[PHP-DB] PHP - Ms SQL Database connection.
Hi All, Has somebody made a connection from PHP (4.2.2) running on Linux os to a Ms SQL database running on a W2K machine. If so how have you done that? Where can I find some kind of tutorial or howto? Cheers, Rene. === R. Groothuis - Aeqis Gruttosingel 50 2496 HX Den Haag - The Netherlands E-mail: [EMAIL PROTECTED] Tel: +31-15-3618824 Mobile: +31-6-51807877 Fax: +31-15-3618823 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: PHP - Ms SQL Database connection.
I guess it is just like connecting to the database at your local machine . Check security on wk2 so you are allowed to connect. Jochem Rene Groothuis wrote: Hi All, Has somebody made a connection from PHP (4.2.2) running on Linux os to a Ms SQL database running on a W2K machine. If so how have you done that? Where can I find some kind of tutorial or howto? Cheers, Rene. === R. Groothuis - Aeqis Gruttosingel 50 2496 HX Den Haag - The Netherlands E-mail: [EMAIL PROTECTED] Tel: +31-15-3618824 Mobile: +31-6-51807877 Fax: +31-15-3618823 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Unlimited Categories
Well, I'm not sure if you actually truly mean unlimited, because if you do, be aware that when you define those fields: # BIGINTUNSIGNED = 8 Byte = = 18446744073709551615 # INT UNSIGNED = 4 Byte = = 4294967295 # MEDIUMINT UNSIGNED = 3 Byte = FF = 16777215 # SMALLINT UNSIGNED = 2 Byte = = 65535 # TINYINT UNSIGNED = 1 Byte = FF = 255 # BIGINTSIGNED = -9223372036854775808 to 9223372036854775807 # INT SIGNED = -2147483648 to 2147483647 # MEDIUMINT SIGNED = -8388608 to 8388607 # SMALLINT SIGNED = -32768 to 32767 # TINYINT SIGNED = -128 to 127 So, while an UNSIGNED BIGINT would be a bahugabyte number, it is in fact finite and not unlimited. :) -Original Message- From: Micah Stevens [mailto:[EMAIL PROTECTED]] Sent: Monday, January 13, 2003 9:45 PM To: Gerard Samuel Cc: php-db Subject: Re: [PHP-DB] Unlimited Categories Use the parent/child relationship thing: CREATE TABLE Categories ( CategoryID int(11) NOT NULL auto_increment, ParentID int(11) NOT NULL default '0', Category_Name tinytext NOT NULL, PRIMARY KEY (CategoryID) ) TYPE=MyISAM; So at each level you can find the subcategories by: SELECT * FROM Category WHERE ParentID = $This_Level_ID; The top level would have a ParentID of 0. *shrug* - One way to do it. -Micah On Mon, 2003-01-13 at 21:32, Gerard Samuel wrote: Im figuring this is more of an sql question than anything else. I'm trying to figure out a table structure to create unlimited depths of categorical data. I've done something for category/subcategories before, but haven't an idea how to create categories at an unlimited depth. Any pointers would be greatly appreciated. Thanks -- Gerard Samuel http://www.trini0.org:81/ http://dev.trini0.org:81/ -- Raincross Technologies Development and Consulting Services http://www.raincross-tech.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
SV: [PHP-DB] LEFT JOIN not working
Hi, You might want to try something like: SELECT ads_displayrate.name, SUM(ads_displayrate.count) as display, SUM( IF( ads_clickrate.date IS NULL, 0, 1 ) ) as click FROM ads_displayrate LEFT JOIN ads_clickrate ON ads_displayrate.name = ads_clickrate.name HAVING YEAR(ads_displayrate.date) = '2003' AND MONTH(ads_displayrate.date) = '01' AND DAYOFMONTH(ads_displayrate.date) = '05' GROUP BY ads_displayrate.name ORDER BY ads_displayrate.name regards Henrik Hornemann -Oprindelig meddelelse- Fra: Lisi [mailto:[EMAIL PROTECTED]] Sendt: 14. januar 2003 18:47 Til: Ignatius Reilly Cc: PHP-DB Emne: Re: [PHP-DB] LEFT JOIN not working Still not working. I made the change, and I'm still getting all results. I even tried it without the leading '0' in front of the 1, no good. Here's my current query, with suggested changes: SELECT ads_displayrate.name, SUM(ads_displayrate.count) as display, SUM( IF( ads_clickrate.date IS NULL, 0, 1 ) ) as click FROM ads_displayrate LEFT JOIN ads_clickrate ON ads_displayrate.name = ads_clickrate.name AND YEAR(ads_displayrate.date) = '2003' AND MONTH(ads_displayrate.date) = '01' AND DAYOFMONTH(ads_displayrate.date) = '05' GROUP BY ads_displayrate.name ORDER BY ads_displayrate.name ads_displayrate.date is a column of date type, so as far as I understand this should work. Is there some typo I'm missing? At 07:15 PM 1/9/03 +0100, you wrote: Oops! I missed again. If you specify conditions pertaining to the right-hand table, such as: ads_clickrate.date = '2001', Then you will lose all result rows for which the right-hand data is NULL. Not the expected result. So your restricting WHERE clauses must apply to the left-hand table only. Therefore: WHERE YEAR(ads_displayrate.date) = '2003' AND MONTH(ads_displayrate.date) = '01' (if your table ads_displayrate has such date fields). HTH Ignatius - Original Message - From: Lisi [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 6:54 PM Subject: Re: [PHP-DB] LEFT JOIN not working Cool! It's mostly working now, the only problem is it's ignoring the other clauses in the ON clause that select the desired date. Perhaps it's not supposed to be connected this way? How would I select specific dates? Thanks again, -Lisi At 01:20 PM 1/9/03 +0100, you wrote: Oops! Sorry, I missed it the first time. Your query should start as: SELECT ads_displayrate.name instead of SELECT ads_clickrate.name then you will always have a non-NULL name (coming from the table on the left of the LEFT JOIN). HTH Ignatius, from Brussels Where the fuck is Belgium? D. Ivester, CEO, Coca Cola - Original Message - From: Lisi [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED]; PHP-DB [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 1:11 PM Subject: Re: [PHP-DB] LEFT JOIN not working Exactly my question - why does it not have a name? How would I modify my query to get a row with a name but null value for date? I thought the join would take care of this, but I'm obviously not doing it right. You mention a unique identifier, there is a separate table with a row for each ad, containing name, URL, and a unique ID number (autoincrement). Should this table be included somehow in the query? How would this help? Thanks, -Lisi At 12:45 PM 1/9/03 +0100, Ignatius Reilly wrote: Your 4th row ought to have an identifier of some sort. From your SELECT statement, this seems to be the name. Why does it not have a name? Probably what you want is a row with a name but a NULL value for ads_clickrate.date. (by the way it is EXTREMELY advisable to use an abstract identifier, such as an id, unique and required, instead of name) Ignatius - Original Message - From: Lisi [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED]; PHP-DB [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 12:18 PM Subject: Re: [PHP-DB] LEFT JOIN not working OK, this helped a bit. Now I have, in addition to the three rows of ads that have ben clicked on, a fourth row with no ad name, 0 clickthroughs, and 24 displays. That plus the other three account for all the displayed ads. However, since it is returning a null value for any ad name that has not been clicked on, and then it's grouped by ad name, it lumps all non-clicked ads into one row. What I need is to see each ad on a separate row, which is what I thought a LEFT JOIN was supposed to do. Any suggestions? Thanks, -Lisi At 11:07 AM 1/9/03 +0100, Ignatius Reilly wrote: problem 1: move the WHERE clauses to the ON
Re: [PHP-DB] LEFT JOIN not working
Should work (provided that the HAVING clauses be written after the GROUP BY) But less efficient. The point of using the WHERE clauses is to restrict during the JOIN. What you propose amounts to joining the full left table, which may be very expensive, especially in the case of web log data. Ignatius - Original Message - From: Henrik Hornemann [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED] Cc: PHP-DB [EMAIL PROTECTED] Sent: Wednesday, January 15, 2003 10:56 AM Subject: SV: [PHP-DB] LEFT JOIN not working Hi, You might want to try something like: SELECT ads_displayrate.name, SUM(ads_displayrate.count) as display, SUM( IF( ads_clickrate.date IS NULL, 0, 1 ) ) as click FROM ads_displayrate LEFT JOIN ads_clickrate ON ads_displayrate.name = ads_clickrate.name HAVING YEAR(ads_displayrate.date) = '2003' AND MONTH(ads_displayrate.date) = '01' AND DAYOFMONTH(ads_displayrate.date) = '05' GROUP BY ads_displayrate.name ORDER BY ads_displayrate.name regards Henrik Hornemann -Oprindelig meddelelse- Fra: Lisi [mailto:[EMAIL PROTECTED]] Sendt: 14. januar 2003 18:47 Til: Ignatius Reilly Cc: PHP-DB Emne: Re: [PHP-DB] LEFT JOIN not working Still not working. I made the change, and I'm still getting all results. I even tried it without the leading '0' in front of the 1, no good. Here's my current query, with suggested changes: SELECT ads_displayrate.name, SUM(ads_displayrate.count) as display, SUM( IF( ads_clickrate.date IS NULL, 0, 1 ) ) as click FROM ads_displayrate LEFT JOIN ads_clickrate ON ads_displayrate.name = ads_clickrate.name AND YEAR(ads_displayrate.date) = '2003' AND MONTH(ads_displayrate.date) = '01' AND DAYOFMONTH(ads_displayrate.date) = '05' GROUP BY ads_displayrate.name ORDER BY ads_displayrate.name ads_displayrate.date is a column of date type, so as far as I understand this should work. Is there some typo I'm missing? At 07:15 PM 1/9/03 +0100, you wrote: Oops! I missed again. If you specify conditions pertaining to the right-hand table, such as: ads_clickrate.date = '2001', Then you will lose all result rows for which the right-hand data is NULL. Not the expected result. So your restricting WHERE clauses must apply to the left-hand table only. Therefore: WHERE YEAR(ads_displayrate.date) = '2003' AND MONTH(ads_displayrate.date) = '01' (if your table ads_displayrate has such date fields). HTH Ignatius - Original Message - From: Lisi [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 6:54 PM Subject: Re: [PHP-DB] LEFT JOIN not working Cool! It's mostly working now, the only problem is it's ignoring the other clauses in the ON clause that select the desired date. Perhaps it's not supposed to be connected this way? How would I select specific dates? Thanks again, -Lisi At 01:20 PM 1/9/03 +0100, you wrote: Oops! Sorry, I missed it the first time. Your query should start as: SELECT ads_displayrate.name instead of SELECT ads_clickrate.name then you will always have a non-NULL name (coming from the table on the left of the LEFT JOIN). HTH Ignatius, from Brussels Where the fuck is Belgium? D. Ivester, CEO, Coca Cola - Original Message - From: Lisi [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED]; PHP-DB [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 1:11 PM Subject: Re: [PHP-DB] LEFT JOIN not working Exactly my question - why does it not have a name? How would I modify my query to get a row with a name but null value for date? I thought the join would take care of this, but I'm obviously not doing it right. You mention a unique identifier, there is a separate table with a row for each ad, containing name, URL, and a unique ID number (autoincrement). Should this table be included somehow in the query? How would this help? Thanks, -Lisi At 12:45 PM 1/9/03 +0100, Ignatius Reilly wrote: Your 4th row ought to have an identifier of some sort. From your SELECT statement, this seems to be the name. Why does it not have a name? Probably what you want is a row with a name but a NULL value for ads_clickrate.date. (by the way it is EXTREMELY advisable to use an abstract identifier, such as an id, unique and required, instead of name) Ignatius - Original Message - From: Lisi [EMAIL PROTECTED] To: Ignatius Reilly [EMAIL PROTECTED]; PHP-DB [EMAIL PROTECTED] Sent: Thursday, January 09, 2003 12:18 PM Subject: Re: [PHP-DB] LEFT JOIN not working OK, this helped a bit. Now I have, in addition to the
[PHP-DB] making my own week? :)
Hi All, I am currently building a website for a cinema - and a cinema week runs from Thursday, to Wednesday. I need some help with how to get PHP to work with one week periods, but having the week starting on a Thursday, instead of Sunday, or Monday. The week will be used to generate dynamic list boxes, and database queries... Anyone done anything like this before that may be able to shed a little light? If anyone is experienced with this sort of date manipulation let me know what info you need and I will provide it where possible. Cheers, Matt
Re: [PHP-DB] making my own week? :)
The simplest seems to store the date normally in the DB. When you need your special Cinema date, just do a modulo 7: myDayOfWeek = ( WEEKDAY( '2003-01-17' ) + 3 ) % 7 HTH Ignatius - Original Message - From: Matthew Nock [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, January 15, 2003 11:45 AM Subject: [PHP-DB] making my own week? :) Hi All, I am currently building a website for a cinema - and a cinema week runs from Thursday, to Wednesday. I need some help with how to get PHP to work with one week periods, but having the week starting on a Thursday, instead of Sunday, or Monday. The week will be used to generate dynamic list boxes, and database queries... Anyone done anything like this before that may be able to shed a little light? If anyone is experienced with this sort of date manipulation let me know what info you need and I will provide it where possible. Cheers, Matt -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Undefined index and variables
Hi to all Being new to PHP I am naturally experiencing some minor problems. I have a form which processes Ref and Chassis data, when I load Add.php which does work but on loading get errors Notice: Undefined index: Ref in C:\Xitami\webpages\add.php on line 22 Notice: Undefined index: Chassis in C:\Xitami\webpages\add.php on line 23 How do I get rid of these undefined errors? Could anyone advise please. Regards Fred printf(FORM action=\add.php\ METHOD=post); printf(TABLE border=\0\ width=\100%%\); printf(TRTDReference:/TD); printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR); printf(TRTDChassis:/TD); printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR); printf(/TABLE); printf(INPUT TYPE=submit name=\submit\); printf(/FORM ); -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Outgoing mail problem.
Friends, I have a RedHat 7.3. I am having problems to send email. I know, I must config the php.ini. What are the configurations, that i need to setup? My php.ini has an smtp entry, but no sendmail or qmail is running on the local machine. Do I have to have one running? I wish I did not have to do it. Any other way to fix this issue? thanks in advanced, Antonio. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] how made php on FreeBSD connect to Oracle 8.1.5 on Windows 2000 server
Thank you!But the oracle client is too big to fix my box.My box is P575+24M RAM and can't run X-windows and don't have a CD-ROM driver.:( Are there any simple ways such as to copy the library needed? Jadiel Flores [EMAIL PROTECTED] дÈëÏûÏ¢ÐÂÎÅ :[EMAIL PROTECTED] You have to install the Oracle Client in your FreeBSD server in order to compile php with the -with-oci8=[ORA_PATH] that's why it didn't work At 09:24 PM 1/14/2003 +0800, Cedar John wrote: Hi!I am a newbie. I've got a system with FreeBSD_4.7_release+apache_1.3.27+php_4.2.3. I want access oracle 8i(8.1.5) on another server with my php. It says that 'can not find the lib' when I complied it with '--with-oci8'. What can I do?Where can I find 'the lib'(oracle oci8 library)? Thanks a lot! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Jadiel Flores - http://www.abargon.com [EMAIL PROTECTED] (52-55) 52-29-80-34 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Undefined index and variables
It does not appear that you have posted the code from the add.php page in your original post (below). From the error messages you provided, that seems to be where the error is happening. You'll need to post the add.php code and identify line #23 for us to be able to help. -Original Message- From: Fred Wright [mailto:[EMAIL PROTECTED]] Sent: Wednesday, January 15, 2003 6:57 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Undefined index and variables Hi to all Being new to PHP I am naturally experiencing some minor problems. I have a form which processes Ref and Chassis data, when I load Add.php which does work but on loading get errors Notice: Undefined index: Ref in C:\Xitami\webpages\add.php on line 22 Notice: Undefined index: Chassis in C:\Xitami\webpages\add.php on line 23 How do I get rid of these undefined errors? Could anyone advise please. Regards Fred printf(FORM action=\add.php\ METHOD=post); printf(TABLE border=\0\ width=\100%%\); printf(TRTDReference:/TD); printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR); printf(TRTDChassis:/TD); printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR); printf(/TABLE); printf(INPUT TYPE=submit name=\submit\); printf(/FORM ); -- 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] Outgoing mail problem.
Hi Antônio Pires de Castro Jr., Friends, I have a RedHat 7.3. I am having problems to send email. I know, I must config the php.ini. What are the configurations, that i need to setup? The path to your sendmailwrapper. I think in the case of qmail you have to write /var/qmail/bin/qmail-inject at the line of the sendmail-configuration. I have to say I've not tested the entry, because at moment I have no mailforms running in my local environment and no specific entry in my php.ini. Regards, Ruprecht -- Ruprecht Helms IT-Service und Softwareentwicklung Tel/Fax.: +49[0]7621 16 99 16 Homepage: http://www.rheyn.de email: [EMAIL PROTECTED] -- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] db connect probs
MySQL cannot find the username in its grant tables.
The error message very clearly states that there is no user named
'x,@host' which is the value of your $user variable. For your example,
$email == 'x' and $femail == '@host'. The comma is included because it
is part of the literal string, i.e., inside the double quotes.
hth,
Doug
On Tue, 14 Jan 2003 22:08:11 -0600, Addison Ellis wrote:
hello,
do you have any idea why the following(line 106)would return
the following
error message ? the login.php has an included form to create an
account and that is what will not connect.
$connection = mysql_connect($host, $user,$password) //this is line 106
Warning: Access denied for user: 'x,@host' (Using password: YES) in
/users/infoserv/web/register/login.php on line 106
Warning: MySQL Connection Failed: Access denied for user: 'x,@host'
(Using password: YES) in /users/infoserv/web/register/login.php on
line 106
Couldn't connect to server.
my required file is as follows:
?
$user=$email,$femail;
$host=$hostname;
$password=password('$password'),password('$fpassword');
$database=$classes;
?
thank you again for your time. addison
--
Addison Ellis
small independent publishing co.
114 B 29th Avenue North
Nashville, TN 37203
(615) 321-1791
[EMAIL PROTECTED]
[EMAIL PROTECTED]
subsidiaries of small independent publishing co.
[EMAIL PROTECTED]
[EMAIL PROTECTED]
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[PHP-DB] 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
RE: [PHP-DB] Concatenate two strings
with a dot $valor=$valor1 . . $valor2 :-) -Original Message- From: Bruno Pereira [mailto:[EMAIL PROTECTED]] Sent: woensdag 15 januari 2003 15:48 Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] 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] Concatenate two strings
-Original Message- From: Bruno Pereira [mailto:[EMAIL PROTECTED]] Sent: 15 January 2003 14:48 How can i join two strings. My code is something like: $valor1=bruno; $valor2=Pereira; $valor=$valor1 + + $valor2 Can someone help me. Thanks. http://www.php.net/manual/en/language.operators.string.php Cheers! Mike - Mike Ford, Electronic Information Services Adviser, Learning Support Services, Learning Information Services, JG125, James Graham Building, Leeds Metropolitan University, Beckett Park, LEEDS, LS6 3QS, United Kingdom Email: [EMAIL PROTECTED] Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Concatenate two strings
On Wed, 15 Jan 2003 14:47:59 - Bruno Pereira [EMAIL PROTECTED] wrote: How can i join two strings. My code is something like: $valor1=bruno; $valor2=Pereira; $valor=$valor1 + + $valor2 $valor=$valor1 . . $valor2; See http://www.php.net/manual/pt_BR/language.operators.string.php hth pier -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Missing LDAP objects
I am using Netware 6 as an LDAP server and running Apache PHP 4.2.2 on a
linux Redhat 7.2.
For some reason when I do a wildcard (ou=*) LDAP_SEARCH I only get part of
the directory.
I then did a LDAP_SEARCH on the CNs in the OUs that the LDAP_SEARCH did not
find, and the
LDAP_SEARCH found all the CNs. Very Strange
Does anyone have any ideas why this is happening?
also
I tried Terrence Miao's LDAPExplorer and his PHP script only shows part of
the tree. ( exactly like mine).
I also tried a java LDAP explorer and I could see the entire tree.
Here is the code.
Our LDAP server has two OUs under the O and only one is displayed.
?php
// Connecting to LDAP server
$ds=ldap_connect(Novell server);
echo connect result is .$ds.p;
if ($ds) {
echo Binding ...;
$r=ldap_bind($ds); // anonymous bind, typically
echo Bind result is .$r.p;
echo Searching for (ou=*) ...;
// Search Organization Unit entry
$sr=ldap_search($ds,o=OurOrg, ou=* );
echo Search result is .$sr.p;
echo Number of entires returned is .ldap_count_entries
($ds,$sr).p;
echo Getting entries ...p;
$info = ldap_get_entries($ds, $sr);
echo Data for .$info[count]. items returned:p;
for($i=0;$i$info[count];$i++)
{
echo $info[$i][dn];
$parts=explode(,,$info[$i][dn]);
$nds=explode(=,$parts[0]);
echo brnds: .$nds[1].br;
}
echo Closing connection;
ldap_close($ds);
} else {
echo h4Unable to connect to LDAP server/h4;
}
?
Thanks!
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[PHP-DB] $user=$email
hello,
i am trying to do the following:
case login:
$connection = mysql_connect($host, $user,$password) //15
or die (Couldn't connect to server.);
$db = mysql_select_db($database, $connection)
or die (Couldn't select database.);
$sql = SELECT email FROM accounts
WHERE email='$email';
$result = mysql_query($sql)
or die(Couldn't execute query.);
$num = mysql_num_rows($result);
if ($num == 1) // login name was found //25
{
$sql = SELECT email FROM accounts
WHERE email='$email'
AND password=password('$password');
$result2 = mysql_query($sql)
or die(Couldn't execute query.);
$num2 = mysql_num_rows($result2);
if ($num2 0) // password is correct
{
$auth=yes;
//35
$user=$email;
$today = date(Y-m-d h:m:s);
$sql = INSERT INTO login (loginName,loginTime)
VALUES ('$email','$today');
mysql_query($sql) or die(Can't execute query.);
header(Location: cat_ad.php);
}
else// password is not correct
{
in my config.php file which is a required file i have
$user=$email, etc.
i continually receive an access denied/couldn't connect to server.
i'm sure i'm leaving something out...
thank you for your time. addison
--
Addison Ellis
small independent publishing co.
114 B 29th Avenue North
Nashville, TN 37203
(615) 321-1791
[EMAIL PROTECTED]
[EMAIL PROTECTED]
subsidiaries of small independent publishing co.
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[PHP-DB] Re: Undefined index and variables
?
//This is lines 22 and 23 of Add.php
require(common.php) ;
GenerateHTMLForm();
$smt2= $_POST['Ref']; //line 22
$smt3=$_POST['Chassis']; //line 23
The form below is in common.php
function GenerateHTMLForm() {
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref/TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE);
printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
}
Fred Wright [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi to all
Being new to PHP I am naturally experiencing some minor problems.
I have a form which processes Ref and Chassis data, when I load Add.php
which does work but on loading get errors
Notice: Undefined index: Ref in C:\Xitami\webpages\add.php on line 22
Notice: Undefined index: Chassis in C:\Xitami\webpages\add.php on line 23
How do I get rid of these undefined errors? Could anyone advise please.
Regards
Fred
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE);
printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
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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,'$CalendarDetailsID')) or die(mysql_error());
My big question is about using the DATE_ADD MySQL function inside
the INSERT statement. Is this allowed? If it is allowed, is it a bad idea
for some reason? Is there a better way of doing this? Thanks in advance.
Scott Nipp
Phone: (214) 858-1289
E-mail: [EMAIL PROTECTED]
Web: http:\\ldsa.sbcld.sbc.com
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[PHP-DB] Mail() and replies
Is there any way to receive mail via PHP? What I am interested in doing is creating a double opt-in e-mail list whereby someone can subscribe to an e-mail service via an on-line form. Upon doing so, they would get a confirmation e-mail. Replying to this e-mail would confirm the subscription and add the sender's e-mail address to the database (MySQL). Doable? Thanks, Jeffrey Baumgartner eBusiness Consultant - ITP Europe http://www.itp-europe.com [EMAIL PROTECTED] +32 2 721 51 00 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date math functions...
[snip]
if ($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 $StopDate in PHP, so you could do it with
some math in PHP, also, but then you'd have to worry about the end of
months, years, etc, whereas DATE_ADD will do this for you.
---John Holmes...
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Re: [PHP-DB] Mail() and replies
I would say this can be done (-: Run a quck search on freshmeat.net, sourceforge.net, of course the manual http://www.php.net/manual/en/ref.mail.php, and http://www.hotscripts.com/, just to name a few places to start. Or just search for mailing list php on Google. There are a ton of mailing list managers out there - check them out and then decide if you need to roll your own. Many are free to use, and would at least offer some idea of where to start. Baumgartner Jeffrey wrote: Is there any way to receive mail via PHP? What I am interested in doing is creating a double opt-in e-mail list whereby someone can subscribe to an e-mail service via an on-line form. Upon doing so, they would get a confirmation e-mail. Replying to this e-mail would confirm the subscription and add the sender's e-mail address to the database (MySQL). Doable? Thanks, Jeffrey Baumgartner eBusiness Consultant - ITP Europe http://www.itp-europe.com [EMAIL PROTECTED] +32 2 721 51 00 -- John Krewson Programmer - SWORPS -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date math functions...
[snip]
if ($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] Mail() and replies
Is there any way to receive mail via PHP? What I am interested in doing is creating a double opt-in e-mail list whereby someone can subscribe to an e-mail service via an on-line form. Upon doing so, they would get a confirmation e-mail. Replying to this e-mail would confirm the subscription and add the sender's e-mail address to the database (MySQL). Doable? Sure, there are a couple modules in PHP that'll let you access and read your POP3 or IMAP email. I'm sure there are some classes on phpclasses.org that can be modified to your needs. There are programs out there that do this already, too, if you don't want to re-create the wheel. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Mail() and replies
Quite doable. Check the IMAP functions. Ignatius - Original Message - From: Baumgartner Jeffrey [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Wednesday, January 15, 2003 5:19 PM Subject: [PHP-DB] Mail() and replies Is there any way to receive mail via PHP? What I am interested in doing is creating a double opt-in e-mail list whereby someone can subscribe to an e-mail service via an on-line form. Upon doing so, they would get a confirmation e-mail. Replying to this e-mail would confirm the subscription and add the sender's e-mail address to the database (MySQL). Doable? Thanks, Jeffrey Baumgartner eBusiness Consultant - ITP Europe http://www.itp-europe.com [EMAIL PROTECTED] +32 2 721 51 00 -- 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] Re: Undefined index and variables
Fred,
I don't have much time today, but here is my initial assessment of what
might be going wrong with your page:
You display add.php which draws a form for data.
You submit that form to add.php (itself).
The form redraws inside the page.
You get errors when add.php looks for data in the $HTTP_POST_VARS array
because the page is fresh and that array is empty.
Problem is that there is no code to handle the data that was entered in the
first place. Once you hit submit, the add.php page is (re)drawn in its
original state - with no values in the fields. Therefore, when the
$_POST['varname'] statement looks in $HTTP_POST_VARS, there's nothing to
display so you get the Undefined Index errors. It's almost like refreshing
the page.
My advice would be to separate the code for the data entry and data handling
parts of your project into independent scripts. Use add.php to collect the
data then send the data to a php script called handleAdd.php that gets the
name/value pairs and submits them to the database. After submitting the data
to the database in handleAdd.php, forward the user back to another page
(add.php again, if you want) where they can get some kind of confirmation
the data was entered.
A further recommendation would be to provide the list with the code you gave
to me so maybe somebody else with a little more free time might be able to
assist you further. Sorry I couldn't provide any more help.
Rich
-Original Message-
From: Fred Wright [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 12:21 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: Undefined index and variables
?
//This is lines 22 and 23 of Add.php
require(common.php) ;
GenerateHTMLForm();
$smt2= $_POST['Ref']; //line 22
$smt3=$_POST['Chassis']; //line 23
The form below is in common.php
function GenerateHTMLForm() {
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref/TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE);
printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
}
Fred Wright [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi to all
Being new to PHP I am naturally experiencing some minor problems.
I have a form which processes Ref and Chassis data, when I
load Add.php
which does work but on loading get errors
Notice: Undefined index: Ref in C:\Xitami\webpages\add.php
on line 22
Notice: Undefined index: Chassis in
C:\Xitami\webpages\add.php on line 23
How do I get rid of these undefined errors? Could anyone
advise please.
Regards
Fred
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE);
printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
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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] Slashed being added..
Hi All.. Quick q here as I cannot remember how to prevent this. Customer is entering their name in a form that get submitted. Their name is: Mike O'Neill But their name is being entered like: Mike O\\\'Neill This is causing his info not to be saved. How do I prevent this? Any thoughts? Thanks! Aaron
[PHP-DB] Re: Undefined index and variables
here are the files to make add.php work. thanks to Richard Hutchins for his answer, that I do not yet understand:-) Fred Wright [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Hi to all Being new to PHP I am naturally experiencing some minor problems. I have a form which processes Ref and Chassis data, when I load Add.php which does work but on loading get errors Notice: Undefined index: Ref in C:\Xitami\webpages\add.php on line 22 Notice: Undefined index: Chassis in C:\Xitami\webpages\add.php on line 23 How do I get rid of these undefined errors? Could anyone advise please. Regards Fred printf(FORM action=\add.php\ METHOD=post); printf(TABLE border=\0\ width=\100%%\); printf(TRTDReference:/TD); printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR); printf(TRTDChassis:/TD); printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR); printf(/TABLE); printf(INPUT TYPE=submit name=\submit\); printf(/FORM ); begin 666 Fred.zip M4$L#!!0(`-L)+RX=#XSLUP0``!$-```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`NL\Y90)[PB@Y0^;3B-'#S#+A1'U@5Y M0EW `^F/'\NY,?-!76Z\!N/DS*ZH#CVJKN)#7UP5931?T*MQZ39?*U771 MH*W!D08=#;H:]#3H:^ W#?(-AKP[?3BC_$$G,,\'XNY*E05N#(PTZG0U MZG0UP#SKY%OD''N^^^\;_[X1\%%!`:/@XG 2 1(C 2@9$(C$1@)(+. M8.2[=OS.$UH*$ L:7YIS/ FP=HG64A$G*4C2Z)DF7$ACZ=S5*RHB-KEEY MV]RVX6QDR;$^;!^BLR#_/,\^)SAA2LELESES*C$V#=^=PD3G^,5EO^F#R$ M:JAC]%WIW95Y QB(*CAI*N8T\ //$LA`#B+^3HAVS%CEWSA'V/Y2.A M!VG[D01 X!:=;0$5$%IN[N/MCRWG7'T+LA%-?+K#^WT9RS47?A2!#PI M4UH)5BU=I?PAFCE%/MC3((*YU QGD@,,K!S*4`VZJ9J$7YXL[;V$'!]$ M=1@TIEFJA:-5S9,;T-\,-ES3\`7C[X_LPZSB=DF*/.J=!7@\4WLU4L MK.\-=2T4GJ2C09X\4\:;@3-,S1^ *QPK5G=0];BC;XFTCU.C\:SF57D3 M:3[,M?-F99MN9:MV9:M69=MRRXZEJU;EJUZEJV:EEUV+5NW+;OH6[9J7/:K MSH67I6EK4,:AMT9%#'H*Y!/8/Z!F$3,] O8=H;*'_VD$[EMAIL PROTECTED] M2?V_]\[N\85!]=VR^UA37$38/FQW$^$`_ZDH1'P31A2/M\DR39_*WU13:6 M8A;2:79)XK1XY_QX(O__`U!+`P04 `@?R\NZ^Z]1L)``Y*0``@`` M`-O;6UO;BYP:']6FUSVD@2_NXJ_X=EL0XZT-^C9T8LX=!-NQBX$!V+K?9 MH@0:0!NA46G$VJZL__MVCR0L0#)RJU2%4L:=3]/OTPSK5'I9V?B;]M;RD* M#/ETRNVBO(757D)HYD]]$QNP@NV8VW6/@3=A4-=NFC!ANL%EP_%JJ9 MB]-H70X6\E,FA#YFN]^VMQS7M+U1(5JJY5:4I:0VNJ9;CA@EGL$5 !J?[$ M!E!2@K'[GE$[D;5*MJW*@77W+7:DOM5K1V%[[DH-IN:6I+P]M7Q5^*GUQS M//'@_=%I?_3'_;\GW)OJIE5$'^SY9(BX:7FB.[+=NTS:.I_-4 M/^[+`S5? M7E?;S7:7[EW)`S4LBW;*I2LD`+W_XC9OPX)I7%S#;UN%:^G?8N/?$/9XSZ M/C5J6AUO'IP=XU5=;5S7B?;)*5Y5FHWK%EYMC6M?4/@[6Y-)6QBTJIHM]U* MDTNFI5KO$LXQ9)VA59X675KELJC#@+@8B4')OMZ$X//CV,LT;HP ^Y MQGY'VM5^D?:#4TD=LC/0ANZBXIFZ52Y?EAB\$27E$GF,\#D[.*I9]!_ M;CFW!_,\B,93\?2!Q1;5D1 ZVDI*X7,#XRE#0ZJBDF\*Q?!.SA6V(5OY9= MM;LW@/E3;]N.NV!I6JUFBW+D@;38L55S1:G5L-M,\=]4+,!E/3Q-O..'F MD*$O[RK-6[K18[H[G$!-1VMTP4*WAIKV@/AG,/G]#50P#=BQU!MSWU, M`1:!NVTN^;S9*$-@!D.U)KK:]=C:!4,'F%HF.OICVFF;J]!%1:I27/%!2 M4-U@YGG+BFHA6+K B!/)XR+?#JV/`,$:I!MNF*-'TU@#Z;P2D!/R[ MK#'A3K )FN\%708QMP.!P0?+$W]*QOB(:^C4L-)A?T'Z?B(,704F06P M(MFR?+LJYGL?1'8*[+70BFL8A,ON !U75OQ+@PG^FP/ 4OF^WJK_^Y;6MJ MUY_2BHI?.J6ZWZE;C\;#Y0J0FJ'@DH40UK+[[82V4H,''-#]^(NU/Z#W3# M4*8RIHJ045GS@W%`O%E1Y/HSHBFH-3N@$CE+Z_%66-#)8]K^]XZ^22 MEOO]-V^6:P56UG9NVS$78/V4?\V5W]8:3:'ZE/'GOP_%^Z$[](H3CZ=:, M70J].M*SRM:=U6/CU:=Z$*8(H+U,M)A`!7(PLLHE]QX?!4$(ZK5TVQDF+ M`:?8MW@J@+/06^,$]#]CHJ72?G#HJR=!%%]BP,?4'W7A'*;G1C$4-$],P ML$!)'^BAKS$T640'K\L7%:E_C/(D_AA,V_#K@#S[248$L@E+G'(L-?B MDPG(;ML;K6Z^%430GQ-K_6B!?.@8F4JHH(YK+E2/)'EPL._#2-E@ M;-3K32;.R\;5G9GBIQ`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`J-2\+1.)SC-WT_--7=F,YI:[ MV45KU[;:\\ZYRYLK?$;W;17-J)H/UYVP_,.V+0-]ASK83]\()]JV:$M*NT M88G,:9O1E!R_Y/VYVD//ETCF:P;9B J$_^7_@G!8PQ33P#1T+_B8@#Z; MH??D8(Y MRSY6N-/W37I$P8!NBOW7+'=-[:W\(G#WDQ]8[@K[] GAU+'0!+ MNZ4Y4E;]2-4`HTCDUFKVZJVXC.W4=_EQ9EV8/I%3%4_BY=V[[:UWX+Y
RE: [PHP-DB] Slashed being added..
stripcslashes (string str) Ric -Original Message- From: Aaron Wolski [mailto:[EMAIL PROTECTED]] Sent: 15 January 2003 16:38 To: [EMAIL PROTECTED] Subject: [PHP-DB] Slashed being added.. Hi All.. Quick q here as I cannot remember how to prevent this. Customer is entering their name in a form that get submitted. Their name is: Mike O'Neill But their name is being entered like: Mike O\\\'Neill This is causing his info not to be saved. How do I prevent this? Any thoughts? Thanks! Aaron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date math functions...
On Thursday 16 January 2003 00:55, NIPP, SCOTT V (SBCSI) wrote: 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? Which probably means your $StopDate is 0. What does it actually contain? You really should assign your query to some variable ($query) THEN plug it into your query function. If and you have any problems you can echo $query to see what you're passing to mysql. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* I'm having BEAUTIFUL THOUGHTS about the INSIPID WIVES of smug and wealthy CORPORATE LAWYERS ... */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Slashed being added..
On Thursday 16 January 2003 01:15, Richard Allinson wrote: stripcslashes (string str) Ric -Original Message- From: Aaron Wolski [mailto:[EMAIL PROTECTED]] Sent: 15 January 2003 16:38 To: [EMAIL PROTECTED] Subject: [PHP-DB] Slashed being added.. Hi All.. Quick q here as I cannot remember how to prevent this. Customer is entering their name in a form that get submitted. Their name is: Mike O'Neill But their name is being entered like: Mike O\\\'Neill This is causing his info not to be saved. How do I prevent this? Any thoughts? Check out the magic_quotes options in php.ini. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Don't SANFORIZE me!! */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Re: Undefined index and variables
Couldn't you use the 'for each' method on $_POST instead?
.b
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: 15 January 2003 16:44
To: 'Fred Wright'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Re: Undefined index and variables
Fred,
I don't have much time today, but here is my initial
assessment of what might be going wrong with your page:
You display add.php which draws a form for data.
You submit that form to add.php (itself).
The form redraws inside the page.
You get errors when add.php looks for data in the
$HTTP_POST_VARS array because the page is fresh and that
array is empty.
Problem is that there is no code to handle the data that was
entered in the first place. Once you hit submit, the add.php
page is (re)drawn in its original state - with no values in
the fields. Therefore, when the $_POST['varname'] statement
looks in $HTTP_POST_VARS, there's nothing to display so you
get the Undefined Index errors. It's almost like refreshing the page.
My advice would be to separate the code for the data entry
and data handling parts of your project into independent
scripts. Use add.php to collect the data then send the data
to a php script called handleAdd.php that gets the name/value
pairs and submits them to the database. After submitting the
data to the database in handleAdd.php, forward the user back
to another page (add.php again, if you want) where they can
get some kind of confirmation the data was entered.
A further recommendation would be to provide the list with
the code you gave to me so maybe somebody else with a little
more free time might be able to assist you further. Sorry I
couldn't provide any more help.
Rich
-Original Message-
From: Fred Wright [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 12:21 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: Undefined index and variables
?
//This is lines 22 and 23 of Add.php
require(common.php) ;
GenerateHTMLForm();
$smt2= $_POST['Ref']; //line 22
$smt3=$_POST['Chassis']; //line 23
The form below is in common.php
function GenerateHTMLForm() {
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref/TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE); printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
}
Fred Wright [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi to all
Being new to PHP I am naturally experiencing some minor problems.
I have a form which processes Ref and Chassis data, when I
load Add.php
which does work but on loading get errors
Notice: Undefined index: Ref in C:\Xitami\webpages\add.php
on line 22
Notice: Undefined index: Chassis in
C:\Xitami\webpages\add.php on line 23
How do I get rid of these undefined errors? Could anyone
advise please.
Regards
Fred
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE);
printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
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Re: [PHP-DB] Re: Undefined index and variables
Of course, but that would solve the problem I have, albeit it would look
more elegant. Have tried everything except the solution which I cannot find,
see the zip file earlier up the list from Fred Wright
Ben Joyce [EMAIL PROTECTED] wrote in message
DF1220D4F915D4119C6B00306E00103E0252EDB7@IDS10004">news:DF1220D4F915D4119C6B00306E00103E0252EDB7@IDS10004...
Couldn't you use the 'for each' method on $_POST instead?
.b
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: 15 January 2003 16:44
To: 'Fred Wright'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Re: Undefined index and variables
Fred,
I don't have much time today, but here is my initial
assessment of what might be going wrong with your page:
You display add.php which draws a form for data.
You submit that form to add.php (itself).
The form redraws inside the page.
You get errors when add.php looks for data in the
$HTTP_POST_VARS array because the page is fresh and that
array is empty.
Problem is that there is no code to handle the data that was
entered in the first place. Once you hit submit, the add.php
page is (re)drawn in its original state - with no values in
the fields. Therefore, when the $_POST['varname'] statement
looks in $HTTP_POST_VARS, there's nothing to display so you
get the Undefined Index errors. It's almost like refreshing the page.
My advice would be to separate the code for the data entry
and data handling parts of your project into independent
scripts. Use add.php to collect the data then send the data
to a php script called handleAdd.php that gets the name/value
pairs and submits them to the database. After submitting the
data to the database in handleAdd.php, forward the user back
to another page (add.php again, if you want) where they can
get some kind of confirmation the data was entered.
A further recommendation would be to provide the list with
the code you gave to me so maybe somebody else with a little
more free time might be able to assist you further. Sorry I
couldn't provide any more help.
Rich
-Original Message-
From: Fred Wright [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 15, 2003 12:21 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: Undefined index and variables
?
//This is lines 22 and 23 of Add.php
require(common.php) ;
GenerateHTMLForm();
$smt2= $_POST['Ref']; //line 22
$smt3=$_POST['Chassis']; //line 23
The form below is in common.php
function GenerateHTMLForm() {
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref/TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE); printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
}
Fred Wright [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi to all
Being new to PHP I am naturally experiencing some minor problems.
I have a form which processes Ref and Chassis data, when I
load Add.php
which does work but on loading get errors
Notice: Undefined index: Ref in C:\Xitami\webpages\add.php
on line 22
Notice: Undefined index: Chassis in
C:\Xitami\webpages\add.php on line 23
How do I get rid of these undefined errors? Could anyone
advise please.
Regards
Fred
printf(FORM action=\add.php\ METHOD=post);
printf(TABLE border=\0\ width=\100%%\);
printf(TRTDReference:/TD);
printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR);
printf(TRTDChassis:/TD);
printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR);
printf(/TABLE);
printf(INPUT TYPE=submit name=\submit\);
printf(/FORM );
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[PHP-DB] export query
I need to essentially take a query and make it into a table. I'll first settle for the ability to export the query into a text file, somehow delimited. Can anyone help me with this? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Date math functions...
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()); 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? Are you sure $StopDate has a value and is in the right format? Echo out your query to the screen when you get an error, so you can look for obvious mistakes. Is this just example code? You're using a function called mysql(), which isn't standard. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] áéíóúÁÉÍÓÚñÑ
How can I query this characters 'áéíóúÁÉÍÓÚñÑ' from a MSSQL DB with PHP??? Actually I obtain this: ' '¡¢£µÖà餥'!!! Thanks Fernando Bernain Senior A Business Process Outsourcing KPMG Argentina Tel: 54 11 4316 5754 Fax: 54 11 4316 5734 [EMAIL PROTECTED] Email Disclaimer The information in this email is confidential and may be legally privileged. It is intended solely for the addressee. Access to this email by anyone else is unauthorised. If you are not the intended recipient, any disclosure, copying, distribution or any action taken or omitted to be taken in reliance on it, is prohibited and may be unlawful. When addressed to our clients any opinions or advice contained in this email are subject to the terms and conditions expressed in the governing KPMG client engagement letter. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Duplicate keys...
OK, I cannot figure out why I keep getting errors about Duplicate
entry for key 1. Someone please come to my rescue here as this is truly
kicking my but.
if ($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] Problem w/ generating png on Linux with more than 250 objects ona single graphic
I have an application that records DNA fingerprints(gel images) in
digital form in a MySQL Database. Each fingerprint consists of an array
of 2000 values between 0-255 that represent a grayscale color. I use the
code below to create an image or fingerprint from the data.
code
---
$y3 = array( 2000 values from MySQL resulta formatted );
//set positions and draw image
$pos1 = 225;
$pos2 = 226;
$pic=ImageCreate(2290,50);
$col1=ImageColorAllocate($pic,200,200,200);
foreach ($y3 as $val)
{
unset( $color );
$color = ( 255 - $val);
//$color = $value;
$col=ImageColorAllocate($pic,$color,$color,$color);
Imageline($pic,$pos1,6,$pos1,44,$col);
$pos1++;
}
ImagePNG($pic,../images/bar$ID.png);
ImageDestroy($pic);
I've check the array for min,max and number of values and the array data
is great. It starts drwing the image which consists of a gray
background and a line being drawn for each value as it loop though the
2000 values in the array, each having a different color value and a
position that places the each line next to the last generates the
correct Imageline() statement on the fly.
Here's the problem, after about drawing 250 or so line objects, it stops
drawing the values and I get nothing but gray or white for the rest of
the image. I reduced the dataset to 400 values to test this and I get
the same result.
Is there a limitation in GD or PHP that would cause this or is there an
error in my code. I've attached copies of the images that PHP is
generating so you can see what is happening. These images contain the
curve and two images PHP draws, the first is a line graph from JPGraph
showing the curve and the second is the gel image below the graph. The
gel image in the fingerprint I am having trouble with. Both example stop
generating line objects after about 250 lines as seen in the examples.
Attachments:
sample1.gif -- zoom in of first 400 points of sample in sample2.gif
containing only 400 values
sample2.gif -- entire DNA fingerprint of 2000 values
My configuration is:
Red Hat 7.3 with all current updates for that release apache 1.3 and PHP
4.1.2
Any ideas on how to get this to work properly?
Thanks in advance for your help and advise!
Robert Trembath
CSE
e| [EMAIL PROTECTED]
Bacterial BarCodes, Inc.
Houston, TX
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[PHP-DB] Links to samples from Problem w/ generating png on Linux
The images were stripped on my post but here's an external link to the images I refer to: Sample1 : http://linux.diversilab.com/sample1.gif Sample2 : http://linux.diversilab.com/sample2.gif Thanks, Robert -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: Undefined index and variables
Problem solved with //Init variables $HTTP_POST_VARS['Ref']=A; etc Thanks guys for your help :-)) Fred Wright [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Hi to all Being new to PHP I am naturally experiencing some minor problems. I have a form which processes Ref and Chassis data, when I load Add.php which does work but on loading get errors Notice: Undefined index: Ref in C:\Xitami\webpages\add.php on line 22 Notice: Undefined index: Chassis in C:\Xitami\webpages\add.php on line 23 How do I get rid of these undefined errors? Could anyone advise please. Regards Fred printf(FORM action=\add.php\ METHOD=post); printf(TABLE border=\0\ width=\100%%\); printf(TRTDReference:/TD); printf(TDINPUT TYPE=text SIZE=5 NAME=Ref /TD/TR); printf(TRTDChassis:/TD); printf(TDINPUT TYPE=text SIZE=25 NAME=Chassis /TD/TR); printf(/TABLE); printf(INPUT TYPE=submit name=\submit\); printf(/FORM ); -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Apache/php will not start after upgrading Oracle from 8.1.7.3 to 8.1.7.4
Hi. Just upgraded my Oracle installation from 8.1.7.3 to 8.1.7.4. From that moment on, apache/php will not start (just hangs). Tried 'make'ing Apache and php again but same result. Checked oracle variables, etc and all seem to be okay (it was working fine). Any help much appreciated since this is a critical application and don't know what to do. Thanks Miguel -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] HTML + PHP
Hello everybody!!!Here am I again with problems... :-( This time I programmed severan php scripts embebed with HTML, but the curiosity comes when I try to see the scripts in a browser. In Opera it's ok, but in Mozilla and Internet Explorer it appears one table on the other, everything mixed up. I have most of the scriptsstructured as follows: a table that contains the name of the enterprise and the site's area and another table (theorically beneath the first one) withe the content of a table of my database. Both of them appear the second one under the first one, but mixed up. I used functions to build the code... What can it be? Was I clear with the problem? Otherwise I can send a screenshoot with the problem, but I don't think it's allowed to attach files in this list. Sabina Alejandra Schneider[EMAIL PROTECTED]
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
today. http://www.phparch.com/
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RE: [PHP-DB] export query
I need to essentially take a query and make it into a table. I'll
first
settle for the ability to export the query into a text file, somehow
delimited.
Can anyone help me with this?
What have you done so far? Do you know how to make a table?
While($row = mysql_fetch_row($result))
{ echo trtd{$row[0]}/tdtd{$row[1]}/td/tr; }
etc...
As for exporting it to a file, use SELECT * INTO OUTFILE ...
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
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RE: [PHP-DB] making my own week? :)
I am currently building a website for a cinema - and a cinema week runs from Thursday, to Wednesday. I need some help with how to get PHP to work with one week periods, but having the week starting on a Thursday, instead of Sunday, or Monday. The week will be used to generate dynamic list boxes, and database queries... Anyone done anything like this before that may be able to shed a little light? If anyone is experienced with this sort of date manipulation let me know what info you need and I will provide it where possible. I think it's all kind of specific to what you want to create. If you want a drop down box filled with dates from the past Thursday to the next Wednesday, for example, you can find the date of the last Thursday with strtotime(last Thursday), I think. Figure out how many seconds are in a day and make a little loop that adds the seconds to what you got from strtotime() and uses date() to format it. What database are you using? There are many functions that most databases provide that can help with this, too. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] HTML + PHP
On Thursday 16 January 2003 11:43, Sabina A. Schneider wrote: Hello everybody!!! Here am I again with problems... :-( This time I programmed severan php scripts embebed with HTML, but the curiosity comes when I try to see the scripts in a browser. In Opera it's ok, but in Mozilla and Internet Explorer it appears one table on the other, everything mixed up. I have most of the scripts structured as follows: a table that contains the name of the enterprise and the site's area and another table (theorically beneath the first one) withe the content of a table of my database. Both of them appear the second one under the first one, but mixed up. I used functions to build the code... What can it be? Was I clear with the problem? Otherwise I can send a screenshoot with the problem, but I don't think it's allowed to attach files in this list. Most likely your HTML is broken -- unclosed/unbalanced tags (/td, /tr, /table). -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* When the going gets tough, the tough go grab a beer. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
