
Code snipped
This section of code creates the Default Shell dropdown list and
the Primary Group dropdown list. One of these lists is derived from the
ENUM values of the field, the other is derived from the values of a
seperate
table.
Code snipped
form name=form1 method=post action=account_final.php
?php
while (isset($_POST['system'][$a])) {
echo td width=\20%\div
align=\center\.$_POST['system'][$a]./div/td;
echo td width=\20%\div
align=\center\.$shell_tmp./div/td;
echo td width=\20%\div
align=\center\.$grp_list./div/td;
echo td width=\20%\div align=\center\input type=\text\
name=\other\ value=\\ size=\15\/div/td/tr;
$tmp = $sbcuid.-.$_POST['system'][$a];
$a++;
array_push( $accnts, $tmp );
array_push( $shells, $shell );
array_push( $grps, $grp );
array_push( $others, $other );
}
?
Code snipped
This section of code actually displays one line for each system
previously selected with additional input items for Default Shell,
Primary Group, and Other Group. This code also creates arrays of the
systems, shells, groups, and others which are passed onto the next page.
This system array is used to display the results as well as populate the
database, and the system array data is the only data actually being
displayed. I do not think that the data for the other arrays is actually
getting populated.
Code snipped
?php
session_register(accnts, shells, grps, others);
?
div align=center
[PHP-DB] Query Help...
I am having a lot of trouble with a query that works fine from a basic SQL command line, but fails in my web script. Here is the portion of code including the query: mysql_select_db($database, $Prod); $query_groups = SELECT name FROM group; $groups = mysql_query($query_groups, $Prod) or die(mysql_error()); Here is the error I am receiving: You have an error in your SQL syntax near 'group' at line 1 I am wondering if for some reason group is trying to be interperetted as the GROUP BY MySQL function? I am running PHP 4.2.3 against a MySQL DB on an Apache 1.3.27 server. Thanks in advance for the help. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Query Help...
Thanks for all the advice, I thought it might have something to do with being a reserved word. I changed the table name from group to grps, and everything works fine now. -Original Message- From: Rob Bryant [mailto:[EMAIL PROTECTED] Sent: Monday, March 10, 2003 11:39 AM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Query Help... - Original Message - From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Monday, March 10, 2003 9:23 AM Subject: [PHP-DB] Query Help... I am having a lot of trouble with a query that works fine from a basic SQL command line, but fails in my web script. Here is the portion of code including the query: mysql_select_db($database, $Prod); $query_groups = SELECT name FROM group; $groups = mysql_query($query_groups, $Prod) or die(mysql_error()); Here is the error I am receiving: You have an error in your SQL syntax near 'group' at line 1 Have you tried using backticks in your query? E.g., $query_groups = SELECT name FROM `group`; -- rob -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Code for drop down lists
I would most definitely recommend pulling them from a DB table
rather than hard coding these into a page. Here is a code snippet that I
use to do this on one of my pages:
mysql_select_db($Prod, $Prod);
$query_systems = SELECT Name FROM systems ORDER BY Name ASC;
$systems = mysql_query($query_systems, $Prod) or die(mysql_error());
# $row_systems = mysql_fetch_assoc($systems);
$totalRows_systems = mysql_num_rows($systems);
$sys_list = select size=\1\ name=\system\\n;
$sys_list .= optionSystem Name/option\n;
$sys_list .= option---/option\n;
while($name = mysql_fetch_row($systems)) {
$sys_list .= option$name[0]/option\n;
}
$sys_list .= /select\n;
snip...HTML portion below
td width=124 valign=topdiv align=rightSystem Name:/div/td
td colspan=2 valign=top
?=$sys_list?
snip
I also do this with Enum field selections using the following
function that someone passed on to me:
function getEnumOptions($table, $field) {
global $finalResult;
$finalResult = array();
if (strlen(trim($table)) 1) return false;
$query = show columns from $table;
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
if ($field != $row[Field]) continue;
//check if enum type
if (ereg('enum.(.*).', $row['Type'], $match)) {
$opts = explode(',', $match[1]);
foreach ($opts as $item)
$finalResult[] = substr($item, 1, strlen($item)-2);
}
else
return false;
}
return $finalResult;
You can then use code similar to the first example to create your
dropdown list from Enum field values. I hope this helps.
-Original Message-
From: Roland Perez [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, February 11, 2003 11:03 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Code for drop down lists
I am a novice to PHP and am trying toget a survey with drop down lists.
I am stuck as to if I should connect to the DB (MySQL DB) to get the
values from a table or do I imbed them in the php file?
Thanks for any help in advance.
Roland Perez
[EMAIL PROTECTED]
--
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To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Weird PHP behavior...
I developed some dynamic content for a larger web page, and once I pasted the code into the appropriate area of the page I am seeing the strangest error. Here is the error: Warning: Failed opening '/www/DW/index1.htm' for inclusion (include_path='.:/usr/local/lib/php') in Unknown on line 0 I have tried this on 2 different servers running 2 different versions of PHP, 4.1.2 and 4.2.3. The code was developed and tested in a blank page and then inserted into a much larger page developed in FrontPage 2002. Any ideas? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Weird PHP behavior...
Here is the working code...
?php
require_once(/www/DW/prod.lib.php);
$hr = date(G);
if ($hr 7) {
$sql = mysql(Calendar2,SELECT B.Title, UNIX_TIMESTAMP(A.StartDate),
UNIX_TIMESTAMP(A.StopDate) FROM `phpCalendar_Details` AS B,
`phpCalendar_Daily` AS A WHERE A.CalendarDetailsID = B.CalendarDetailsID
AND CURDATE() BETWEEN A.StartDate AND A.StopDate) or die(mysql_error());
$mid = mktime (0,0,0,date(m) ,date(d)+1,date(Y));
} else {
$sql = mysql(Calendar2,SELECT B.Title, UNIX_TIMESTAMP(A.StartDate),
UNIX_TIMESTAMP(A.StopDate) FROM `phpCalendar_Details` AS B,
`phpCalendar_Daily` AS A WHERE A.CalendarDetailsID = B.CalendarDetailsID
AND CURDATE()-2 BETWEEN A.StartDate AND A.StopDate) or die(mysql_error());
$sql2 = mysql(Calendar2,SELECT B.Title FROM `phpCalendar_Details` AS
B, `phpCalendar_Daily` AS A WHERE A.CalendarDetailsID =
B.CalendarDetailsID AND CURDATE() BETWEEN A.StartDate AND A.StopDate) or
die(mysql_error());
$res2 = mysql_fetch_row($sql2);
$new = $res2[0];
$mid = mktime (0,0,0,date(m) ,date(d)-2,date(Y));
}
$res = mysql_fetch_row($sql);
$sa = $res[0];
$start = $res[1];
$stop = $res[2];
$day1 = date(l, $start);
$date1 = date(jS, $start);
$month1 = date(F, $start);
$day3 = date(l, $stop);
$date3 = date(jS, $stop);
$month3 = date(F, $stop);
if ($day1 == Friday) {
$day2 = date(l, $mid);
$date2 = date(jS, $mid);
$month2 = date(F, $mid);
if ($month1 != $month2) {
$line2 = $month1 $date1, $month2 $date2 and $date3;
} elseif ($month2 != $month3) {
$line2 = $month1 $date1 and $date2, and $month3 $date3;
} else {
$line2 = $month1 $date1, $date2, and $date3;
}
$line1 = ONCALL SA for Friday, Saturday, and Sunday;
} else {
if ($month1 != $month3) {
$line2 = $month1 $date1 and $date3;
} else {
$line2 = $month1 $date1 and $month3 $date3;
}
$line1 = ONCALL SA for $day1 and $day3;
}
mysql_select_db($database, $Prod);
$query = SELECT name, pager FROM contacts_sa WHERE sbcuid='$sa';
$said = mysql_query($query, $Prod) or die(mysql_error());
$row = mysql_fetch_row($said);
$name = $row[0];
$pager = $row[1];
if ($date3 == date(jS, $stop) $hr 8) {
$query2 = SELECT name FROM contacts_sa WHERE sbcuid='$new';
$said2 = mysql_query($query2, $Prod) or die(mysql_error());
$row2 = mysql_fetch_row($said2);
$name2 = $row2[0];
$line4 = Please be aware the On-Call turnover occurs at 8AM. brThe
oncoming SA after 8AM will be $name2.;
} else {
$line4 = $name begins On-Call at 8AM on $day1.;
}
?
html
head
titleUntitled Document/title
meta http-equiv=Content-Type content=text/html; charset=iso-8859-1
/head
body
h2 align=centerspan lang=en-usCALL 214-741-6961 br
FOR CURRENT ONCALL STATUS./span/h2
hr color=#6363C9 size=3 width=50%
p align=center?php echo $line1; ?br
/span(span lang=en-us?php echo $line2; ?/span) br
span lang=en-us?php echo $name; ? /span(pager span
lang=en-us
?php echo $pager; ?)br
?php echo $line4; ?/span/p
hr color=#6363C9 size=3 width=50%
/body
/html
The portion of PHP code prior to the HTML is inserted at the top of the
file in the broken page. Here is how the HTML portion of the PHP is
inserted...
snip...
td background=images/bg_leftside.gif width=20nbsp;/td
td width=20 background=images/bg_main.gifnbsp;/td
td width=550 background=images/bg_main.gif valign=top
p align=left dynamicanimation=fpAnimOutflyBottomRightFP1
id=fpAnimOu
tflyBottomRightFP1 style=position: relative !important
onclick=dynAnimOut(th
is) language=Javascript1.2nbsp;/p
h2 align=centerspan lang=en-usCALL 214-741-6961 br
FOR CURRENT ONCALL STATUS./span/h2
hr color=#6363C9 size=3 width=50%
p align=center?php echo $line1; ?br
/span(span lang=en-us?php echo $line2; ?/span) br
span lang=en-us?php echo $name; ? /span(pager span
lang=en-us
?php echo $pager; ?)br
?php echo $line4; ?/span/p
hr color=#6363C9 size=3 width=50%
h4 align=centerspan lang=en-usFor escalation procedures,
a href=on-call/escalation.htmclick here./a/span/h4
h4 align=centerspan lang=en-usFor restore requests call the
Help
Desk/span/h4
...snip
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 06, 2003 11:09 AM
To: NIPP, SCOTT V (SBCSI)
Subject: RE: [PHP-DB] Weird PHP behavior...
well show the working code and then show just a portion of the code that is
broken specifically where you are trying to inslucde the working code.
K.
NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
02/06/2003 12:09 PM
To:'[EMAIL PROTECTED]'
[EMAIL PROTECTED]
cc:
Subject:RE: [PHP-DB] Weird PHP behavior...
I can show the code, but do I show the working code? If I show the
broken code, it is REALLY long.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL
RE: [PHP-DB] Weird PHP behavior...
It turned out to be just a simple file permissions issue.
-Original Message-
From: Mohammad Saad [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 06, 2003 12:31 PM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED];
[EMAIL PROTECTED]
Subject: Re: [PHP-DB] Weird PHP behavior...
Check path on this require
require_once(/www/DW/prod.lib.php);
something is wrong with this
- Original Message -
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Thursday, February 06, 2003 10:17 PM
Subject: RE: [PHP-DB] Weird PHP behavior...
Here is the working code...
?php
require_once(/www/DW/prod.lib.php);
$hr = date(G);
if ($hr 7) {
$sql = mysql(Calendar2,SELECT B.Title, UNIX_TIMESTAMP(A.StartDate),
UNIX_TIMESTAMP(A.StopDate) FROM `phpCalendar_Details` AS B,
`phpCalendar_Daily` AS A WHERE A.CalendarDetailsID = B.CalendarDetailsID
AND CURDATE() BETWEEN A.StartDate AND A.StopDate) or die(mysql_error());
$mid = mktime (0,0,0,date(m) ,date(d)+1,date(Y));
} else {
$sql = mysql(Calendar2,SELECT B.Title, UNIX_TIMESTAMP(A.StartDate),
UNIX_TIMESTAMP(A.StopDate) FROM `phpCalendar_Details` AS B,
`phpCalendar_Daily` AS A WHERE A.CalendarDetailsID = B.CalendarDetailsID
AND CURDATE()-2 BETWEEN A.StartDate AND A.StopDate) or
die(mysql_error());
$sql2 = mysql(Calendar2,SELECT B.Title FROM `phpCalendar_Details`
AS
B, `phpCalendar_Daily` AS A WHERE A.CalendarDetailsID =
B.CalendarDetailsID AND CURDATE() BETWEEN A.StartDate AND A.StopDate) or
die(mysql_error());
$res2 = mysql_fetch_row($sql2);
$new = $res2[0];
$mid = mktime (0,0,0,date(m) ,date(d)-2,date(Y));
}
$res = mysql_fetch_row($sql);
$sa = $res[0];
$start = $res[1];
$stop = $res[2];
$day1 = date(l, $start);
$date1 = date(jS, $start);
$month1 = date(F, $start);
$day3 = date(l, $stop);
$date3 = date(jS, $stop);
$month3 = date(F, $stop);
if ($day1 == Friday) {
$day2 = date(l, $mid);
$date2 = date(jS, $mid);
$month2 = date(F, $mid);
if ($month1 != $month2) {
$line2 = $month1 $date1, $month2 $date2 and $date3;
} elseif ($month2 != $month3) {
$line2 = $month1 $date1 and $date2, and $month3 $date3;
} else {
$line2 = $month1 $date1, $date2, and $date3;
}
$line1 = ONCALL SA for Friday, Saturday, and Sunday;
} else {
if ($month1 != $month3) {
$line2 = $month1 $date1 and $date3;
} else {
$line2 = $month1 $date1 and $month3 $date3;
}
$line1 = ONCALL SA for $day1 and $day3;
}
mysql_select_db($database, $Prod);
$query = SELECT name, pager FROM contacts_sa WHERE sbcuid='$sa';
$said = mysql_query($query, $Prod) or die(mysql_error());
$row = mysql_fetch_row($said);
$name = $row[0];
$pager = $row[1];
if ($date3 == date(jS, $stop) $hr 8) {
$query2 = SELECT name FROM contacts_sa WHERE sbcuid='$new';
$said2 = mysql_query($query2, $Prod) or die(mysql_error());
$row2 = mysql_fetch_row($said2);
$name2 = $row2[0];
$line4 = Please be aware the On-Call turnover occurs at 8AM. brThe
oncoming SA after 8AM will be $name2.;
} else {
$line4 = $name begins On-Call at 8AM on $day1.;
}
?
html
head
titleUntitled Document/title
meta http-equiv=Content-Type content=text/html; charset=iso-8859-1
/head
body
h2 align=centerspan lang=en-usCALL 214-741-6961 br
FOR CURRENT ONCALL STATUS./span/h2
hr color=#6363C9 size=3 width=50%
p align=center?php echo $line1; ?br
/span(span lang=en-us?php echo $line2; ?/span) br
span lang=en-us?php echo $name; ? /span(pager span
lang=en-us
?php echo $pager; ?)br
?php echo $line4; ?/span/p
hr color=#6363C9 size=3 width=50%
/body
/html
The portion of PHP code prior to the HTML is inserted at the top of
the
file in the broken page. Here is how the HTML portion of the PHP is
inserted...
snip...
td background=images/bg_leftside.gif width=20nbsp;/td
td width=20 background=images/bg_main.gifnbsp;/td
td width=550 background=images/bg_main.gif valign=top
p align=left dynamicanimation=fpAnimOutflyBottomRightFP1
id=fpAnimOu
tflyBottomRightFP1 style=position: relative !important
onclick=dynAnimOut(th
is) language=Javascript1.2nbsp;/p
h2 align=centerspan lang=en-usCALL 214-741-6961 br
FOR CURRENT ONCALL STATUS./span/h2
hr color=#6363C9 size=3 width=50%
p align=center?php echo $line1; ?br
/span(span lang=en-us?php echo $line2; ?/span) br
span lang=en-us?php echo $name; ? /span(pager span
lang=en-us
?php echo $pager; ?)br
?php echo $line4; ?/span/p
hr color=#6363C9 size=3 width=50%
h4 align=centerspan lang=en-usFor escalation procedures,
a href=on-call/escalation.htmclick here./a/span/h4
h4 align=centerspan lang=en-usFor restore requests call the
Help
Desk/span/h4
...snip
-Original Message-
From: [EMAIL PROTECTED
RE: [PHP-DB] Date Range Question...
WOW... I have never even heard of BETWEEN before. How does this look to you? SELECT StartDate, StopDate, LocationID FROM phpCalendar_Daily WHERE CURDATE() BETWEEN StopDate AND StopDate Would the query work like this? This would mean I don't even have to care about the difference of 2 or 3 days? -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Tuesday, February 04, 2003 10:04 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Subject: Re: [PHP-DB] Date Range Question... I am working on an app that needs to post information to a website based on date. The tricky part for me is that the date is a range that spans either 2 or 3 days. I want the web page to dynamically populate this information based on a query. The query will look something like this: mysql(Calendar2,SELECT StartDate, StopDate, LocationID FROM phpCalendar_Daily WHERE StartDate=CURDATE() ) My question is say for example the StartDate is actually Jan. 1. The page is accessed on Jan. 2 or Jan. 3... The StartDate for the next entry is Jan. 4. How do I write this query to select the proper entry? In doing some research, it appears that a MySQL SELECT CASE might do the trick, but I have never used this before. The other option I can envision is using PHP to handle the login and simply running the query, testing the result, if NULL, run the query again with CURDATE(-1). Any hints, ideas, or suggestions would be most appreciated. Thanks in advance. If I understand you correctly, I think you could use something like this: SELECT StartDate, StopDate, LocationID FROM phpCalendar_Daily WHERE StartDate BETWEEN CURDATE() AND CURDATE() - INTERVAL $x DAY Where $x is 2 or 3, depending on your case. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] matching data from one table with another and displaying results?
I have a JOIN that is not working, and figured I would piggie-back on this thread to see if someone could give me a quick answer... Here is the JOIN syntax. This is my very first attempt at a JOIN statement, so I may be doing something blatantly wrong here. SELECT 'A.Title' FROM `phpCalendar_Daily` AS 'B' JOIN 'phpCalendar_Details' AS 'A' ON 'B.CalendarDetailsID' = 'A.CalendarDetailsID' WHERE CURDATE( ) BETWEEN 'StartDate' AND 'StopDate' I am attempting this from the SQL window of phpMyAdmin, so this is not embedded somewhere in a script and failing. Thanks in advance. -Original Message- From: SELPH,JASON (HP-Richardson,ex1) [mailto:[EMAIL PROTECTED]] Sent: Tuesday, February 04, 2003 10:59 AM To: Aaron Wolski; [EMAIL PROTECTED] Subject: RE: [PHP-DB] matching data from one table with another and displa ying results? Seek out LEFT JOIN on mysql web site, thats what you want. Cheers Jason -Original Message- From: Aaron Wolski [mailto:[EMAIL PROTECTED]] Sent: Tuesday, February 04, 2003 10:43 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] matching data from one table with another and displaying results? Hi All, Having a problem figuring out the logic of something here and hoping someone can point me in the right direction. I have 2 tables. First table holds a listing of all Provinces and States. Here a brief of what it looks like: value label AB Alberta BC British Columbia MB Manitobia NF Newfoundland NT Northwest Territories NS Nova Scotia Second Table is an OrderTable which holds all order and customer info including the province/state. The province/state is dumped from the value column of the ProvinceTable. Meaning.. an order record would show the province as AB or NS instead of Alberta or Nova Scotia. I am creating an order report and in this report I want to display the LABEL name of the Province/State that relates to an the province column in the OrderTable. I could simply just display all provinces/states from the ProvinceTable but then there would be entries for selection that had NO orders. Any idea on this? Thanks guys! Aaron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] A little JOIN help please...
This is more of an SQL question, but I am not sure where else to turn for help. I am trying a JOIN, for the first time ever, and it is NOT working for me. The following is what I am attempting; SELECT 'A.Title' FROM `phpCalendar_Daily` AS 'B' JOIN 'phpCalendar_Details' AS 'A' ON 'A.CalendarDetailsID' = 'B.CalendarDetailsID' WHERE CURDATE( ) BETWEEN 'B.StartDate' AND 'B.StopDate' This is simply in a SQL command window in phpMyAdmin, and it is returning an error. I have tried messing around with the quotes, and everything else I can think of. Someone please let me know what I am doing wrong. Thanks. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] A little JOIN help please...
Thanks... I was able to figure it out using what my book calls the old method. Here is what ended up working for me: SELECT B.Title FROM `phpCalendar_Details` AS B, `phpCalendar_Daily` AS A WHERE A.CalendarDetailsID = B.CalendarDetailsID AND CURDATE( ) BETWEEN A.StartDate AND A.StopDate -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Tuesday, February 04, 2003 1:36 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Subject: Re: [PHP-DB] A little JOIN help please... SELECT 'A.Title' FROM `phpCalendar_Daily` AS 'B' JOIN 'phpCalendar_Details' AS 'A' ON 'A.CalendarDetailsID' = 'B.CalendarDetailsID' WHERE CURDATE( ) BETWEEN 'B.StartDate' AND 'B.StopDate' Take out all of those quotes! You're making strings out of everything. WHERE CURDATE() BETWEEN 'string' AND 'anotherstring' etc... ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Send mail with a file
Under Unix I don't know that this is possible, but hopefully someone else will know of a way. I would love to be able to e-mail some Excel files that I generate with a Perl cron job, but there does not appear to be any practical way to do this in Unix. -Original Message- From: Bruno Pereira [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 16, 2003 3:47 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] Send mail with a file How can i insert a file on a mail, with the command mail()? It is possible? Cumprimentos Bruno Pereira [EMAIL PROTECTED] -Original Message- From: Bruno Pereira [mailto:[EMAIL PROTECTED]] Sent: quarta-feira, 15 de Janeiro de 2003 14:48 Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: Concatenate two strings How can i join two strings. My code is something like: $valor1=bruno; $valor2=Pereira; $valor=$valor1 + + $valor2 Can someone help me. Thanks. Cumprimentos Bruno Pereira [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Duplicate keys...
Thanks once again. I finally figured this out late yesterday. I
have no idea as to why whoever wrote this calendar to begin with did it this
way. On suggestion I have for anyone looking at a calendar system to use or
modify is to absolutely NOT use phpCommunityCalendar. This is a very poorly
coded piece of software, and if I as a relative PHP newbie know this, it is
really bad.
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 10:07 PM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Duplicate keys...
OK, I cannot figure out why I keep getting errors about
Duplicate
entry for key 1. Someone please come to my rescue here as this is
truly
kicking my but.
You have a key or unique column in your table that you're trying to
insert a duplicate value into.
if ($dateDiff == 2) {
mysql($DBName,INSERT INTO phpCalendar_Details VALUES (
'$said','Comp day for weekend
On-Call','','','','','','','','','',2,'$CalendarDetailsID')) or
die(mysql_error());
mysql($DBName,UPDATE Balances SET CompEarned=CompEarned+8,
CompTaken=CompTaken+8 WHERE said='$said') or die(mysql_error());
$result = mysql($DBName,SELECT LAST_INSERT_ID() FROM
phpCalendar_Details) or die(mysql_error());
while ($row = mysql_fetch_row($result)) {
$CalendarDetailsID = $row[0];
}
mysql($DBName,INSERT INTO Log VALUES(
'$entryid','$said','Comp',DATE_ADD('$StopDate',INTERVAL 1
DAY),'Y','','Awarded for weekend On-Call','8','$CalendarDetailsID'))
or
die(mysql_error());
mysql($DBName,INSERT INTO phpCalendar_Daily VALUES(
DATE_ADD('$StopDate',INTERVAL 1
DAY),'',1,2,'$CalendarDetailsID'))
or die(mysql_error());
}
if ($dateDiff 0) {
commonHeader($glbl_SiteTitle);
echo centerfont color=\red\bYour END DATE COMES BEFORE YOUR
START
DATE. Please complete the form below./b/font/centerp\n\n;
} elseif ($Stop != 1) {
if ($CalendarDetailsID) {
$CalendarDetailsID = $CalendarDetailsID + 1;
} else {
$result = mysql($DBName,SELECT LAST_INSERT_ID() FROM
phpCalendar_Details) or die(mysql_error());
while ($row = mysql_fetch_row($result)) {
$CalendarDetailsID = $row[0];
}
}
mysql($DBName,INSERT INTO phpCalendar_Details VALUES (
'$said','On-Call for $StartDate to
$StopDate','','','','','','','','','',1,'$CalendarDetailsID')) or
die(mysql_error());
mysql($DBName,INSERT INTO phpCalendar_Daily VALUES(
'$StartDate','$StopDate','1','1','$CalendarDetailsID')) or
die(mysql_error());
This is the only place in the script that is attempting to enter
data into the CalendarDetailsID field. This is the auto_increment
field
that is having problems, I think. The other strange part of this is
that
thedata in the auto_increment field has two separate ranges (10 - 40
with
gaps, and 70 - 78 with gaps). I have been doing a lot of adding and
removing of data to test during development, but I thought an
auto_increment
field would still track and deal with this correctly.
If you're trying to insert a number into an auto_increment field that's
already there, you'll get this error. You should always insert a NULL
value into an auto_increment column so MySQL will handle giving it the
number itself.
Also, don't worry about the gaps. They are irrelevant. If the gaps
matter to your program, then you are coding incorrectly. The
auto_increment column is there _only_ to provide a unique identifier for
each row and nothing else.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
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[PHP-DB] Date math functions...
I have a question about using DATE_ADD in MySQL. I am not sure if
this is the most effective way of dealing with this problem so I am asking
here. The situation is that I am working on modifying a calendar system for
our use. Our On-Call rotation is such that when someone is On-Call over the
weekend they automatically get Monday off. I have created a page to
simplify inputting the On-Call schedule into the calendar system, and now I
want the Monday off to be automagically scheduled when a weekend On-Call is
entered. The way that I see for doing this is adding 1 day to the end date
of the weekend On-Call and using that as the input for the day off
scheduling. Here is a bit of what I have at this point...
if ($DailyStopYear) {
$result = mysql($DBName,SELECT (TO_DAYS('$StopDate') -
TO_DAYS('$StartDate'))) or die(mysql_error());
while ($row = mysql_fetch_row($result)) {
$dateDiff = $row[0];
}
} else {
$dateDiff = 0;
}
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
My big question is about using the DATE_ADD MySQL function inside
the INSERT statement. Is this allowed? If it is allowed, is it a bad idea
for some reason? Is there a better way of doing this? Thanks in advance.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] Date math functions...
Thanks. I was noticing that it was not working. Let me give this a
try and see how things go.
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 10:32 AM
To: 1LT John W. Holmes; NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date math functions...
[snip]
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
My big question is about using the DATE_ADD MySQL function inside
the INSERT statement. Is this allowed? If it is allowed, is it a bad
idea
for some reason? Is there a better way of doing this? Thanks in
advance.
Sure, that's allowed. You have $StopDate in PHP, so you could do it with
some math in PHP, also, but then you'd have to worry about the end of
months, years, etc, whereas DATE_ADD will do this for you.
---John Holmes...
Wait... just noticed your syntax error. Don't enclose the function in single
quotes, otherwise you're trying to insert a string.
Should be:
... VALUES (DATE_ADD($StopDate,INTERVAL 1 DAY), ...
---John Holmes...
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RE: [PHP-DB] Date math functions...
Actually this is generating another error. Now, without the single
quotes I am getting the following error:
Column 'StartDate' cannot be null
It looks like for some reason the DATE_ADD is returning a NULL
value. Any more ideas?
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 10:32 AM
To: 1LT John W. Holmes; NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date math functions...
[snip]
if ($dateDiff == 3) {
mysql($DBName,UPDATE Balances SET CompEarned=CompTaken+8 WHERE
said='$said') or die(mysql_error());
mysql($DBName,INSERT INTO Log VALUES('DATE_ADD($StopDate,
INTERVAL 1 DAY','',1,2,'$CalendarDetailsID')) or die(mysql_error());
My big question is about using the DATE_ADD MySQL function inside
the INSERT statement. Is this allowed? If it is allowed, is it a bad
idea
for some reason? Is there a better way of doing this? Thanks in
advance.
Sure, that's allowed. You have $StopDate in PHP, so you could do it with
some math in PHP, also, but then you'd have to worry about the end of
months, years, etc, whereas DATE_ADD will do this for you.
---John Holmes...
Wait... just noticed your syntax error. Don't enclose the function in single
quotes, otherwise you're trying to insert a string.
Should be:
... VALUES (DATE_ADD($StopDate,INTERVAL 1 DAY), ...
---John Holmes...
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[PHP-DB] Duplicate keys...
OK, I cannot figure out why I keep getting errors about Duplicate
entry for key 1. Someone please come to my rescue here as this is truly
kicking my but.
if ($dateDiff == 2) {
mysql($DBName,INSERT INTO phpCalendar_Details VALUES (
'$said','Comp day for weekend
On-Call','','','','','','','','','',2,'$CalendarDetailsID')) or
die(mysql_error());
mysql($DBName,UPDATE Balances SET CompEarned=CompEarned+8,
CompTaken=CompTaken+8 WHERE said='$said') or die(mysql_error());
$result = mysql($DBName,SELECT LAST_INSERT_ID() FROM
phpCalendar_Details) or die(mysql_error());
while ($row = mysql_fetch_row($result)) {
$CalendarDetailsID = $row[0];
}
mysql($DBName,INSERT INTO Log VALUES(
'$entryid','$said','Comp',DATE_ADD('$StopDate',INTERVAL 1
DAY),'Y','','Awarded for weekend On-Call','8','$CalendarDetailsID')) or
die(mysql_error());
mysql($DBName,INSERT INTO phpCalendar_Daily VALUES(
DATE_ADD('$StopDate',INTERVAL 1 DAY),'',1,2,'$CalendarDetailsID'))
or die(mysql_error());
}
if ($dateDiff 0) {
commonHeader($glbl_SiteTitle);
echo centerfont color=\red\bYour END DATE COMES BEFORE YOUR START
DATE. Please complete the form below./b/font/centerp\n\n;
} elseif ($Stop != 1) {
if ($CalendarDetailsID) {
$CalendarDetailsID = $CalendarDetailsID + 1;
} else {
$result = mysql($DBName,SELECT LAST_INSERT_ID() FROM
phpCalendar_Details) or die(mysql_error());
while ($row = mysql_fetch_row($result)) {
$CalendarDetailsID = $row[0];
}
}
mysql($DBName,INSERT INTO phpCalendar_Details VALUES (
'$said','On-Call for $StartDate to
$StopDate','','','','','','','','','',1,'$CalendarDetailsID')) or
die(mysql_error());
mysql($DBName,INSERT INTO phpCalendar_Daily VALUES(
'$StartDate','$StopDate','1','1','$CalendarDetailsID')) or
die(mysql_error());
This is the only place in the script that is attempting to enter
data into the CalendarDetailsID field. This is the auto_increment field
that is having problems, I think. The other strange part of this is that
thedata in the auto_increment field has two separate ranges (10 - 40 with
gaps, and 70 - 78 with gaps). I have been doing a lot of adding and
removing of data to test during development, but I thought an auto_increment
field would still track and deal with this correctly.
Thank again for the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Stumped...
I am getting an error that is proving very difficult to isolate and was hoping for help. The error is: Column count doesn't match value count at row 1. I would include the code, but it is about 350 lines, and I am not sure where to narrow it down at. Any ideas? Thanks. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Stumped...
Nevermind. I just stumbled across the nature of my problem on Deja.com. Thanks anyway and sorry for bugging you guys about what really was a simple problem. -Original Message- From: Hutchins, Richard [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 09, 2003 12:36 PM To: NIPP, SCOTT V (SBCSI); '[EMAIL PROTECTED]' Subject: RE: [PHP-DB] Stumped... Post your SQL statement. -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 09, 2003 1:37 PM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] Stumped... I am getting an error that is proving very difficult to isolate and was hoping for help. The error is: Column count doesn't match value count at row 1. I would include the code, but it is about 350 lines, and I am not sure where to narrow it down at. Any ideas? Thanks. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Mouseover question...
OK, this isn't really a DB question, but it is definitely a PHP
question. I am working on customizing a calendar application for my group's
use and noticed that there is a function that has to do with a mouseover
effect. The problem is that this mouseover does not appear to do anything.
Could one of you gurus take a look at this function and see if you can spot
anything wrong? Thanks.
// getEvent() displays events on the calendar for IE browsers //
// This distinction is made because IE is currently the only //
// browser on which the javascript mouseover effects work.//
function getEventIE($ID, $LinkTitle, $AbbrTitle, $Date) {
$LinkTitle =ereg_replace(',#146;,$LinkTitle);
$AbbrTitle =ereg_replace(',#146;,$AbbrTitle);
echo script
dLink('./event.php?ID=$IDDate=$Date','$LinkTitle','$AbbrTitle');
/script\n\n;
}
This function is called from another page as such:
if ($ver = 4) {
echo br\n;
getEventIE($CDCI, $CDTi, $showName, $queryDate);
Thanks again.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] Mouseover question...
You are correct. I have found the JS that the dLink references.
Here it is if there are any JS experts out there who want to take a gander.
//
// expandItems() calls a javascript that replaces one string of //
// text with another on mouseover (IE only) //
//
function expandItems() {
?
script language=JavaScript!--
var no=0;
function mover(object,text) {
eval(object + '.innerText = text');
}
function mout(object,text) {
eval(object + '.innerText = text');
}
function dLink(href,text,txet) {
document.write('a href='+href+'
onMouseOut=mout(\'link'+no+'\',\''+txet+'\')
onMouseOver=mover(\'link'+no+'\',\''+text+'\') id=link'+no+'
onClick=return(openSmallWindow(\''+href+'\'))'+txet+'\/a');
no+=1;
}
//--/script
Thanks again.
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 08, 2003 11:56 AM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Mouseover question...
OK, this isn't really a DB question, but it is definitely a PHP
question. I am working on customizing a calendar application for my
group's
use and noticed that there is a function that has to do with a
mouseover
effect. The problem is that this mouseover does not appear to do
anything.
Could one of you gurus take a look at this function and see if you can
spot
anything wrong? Thanks.
// getEvent() displays events on the calendar for IE browsers //
// This distinction is made because IE is currently the only //
// browser on which the javascript mouseover effects work.//
function getEventIE($ID, $LinkTitle, $AbbrTitle, $Date) {
$LinkTitle =ereg_replace(',#146;,$LinkTitle);
$AbbrTitle =ereg_replace(',#146;,$AbbrTitle);
echo script
dLink('./event.php?ID=$IDDate=$Date','$LinkTitle','$AbbrTitle');
/script\n\n;
}
This function is called from another page as such:
if ($ver = 4) {
echo br\n;
getEventIE($CDCI, $CDTi, $showName, $queryDate);
That function has nothing to do with mouseovers. Maybe the dLink()
function does, but this is all user functions, so there's no telling
what they do. Mouseovers have nothing to do with PHP. PHP can simply
echo the javascript code required to create the mouseover effect.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
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[PHP-DB] Sore head...
My head is sore from banging my head on my desk, so I will now ask
the gurus...
I am trying to get a canned Calendar application working, and I am
running into problems. The application is phpCommunityCalendar by
AppIdeas.com. The calendar is currently semi-functional, but for some
reason the approval system is not working for me. I am trying to
troubleshoot why this is the case and this is where I am running into
problems. The following is a bit of code that I do not understand and was
hoping that someone could clear up for me.
** Not sure exactly what this code does, especially the %% portion**
if (!$glbl_LocationID) {
$glbl_LocationID = 0;
$LocQuery = LocationID LIKE '%%';
} else {
$LocQuery = LocationID = '$glbl_LocationID';
}
** SNIP This section queries the database for events that have been
submitted but NOT approved, this is the Active = '0'**
$result = mysql($DBName,SELECT CalendarDetailsID FROM phpCalendar_Daily
WHERE Active = '0') or die(mysql_error());
while ($row = mysql_fetch_row($result)) {
$ID[$i] = $row[0];
$i++;
}
** SNIP Based on looking at the DB, this section associates the matched
item descriptions for display**
for ($j = 0; $ID[$j]; $j++) {
$query .= OR CalendarDetailsID = '$ID[$j]';
}
$query = ereg_replace(^ OR ,,$query);
if (!$query) {
echo centerbThere are no results to display/b/center;
commonFooter();
exit;
** SNIP Even though I see entries in the DB that appear as they should
display (Active = 0), nothing actually appears**
$result = mysql($DBName,SELECT *
FROM phpCalendar_Details
WHERE .$LocQuery. AND (.$query.)
ORDER BY Title) or die(mysql_error());
OK. I have tested the basics of the SQL in both the second section
and the last section. If I manually query the database as in the second
section, and then use that output in the query of the last section I am
returned a record as expected. This is the reason why I am lost as to why
this is not working here.
Thanks in advance for the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] Newbie PHP/MySQL question
I would say his problem is the IIS 5.0. Switch to some non-MS operating system running Apache as a web server and that should clear up your problems. :) -Original Message- From: Jason Wong [mailto:[EMAIL PROTECTED]] Sent: Friday, December 20, 2002 11:24 AM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Newbie PHP/MySQL question On Saturday 21 December 2002 01:11, Matt Matijevich wrote: I am just getting started with php an mysql. When I try to insert a record php get stuck in some kind of loop and about 25 records are created. Any help would be greatly appreciated. Windows 2000 server latest service pack. PHP 4.2.3 MySql 3.23 IIS 5.0 My mind reading skills aren't very good, but I think there's something wrong with line 13 of your program. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* The greatest disloyalty one can offer to great pioneers is to refuse to move an inch from where they stood. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Oracle connectivity question...
OK. I think I now have Oracle support built into PHP/Apache. The biggest hiccup for HP-UX seems to be that once you add Oracle support to PHP, you CANNOT built PHP as a DSO, PHP MUST be compiled into Apache. Once again, I think this is the case, I am still not completely sure at this point. I am now receiving an error on trying to connect to the Oracle database that I do not understand, but leads me to believe that the PHP Oracle support is functioning. Here is the code I am attempting: ?php $connection = OCILogon(userid, password) or die (Unable to logon to database.); # phpinfo(); ? Here is the error I am receiving: Warning: OCISessionBegin: ORA-01034: ORACLE not available in /www/DW/oratest.php on line 10 Unable to logon to database. The code I am attempting is simply to verify successful connectivity to the database. I know that this doesn't really do anything, but I am trying this simply to test for a successful connection. Please let me know if you have any ideas or suggestions. Thanks again. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Oracle/PHP question...
Correct. I am working on Unix, specifically HP-UX 11.00. -Original Message- From: Andrey Hristov [mailto:[EMAIL PROTECTED]] Sent: Tuesday, December 17, 2002 8:38 AM To: xxx ; [EMAIL PROTECTED] Subject: Re: [PHP-DB] Oracle/PHP question... AFAIK he uses *nix, not windows. (.so-s are possible sollution but prebuilt ones are not shipped with php). Andrey - Original Message - From: xxx [EMAIL PROTECTED] To: Anthony Carlos [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Tuesday, December 17, 2002 4:32 PM Subject: Re: [PHP-DB] Oracle/PHP question... Hy, in php.ini you have to uncomment the ;extension=php_oci8.dll ;extension=php_oracle.dll ; is a comment and restart the server cybercop78 On Tue, 17 Dec 2002 08:29:36 -0500, Anthony Carlos wrote Yes, you need to recompile PHP: ./configure --with-oci8=[your_oracle_home] -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 1:58 PM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] Oracle/PHP question... I am now trying to setup connectivity to a remote Oracle database for my PHP/Apache web server. I am attempting to verify connectivity to the database by: ?php $connection = OCILogon(user, password) or die (Unable to logon to database.); ? I have installed the Oracle client on the system, and I have also setup the ORACLE_HOME and ORACLE_SID environment variables. Unfortunately at this point I am not even able to make a successful connection to the Oracle database. I am currently receiving the following error: Fatal error: Call to undefined function: ocilogon() in /www/DW/oratest.php on line 10 Do I need to recompile PHP to be Oracle aware or something? What is it I am missing at this point? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php http://www.idilis.ro - Stiri, e-mail gratuit, download, SMS, server de counter-strike, hosting gratuit, servicii internet... Fii cu un pas inaintea celorlati! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] PHP/Oracle Install questions...
I am attempting to recompile PHP 4.2.3 on HP-UX 11.00 for connectivity to both a MySQL and a remote Oracle server. I am not having much luck with this right now. Currently I am getting a couple of different errors that are of concern for me. First of all, in the Apache error log I am seeing the following entry every time I start the Apache server: [Tue Dec 17 09:42:39 2002] [warn] module php4_module is already loaded, skipping I am not sure as to why I am receiving this error. I know that PHP is only compiled as a module, because it does not show up in a httpd -l. I am wondering if this is a result of some kind of syntax error in my httpd.conf file. The big problem though is getting PHP to compile completely again. I know that on HP there is an issue with the extension on the php library. It has something to do with .sl as opposed to .so, but I cannot remember exactly what it is and for some reason I can't seem to find the documentation on it this time around. Any help in the right direction here would be most appreciated. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] FW: PHP/Oracle Install questions...
More information for someone to hopefully come to my rescue on this... root@torvalds:/usr/local/apache/bin ./apachectl startssl /usr/lib/dld.sl: Can't shl_load() a library containing Thread Local Storage: /us r/lib/libcl.2 /usr/lib/dld.sl: Exec format error Syntax error on line 207 of /usr/local/apache/conf/httpd.conf: Cannot load /usr/local/apache/libexec/libphp4.so into server: Exec format error ./apachectl startssl: httpd could not be started This is what I get when I attempt to start Apache once I think I have compiled the new PHP module properly. -Original Message- From: NIPP, SCOTT V (SBCSI) Sent: Tuesday, December 17, 2002 9:56 AM To: '[EMAIL PROTECTED]' Subject: PHP/Oracle Install questions... I am attempting to recompile PHP 4.2.3 on HP-UX 11.00 for connectivity to both a MySQL and a remote Oracle server. I am not having much luck with this right now. Currently I am getting a couple of different errors that are of concern for me. First of all, in the Apache error log I am seeing the following entry every time I start the Apache server: [Tue Dec 17 09:42:39 2002] [warn] module php4_module is already loaded, skipping I am not sure as to why I am receiving this error. I know that PHP is only compiled as a module, because it does not show up in a httpd -l. I am wondering if this is a result of some kind of syntax error in my httpd.conf file. The big problem though is getting PHP to compile completely again. I know that on HP there is an issue with the extension on the php library. It has something to do with .sl as opposed to .so, but I cannot remember exactly what it is and for some reason I can't seem to find the documentation on it this time around. Any help in the right direction here would be most appreciated. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Oracle/PHP question...
I am now trying to setup connectivity to a remote Oracle database for my PHP/Apache web server. I am attempting to verify connectivity to the database by: ?php $connection = OCILogon(user, password) or die (Unable to logon to database.); ? I have installed the Oracle client on the system, and I have also setup the ORACLE_HOME and ORACLE_SID environment variables. Unfortunately at this point I am not even able to make a successful connection to the Oracle database. I am currently receiving the following error: Fatal error: Call to undefined function: ocilogon() in /www/DW/oratest.php on line 10 Do I need to recompile PHP to be Oracle aware or something? What is it I am missing at this point? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Oracle/PHP question...
Thanks, but my web server is HP-UX. Didn't anyone ever tell you Win-blows sucks? :) -Original Message- From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 3:07 PM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Oracle/PHP question... If your webserver is win32 then you need to uncomment the line in php.ini that says : ;extension=php_oci8.dll if it is unix based I am not sure what you need to do. Ryan -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 1:58 PM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] Oracle/PHP question... I am now trying to setup connectivity to a remote Oracle database for my PHP/Apache web server. I am attempting to verify connectivity to the database by: ?php $connection = OCILogon(user, password) or die (Unable to logon to database.); ? I have installed the Oracle client on the system, and I have also setup the ORACLE_HOME and ORACLE_SID environment variables. Unfortunately at this point I am not even able to make a successful connection to the Oracle database. I am currently receiving the following error: Fatal error: Call to undefined function: ocilogon() in /www/DW/oratest.php on line 10 Do I need to recompile PHP to be Oracle aware or something? What is it I am missing at this point? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Oracle/PHP question...
Figuring how shouldn't be too tough. Compiling software for the most part is a breeze. My concern was whether I needed to recompile or not. I am now trying that, and hopefully this will address the problem for me. As to the recompiling with every release, that is no big deal either as the recompile is all of 4 entries on the command line. Putting this together in a shell script is simplicity in itself, and requires no polishing up even though most of my system scripting I do in Perl. It is nice to see though that you are not trying to tell me how Win-blows is better than Unix. Sure Windows is nice in that virtually everything is simple enough for a 3rd grader to do stuff on it, but I like to think I have a little more intelligence than your average 3rd grader. Oh well, I don't want this to degenerate any further so we can agree to disagree about platforms or you can admit to possibly being a closet Unix fan. :) -Original Message- From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 3:19 PM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Oracle/PHP question... BTW... I'm not the one stuck trying to figure out how to recompile my current version of PHP to include the oci functions... haha! And guess what, with every release you get to do it all over again... so polish up on the shell scripting unless you want to reinvent the wheel every month or two. :-D Ryan -Original Message- From: Ryan Jameson (USA) Sent: Monday, December 16, 2002 2:15 PM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Oracle/PHP question... :-) ... Windows is the best at running the C# .NET applications I have to build to make everyone happy that we are industry standard ... when big brother is not looking I use PHP for everything I can get away with. I'm actually very impressed with how easily PHP integrates into an IIS / MSSQL environment (in CGI mode that is) ... it has been quite stable and believe it or not upgrades don't require a reboot!!! I have even called the php parser directly from sql server dts packages agent jobs... PHP is much more powerful then anything m$ offers. Ryan -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 2:08 PM To: Ryan Jameson (USA); [EMAIL PROTECTED] Subject: RE: [PHP-DB] Oracle/PHP question... Thanks, but my web server is HP-UX. Didn't anyone ever tell you Win-blows sucks? :) -Original Message- From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 3:07 PM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Oracle/PHP question... If your webserver is win32 then you need to uncomment the line in php.ini that says : ;extension=php_oci8.dll if it is unix based I am not sure what you need to do. Ryan -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Monday, December 16, 2002 1:58 PM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] Oracle/PHP question... I am now trying to setup connectivity to a remote Oracle database for my PHP/Apache web server. I am attempting to verify connectivity to the database by: ?php $connection = OCILogon(user, password) or die (Unable to logon to database.); ? I have installed the Oracle client on the system, and I have also setup the ORACLE_HOME and ORACLE_SID environment variables. Unfortunately at this point I am not even able to make a successful connection to the Oracle database. I am currently receiving the following error: Fatal error: Call to undefined function: ocilogon() in /www/DW/oratest.php on line 10 Do I need to recompile PHP to be Oracle aware or something? What is it I am missing at this point? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Remote Oracle database questions...
I have now been granted access to a remote Oracle database that I need to query. All of the work that I have done with databases and PHP has been focused on MySQL, does anyone know of a good tutorial for PHP with Oracle? My immediate question is how do I get logged into the remote database to perform my query? I notice that there is a Oracle function in PHP called ora_logon. This function description mentions something about logging in using 'user@TNSNAME'. I am pretty confused by what this means. Thanks in advance for the help. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Remote Oracle database questions...
DOH!!! I am seriously hoping that I do not need to have the Oracle client installed on the webserver running the PHP queries. I have no idea about the client licensing, and if it involves money in any way it is pretty much out of the question. Oh well... Hopefully I can find out about this stuff from one of our DBAs. -Original Message- From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]] Sent: Wednesday, December 11, 2002 10:58 AM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Remote Oracle database questions... One thing I can help with is TNSNAME. It was very confusing for me at first, it's kind of like an alias on steroids... It is a necessary part of connecting to an Oracle server. TNSNAMES are declared in a file called... go figure... tnsnames. Even good Oracle DBAs run into confusion because the tnsnames file (on the connecting client) can get huge. As far as how PHP does it, I'd ASSUME that the Oracle client would have to be installed on the server PHP is running on. Upon installing the Oracle client it will create the tnsnames file and set the ORACLE_HOME environment variable, the file needs a reference to the server you wish to connect to. I haven't used Oracle since 8i came out... but I'm pretty sure this is accurate. Except the guess about how PHP does it. Ryan -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Wednesday, December 11, 2002 9:24 AM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] Remote Oracle database questions... I have now been granted access to a remote Oracle database that I need to query. All of the work that I have done with databases and PHP has been focused on MySQL, does anyone know of a good tutorial for PHP with Oracle? My immediate question is how do I get logged into the remote database to perform my query? I notice that there is a Oracle function in PHP called ora_logon. This function description mentions something about logging in using 'user@TNSNAME'. I am pretty confused by what this means. Thanks in advance for the help. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Form Best Practice/Help...
I am working on an application and I am not having a lot of luck
getting a form to behave the way I want. I am trying to keep the form self
referencing, not sure if that is the correct terminology. The form is an
approval form that queries a database to create a list of checkboxes. These
checkbox selections shall then, upon submit, update the database table with
an approval timestamp and fire off an e-mail to one of our administrators to
take action. The problem I am having is that once the submit button is
clicked the page is not displaying the results I am expecting. This leads
me to the obvious conclusion that something in the flow control of my page
is incorrect.
My questions are regarding the way in which I am laying out this
page. I am trying to keep the page self referencing such that there are
fewer pages to keep track of, and I also feel this looks more professional
since the approval process will only involve a single URL. Currently I am
simply trying to get a click on the submit button to display a Done
message to test the flow control. Now, on to the questions... Is this type
of flow control logic a good idea? Is there a better way to implement this
process? What am I missing that is causing this flow control logic to fail?
OK. Here is the code:
?php require_once('useraccounts.lib.php');
$safe_verify_string = urldecode($_REQUEST['verify_string']);
mysql_select_db($database, $Prod);
$query = SELECT * FROM accounts WHERE verifyurl='{$safe_verify_string}';
$result = mysql_query($query, $Prod) or die(mysql_error());
# $test = mysql_num_rows($result);
$list = mysql_fetch_assoc($result);
$id = split('-', $list['id-sys']);
if (isset($cnt)) {
echo p align=\center\font color=\#0033FF\ size=\5\Done./p;
}
?
html
head
titleUntitled Document/title
meta http-equiv=Content-Type content=text/html; charset=iso-8859-1
/head
body
p align=centerfont color=#0033FF size=5Welcome to the New User
Account Approval system./font
/p
?php
if (mysql_num_rows($result) == 0 ) {
echo p align=\center\No accounts pending approval were found for this
request. /p;
echo p align=\center\Please contact the SA Team if you are having
problems with this request system./p;
} else {
echo p align=\center\The following accounts for $id[0] have been
requested and require your approval prior to creation. /p;
echo p align=\center\;
$test1 = CheckboxList($query, $CheckboxText, $link);
}
?
/p
/body
/html
The function CheckboxList is called from the useraccounts.lib.php
script and is returning the $cnt variable. This $cnt variable is only used
as a mechanism to create the checkbox list of systems in four columns. I
decided to check for the existence of this variable since it must be set if
the CheckboxList function is called. Below is the CheckboxList
function:
function CheckboxList($query, $CheckboxText, $link) {
global $cnt;
$cnt = 1;
$result1 = mysql_query($query);
echo form name=\form1\ method=\get\ action=\$link\;
echo table border=\0\ align=\center\;
while ($list = mysql_fetch_assoc($result1)) {
$sys = split('-', $list['id-sys']);
if ($cnt == 1) {
$cnt = 2;
echo trtd width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td;
} elseif ($cnt == 2) {
$cnt = 3;
echo td width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td;
} elseif ($cnt == 3) {
$cnt = 4;
echo td width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td;
} elseif ($cnt == 4) {
$cnt = 1;
echo td width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td/tr;
}
}
echo /table;
if ($CheckboxText) {
echo p align=\center\font size=\4\$CheckboxText/font/p;
}
echo div align=\center\input type=\submit\ name=\Submit\
value=\Submit\/div/form;
return $cnt;
}
Thanks in advance, and again, for the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Update Query Help...
I am attempting to UPDATE a table and I have having absolutely ZERO success. Here is the query: $tmp = $_POST['sbcuid'].-.$_POST['system'][$a]; $update = UPDATE accounts SET atime='NOW()' WHERE \id-sys\='.$tmp.'; echo $update; $result1 = mysql_query($update, $Prod) or die(mysql_error()); echo mysql_affected_rows(); Please help me figure out why this is not working. I have tried quoting the column name 'id-sys' every way I can think of, but nothing works. The column is initially NULL, and I am attempting to update just the single column with a timestamp. Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Update Query Help...
I have definitely isolated the problem. If I change the column name that the match is being performed on from 'id-sys' to 'id_sys' I can execute the query without having to worry about quoting the column name and everything works. I think that I am just going to leave the column name changed and take this as a lesson never to use '-' in a column name again. Thanks again for the help. Maybe we all learned something from this. -Original Message- From: DL Neil [mailto:[EMAIL PROTECTED]] Sent: Tuesday, December 10, 2002 2:00 PM To: NIPP, SCOTT V (SBCSI); '1LT John W. Holmes'; [EMAIL PROTECTED] Subject: Re: [PHP-DB] Update Query Help... SCOTT, The list's crystal ball filter is down for maintenance. Show us the tbl schema and the debug print of $update. Then we won't be firing blind! =dn I understand that the column name does not normally need quotes, but without quotes on the column name I get a mysql error message about no column named 'id'. Unfortunately, I cannot even get the UPDATE statement working outside of PHP through an direct SQL statement. Everything looks like it should work, but the affected columns is always zero, and obviously the atime column is not getting the timestamp. This is quite frustrating. Thanks again for the help. Hopefully someone will come up with something to help me get this working. -Original Message- From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] Sent: Tuesday, December 10, 2002 1:45 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Subject: Re: [PHP-DB] Update Query Help... $tmp = $_POST['sbcuid'].-.$_POST['system'][$a]; $update = UPDATE accounts SET atime='NOW()' WHERE \id-sys\='.$tmp.'; echo $update; $result1 = mysql_query($update, $Prod) or die(mysql_error()); echo mysql_affected_rows(); Try: $update = UPDATE accounts SET atime=NOW() WHERE id-sys='$tmp'; NOW() is a function, don't enclose it within quotes and make it a string. You don't put quotes around column names, either, only string values. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Function questions...
I am trying to convert something that I have working in a PHP script
to be a function in a library type of file. The code contained in the
function works fine in another script, but I am currently getting no output
from the function. I have a question about the nature of functions that may
help clear up my problems.
Do 'echo' commands within functions actually work to directly
display information? I have only used functions up to this point that use
the 'return' command to pass a variable back to the script that called the
function. I am pretty confused on how to actually use a function that
formats data to be displayed. This is my primary source of confusion.
Below is the function that is giving me trouble:
function CheckboxList($query) {
$result1 = mysql_query($query);
do {
$sys = split('-', $list['id-sys']);
if ($cnt == 1) {
$cnt = 2;
echo trtd width=\161\input name=\system[]\ type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 2) {
$cnt = 3;
echo trtd width=\161\input name=\system[]\ type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 3) {
$cnt = 4;
echo trtd width=\161\input name=\system[]\ type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 4) {
$cnt = 1;
echo trtd width=\161\input name=\system[]\ type=\checkbox\
value=\$sys\;
echo $sys./td/tr;
}
} while ($list = mysql_fetch_assoc($result1));
}
Here is also the way I am calling this function:
snip...
$query = SELECT * FROM accounts WHERE verifyurl='$safe_verify_string';
snip...
CheckboxList($query);
snip...
Is there something wrong with the way in which I am calling this
function? Thanks in advance for all the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] Function questions...
Thanks guys. I was able to work through this issue pretty much on
my own. I got this portion of things working nicely except for a form
button that is appearing at the top rather than the bottom of the page.
-Original Message-
From: Mark [mailto:[EMAIL PROTECTED]]
Sent: Friday, December 06, 2002 1:58 PM
To: '[EMAIL PROTECTED]'
Subject: Re: [PHP-DB] Function questions...
$sys is an array, so echoing $sys will cause problems.
Are you trying to get a 4xN table of values and checkboxes? There's
got to be an easier way...
--- NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] wrote:
I am trying to convert something that I have working in a PHP
script
to be a function in a library type of file. The code contained in
the
function works fine in another script, but I am currently getting
no output
from the function. I have a question about the nature of functions
that may
help clear up my problems.
Do 'echo' commands within functions actually work to directly
display information? I have only used functions up to this point
that use
the 'return' command to pass a variable back to the script that
called the
function. I am pretty confused on how to actually use a function
that
formats data to be displayed. This is my primary source of
confusion.
Below is the function that is giving me trouble:
function CheckboxList($query) {
$result1 = mysql_query($query);
do {
$sys = split('-', $list['id-sys']);
if ($cnt == 1) {
$cnt = 2;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 2) {
$cnt = 3;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 3) {
$cnt = 4;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 4) {
$cnt = 1;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td/tr;
}
} while ($list = mysql_fetch_assoc($result1));
}
Here is also the way I am calling this function:
snip...
$query = SELECT * FROM accounts WHERE
verifyurl='$safe_verify_string';
snip...
CheckboxList($query);
snip...
Is there something wrong with the way in which I am calling this
function? Thanks in advance for all the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
--
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To unsubscribe, visit: http://www.php.net/unsub.php
=
Mark Weinstock
[EMAIL PROTECTED]
***
You can't demand something as a right unless you are willing to fight to
death to defend everyone else's right to the same thing.
-Stolen from the now-defunct Randy's Random mailing list.
***
__
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RE: [PHP-DB] Function questions...
OK. Stumped once again. This function is now properly generating
and displaying the form as I intend it to. The only problem is that the
optional text and button at the end of the function is actually displaying
above the choices provided in the while loop from the database. Please let
me know what you think.
function CheckboxList($query, $CheckboxText) {
$cnt = 1;
$result1 = mysql_query($query);
echo form name=\form1\ method=\post\ action=\test_display.php\;
echo table border=\0\ align=\center\;
while ($list = mysql_fetch_assoc($result1)) {
$sys = split('-', $list['id-sys']);
if ($cnt == 1) {
$cnt = 2;
echo trtd width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td;
} elseif ($cnt == 2) {
$cnt = 3;
echo td width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td;
} elseif ($cnt == 3) {
$cnt = 4;
echo td width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td;
} elseif ($cnt == 4) {
$cnt = 1;
echo td width=\161\input name=\system[]\ type=\checkbox\
value=\$sys[1]\;
echo $sys[1]./td/tr;
}
}
if ($CheckboxText) {
echo p align=\center\font size=\4\$CheckboxText/font/p;
}
echo div align=\center\input type=\submit\ name=\Submit\
value=\Submit\/div/form;
}
One other question while I am at it... I am wanting to provide a
button that selects all of the checkboxes automatically and refreshes the
page. You know, a select all button. I know nothing of Javascript, but
this is what should do this, correct? Once again, any and all help is most
appreciated.
-Original Message-
From: Mark [mailto:[EMAIL PROTECTED]]
Sent: Friday, December 06, 2002 1:58 PM
To: '[EMAIL PROTECTED]'
Subject: Re: [PHP-DB] Function questions...
$sys is an array, so echoing $sys will cause problems.
Are you trying to get a 4xN table of values and checkboxes? There's
got to be an easier way...
--- NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] wrote:
I am trying to convert something that I have working in a PHP
script
to be a function in a library type of file. The code contained in
the
function works fine in another script, but I am currently getting
no output
from the function. I have a question about the nature of functions
that may
help clear up my problems.
Do 'echo' commands within functions actually work to directly
display information? I have only used functions up to this point
that use
the 'return' command to pass a variable back to the script that
called the
function. I am pretty confused on how to actually use a function
that
formats data to be displayed. This is my primary source of
confusion.
Below is the function that is giving me trouble:
function CheckboxList($query) {
$result1 = mysql_query($query);
do {
$sys = split('-', $list['id-sys']);
if ($cnt == 1) {
$cnt = 2;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 2) {
$cnt = 3;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 3) {
$cnt = 4;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td;
} elseif ($cnt == 4) {
$cnt = 1;
echo trtd width=\161\input name=\system[]\
type=\checkbox\
value=\$sys\;
echo $sys./td/tr;
}
} while ($list = mysql_fetch_assoc($result1));
}
Here is also the way I am calling this function:
snip...
$query = SELECT * FROM accounts WHERE
verifyurl='$safe_verify_string';
snip...
CheckboxList($query);
snip...
Is there something wrong with the way in which I am calling this
function? Thanks in advance for all the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
=
Mark Weinstock
[EMAIL PROTECTED]
***
You can't demand something as a right unless you are willing to fight to
death to defend everyone else's right to the same thing.
-Stolen from the now-defunct Randy's Random mailing list.
***
__
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RE: [PHP-DB] Function questions...
Duh, OK. That fixed me right up. Thanks. Now if I can just figure out the 'Select All' option. -Original Message- From: Jason Wong [mailto:[EMAIL PROTECTED]] Sent: Friday, December 06, 2002 2:28 PM To: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Function questions... On Saturday 07 December 2002 04:12, NIPP, SCOTT V (SBCSI) wrote: OK. Stumped once again. This function is now properly generating and displaying the form as I intend it to. The only problem is that the optional text and button at the end of the function is actually displaying above the choices provided in the while loop from the database. Please let me know what you think. Your table doesn't seem to be closed. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Never volunteer for anything. -- Lackland */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Data not entering DB
Yes, and yes. -Original Message- From: John [mailto:[EMAIL PROTECTED]] Sent: Friday, December 06, 2002 2:41 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Data not entering DB Questions: Is this the right email addy to post help questions, and may I post the php code within the email html? Thanks, John --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.426 / Virus Database: 239 - Release Date: 12/2/2002 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Special URL Verification String... PS
Thanks for the responses. I chose to use this method as it appears
to be the most simple and effective. For anyone else I am also using the
substr function to cut the random string down to something more manageable
for the database. Thanks again for the feedback.
-Original Message-
From: Peter Lovatt [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, December 04, 2002 4:45 PM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Special URL Verification String... PS
Hi
PS
I would use
$verify_string = md5(uniqid(rand()));
to generate a random identifier of the sort you are looking for
Peter
---
Excellence in internet and open source software
---
Sunmaia
Birmingham
UK
www.sunmaia.net
tel. 0121-242-1473
International +44-121-242-1473
---
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]]
Sent: 04 December 2002 22:07
To: '[EMAIL PROTECTED]'
Subject: [PHP-DB] Special URL Verification String...
I am trying to implement an approval system that sends an e-mail to
a user for approval linking them back to a special URL that matches database
entries for a verification string. The verification string sent via the
e-mail does not match that in the database and I am looking for help in
figuring out why. Here is what generates the verification string:
$verify_string = $sbcuid;
for ($i = 0 ; $i 14; $i++) {
$verify_string .= chr(mt_rand(32,126));
}
Here is what this results in inside the database table:
sn4265e~zD|W(XTMxO+
Here is what is sent in the e-mail:
sn4265e%7EzD%7CW%28XTMxO%22%2B
I know that I am doing something wrong here, but what. Thanks in
advance.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Newline help...
I must be stupid, because I cannot for the life of me figure out why
I am not getting newlines from some very simply 'echo' statements. Here is
the portion of the code below, and I am getting everything run together on a
single line:
?php do {
echo $list['id-sys'] . \n;
$sys = split('-', $list['id-sys']);
echo $sys[1]\n;
} while ($list = mysql_fetch_assoc($result));
I am truly stumped by this. It is quite frustrating that something
as simple as a newline can misbehave. I have been searching for this on the
PHP site, newsgroups, and in a couple of books I have and yet I am still
stumped. Help please. Thanks.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] Newline help...
Why does the documentation everywhere seem to imply that you can use
the typical newline character? This is most confusing. I seem code
examples all over the place using newlines. Oh well, now I know. Thanks.
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Thursday, December 05, 2002 10:34 AM
To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Newline help...
HTML doesn't render newlines, only BR. Look at your HTML source code and
the newlines are there.
or,
if you're on windows, use \r\n for newlines.
---John Holmes...
- Original Message -
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, December 05, 2002 11:28 AM
Subject: [PHP-DB] Newline help...
I must be stupid, because I cannot for the life of me figure out why
I am not getting newlines from some very simply 'echo' statements. Here
is
the portion of the code below, and I am getting everything run together on
a
single line:
?php do {
echo $list['id-sys'] . \n;
$sys = split('-', $list['id-sys']);
echo $sys[1]\n;
} while ($list = mysql_fetch_assoc($result));
I am truly stumped by this. It is quite frustrating that something
as simple as a newline can misbehave. I have been searching for this on
the
PHP site, newsgroups, and in a couple of books I have and yet I am still
stumped. Help please. Thanks.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Special URL Verification String...
I am trying to implement an approval system that sends an e-mail to
a user for approval linking them back to a special URL that matches database
entries for a verification string. The verification string sent via the
e-mail does not match that in the database and I am looking for help in
figuring out why. Here is what generates the verification string:
$verify_string = $sbcuid;
for ($i = 0 ; $i 14; $i++) {
$verify_string .= chr(mt_rand(32,126));
}
Here is what this results in inside the database table:
sn4265e~zD|W(XTMxO+
Here is what is sent in the e-mail:
sn4265e%7EzD%7CW%28XTMxO%22%2B
I know that I am doing something wrong here, but what. Thanks in
advance.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] INSERT question...
I have an INSERT statement that I cannot quiet get working. The
table that this data is being put into has 7 columns, but I only care about
putting in the data from the first two columns at this time. The first
column is an array element, and the second column needs to be a timestamp.
Here is the INSERT statement:
$query = INSERT INTO accounts VALUES('.$accnts[0].', 'TIMESTAMP(10)', , ,
, ,);
Here is the error being displayed on the web page:
You have an error in your SQL syntax near ' , , ,)' at line 1
Thanks in advance for the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] INSERT question...
OK. This sounds great, but now I am getting a completely different
error message.
You have an error in your SQL syntax near '-sys) VALUES('sn4265-turner')' at
line 1
Here is the INSERT statement:
$query = INSERT INTO accounts (id-sys) VALUES('.$accnts[0].');
Do I have to rename the column? I would rather not if I can avoid
it. If I must, is this a problem with PHP or MySQL not understanding '-' in
a column name? Thanks again.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, November 20, 2002 1:28 PM
To: '1LT John W. Holmes'; NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED]
Subject: RE: [PHP-DB] INSERT question...
Hmm
You learn something new everyday :)
Aaron
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: November 20, 2002 2:27 PM
To: Aaron Wolski; 'NIPP, SCOTT V (SBCSI)'; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] INSERT question...
Try... $query = INSERT INTO accounts (accnts,timestamp)
VALUES(''.$accnts[0].'','TIMESTAMP(10)');
Might work?
Aaron
For the OP, you don't really want to insert the text TIMESTAMP(10) into
a
column, do you? If you want the current 10-digit timestamp, then you can
just use NOW() or CURRENT_DATE, so long as your column is declared with
a
width of 10. In fact, if it's a timestamp, you don't need to do
anything,
it'll be updated with the current time upon insert. so you could
probably
just do this:
$query = INSERT INTO accounts (accnts) VALUES (.$accnts[0].);
---John Holmes...
-Original Message-
From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]]
Sent: November 20, 2002 2:16 PM
To: '[EMAIL PROTECTED]'
Subject: [PHP-DB] INSERT question...
I have an INSERT statement that I cannot quiet get working. The
table that this data is being put into has 7 columns, but I only care
about
putting in the data from the first two columns at this time. The
first
column is an array element, and the second column needs to be a
timestamp.
Here is the INSERT statement:
$query = INSERT INTO accounts VALUES('.$accnts[0].',
'TIMESTAMP(10)',
, ,
, ,);
Here is the error being displayed on the web page:
You have an error in your SQL syntax near ' , , ,)' at line 1
Thanks in advance for the help.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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RE: [PHP-DB] INSERT question...
Great. That is perfect. The last minor issue is that the timestamp
is a datetime column, and needs to be such since I don't want this to change
when the record is updated. This is correct, right? If the column were a
timestamp data type, the value would change every time the record is
updated? I am trying:
$query = INSERT INTO accounts (`id-sys`, rtime) VALUES('.$accnts[0].',
'NOW()');
This is entering the array element properly, but not the time. The
datetime field is coming up with all zeroes.
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, November 20, 2002 2:21 PM
To: NIPP, SCOTT V (SBCSI); 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] INSERT question...
OK. This sounds great, but now I am getting a completely different
error message.
You have an error in your SQL syntax near '-sys) VALUES('sn4265-turner')'
at
line 1
Here is the INSERT statement:
$query = INSERT INTO accounts (id-sys) VALUES('.$accnts[0].');
Do I have to rename the column? I would rather not if I can avoid
it. If I must, is this a problem with PHP or MySQL not understanding '-'
in
a column name? Thanks again.
Try putting backtics (not single quotes) around the column name.
$query = INSERT INTO accounts (`id-sys`) VALUES ('.$accnts[0].');
if that don't work, you'll have to rename the column.
---John Holmes...
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RE: [PHP-DB] INSERT question...
Thanks a bunch. That fixed me right up. Still learning a lot about
PHP and MySQL. Hopefully I will be able to remember this as I am sure this
issue will come up for me again in the future.
-Original Message-
From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, November 20, 2002 2:47 PM
To: NIPP, SCOTT V (SBCSI); 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] INSERT question...
$query = INSERT INTO accounts (`id-sys`, rtime) VALUES('.$accnts[0].',
'NOW()');
Remove the quotes around NOW(). It's a function, not a string.
---John Holmes...
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[PHP-DB] General How-to Question...
Once again I am posting a bit off topic, so tell me to go away if this is bugging people... Thanks for all of the assistance so far in creating what will be a completely automated user account request system. I am now trying to figure out the mechanics of the account approval process. So far I have created a set of web pages that allows a user to login and select systems from a database populated list. These selections in turn populate another database table and include a timestamp as to when the request was made. I am working on a Perl script to query the database looking for new requests, and from this an e-mail will be sent to the user's supervisor for approval. The area where I am needing more help is getting the supervisor back to the approval web page. My intention is to provide the supervisor with a web link in the e-mail to take him directly to the approval page. The question is how do I actually make this occur. I am pretty sure I can figure out the Perl scripting to send the e-mail, but what is the URL that I am having him come back to? How do I create this new URL? Any suggestions ideas or pointers would be most helpful. Thanks again for all the help and sorry for the off-topic post. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] HTML Forms question...
I know that this is not the forum for this question, I am only looking for a pointer in the right direction here. I need to gain a better understanding of HTML forms, specifically checkboxes. What I am looking for is once I present a list of checkboxes and the user makes his selections, how is this presented to the subsequent pages? Does anyone know of a website that has a tutorial or good explanation of how this works? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] HTML Forms question...
OK. If I am using the POST method and the checkbox 'name' for all of the checkboxes is 'system', how do I access the results? I am researching on the PHP site trying to figure out what this array is, and how I access this if the index name is the same for all elements. I am hoping that this data will be in an array and I can simply loop through the contents of the array. If this data is stored in a hash (Perl word for this type of array, not sure if it's the same in PHP), how do I access this data since the index for every element would be the same? I am SURE that I am probably missing some important conceptual issues here, but still learning as I go. Unfortunately, this probably means that I will be completely rewriting this application at least once in the future. Oh well, live and learn. Thanks again for the help. -Original Message- From: Ryan Jameson (USA) [mailto:[EMAIL PROTECTED]] Sent: Tuesday, November 19, 2002 10:35 AM To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] HTML Forms question... input type=checkbox name=checkbox1 value=scoobydoo if this is checked the value scoobydoo will be put into the array with index checkbox1 which array is dependant on which method your form uses. :-) If it is not checked then there is no entry. in PHP with register_globals on the variable $checkbox1 will equal scoobydoo and not be set if the box is not checked. hope this helps... Ryan -Original Message- From: NIPP, SCOTT V (SBCSI) [mailto:[EMAIL PROTECTED]] Sent: Tuesday, November 19, 2002 9:28 AM To: '[EMAIL PROTECTED]' Subject: [PHP-DB] HTML Forms question... I know that this is not the forum for this question, I am only looking for a pointer in the right direction here. I need to gain a better understanding of HTML forms, specifically checkboxes. What I am looking for is once I present a list of checkboxes and the user makes his selections, how is this presented to the subsequent pages? Does anyone know of a website that has a tutorial or good explanation of how this works? Thanks in advance. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
