[PHP-DB] MSSQL: PHP fails to connect
Hello community! After reinstalling our Active Directory domain from scratch, PHP 4.0.6 does no longer connect to MS SQL Server 2000 by mssql_connect(). PHP itself resides on our master DC (W2K Small Business Server, 192.168.1.1), while the SQL Server is installed on the 2nd DC (W2K Server, 192.168.1.2) -- which is the same constellation as 7 days ago, when PHP an MSSQL happily worked together. Does anybody have an idea what might have been going wrong during installation of either part of them? Or thereafter? I've already checked the following: extension directory entry in php.ini, read/execution grants for `IUSR_xyz´ on php's exes and dlls, mssql_connect() with: FQDN *or* host name *or* IP address, several different database users including `sa´, MSSQL Server *in* the Active Directory as well as *not in*, MSSQL's network configuration w/ Named Pipes as well as w/o, DNS entries and several successful access tests with: Enterprise Manager, Query Analyzer and ColdFusion 5 (ODBC). BTW: both, php as well as MSSQL, have been installed using the `default´ settings of both installers (except for the installation dirs). The baddest thing at last: mssql_connect() to the *same* Server from a different, arbitrarily chosen workstation (W2K Pro) with PHP 4.0.6 *does* work! Any hints? *Please*! TIA! Rodya Rodya Jörn Koester B2A MEDIA GRAFIC DESIGN mailto:[EMAIL PROTECTED] Kassel,Germany,Europe,Earth -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Postgress+ODBC+BLOB
Hello Sorry for previous empty message As I understand there is no BLOB type in Postgress for storing large binary data ( in my case image files) I can't use Large Object cause I connecting to Postgres via ODBC ( there is no support for large objects ) Have anybody solved this problem? is there any alternative datatypes or ways of solution? best regards Ling -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Re: Postgress+ODBC+BLOB
Ling wrote: Hello Sorry for previous empty message As I understand there is no BLOB type in Postgress for storing large binary data ( in my case image files) What do you mean no BLOB suuport for PostgreSQL? You can use lo (Large Object) interface Read manual pages for that I can't use Large Object cause I connecting to Postgres via ODBC ( there is no support for large objects ) Have anybody solved this problem? is there any alternative datatypes or ways of solution? If you are using PostgreSQL and PHP, just use pgsql extension -- Yasuo Ohgaki Please CC me when you reply to news/list messages Do not reply only to me :) -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Re: Postgress+ODBC+BLOB
Ling wrote: Hello Sorry for previous empty message As I understand there is no BLOB type in Postgress for storing large binary data ( in my case image files) What do you mean no BLOB suuport for PostgreSQL? You can use lo (Large Object) interface Read manual pages for that I can't use Large Object cause I connecting to Postgres via ODBC ( there is no support for large objects ) Have anybody solved this problem? is there any alternative datatypes or ways of solution? If you are using PostgreSQL and PHP, just use pgsql extension -- Yasuo Ohgaki Please CC me when you reply to news/list messages Do not reply only to me :) -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Re: pg_pconnect Causing SegFaults?!?!
Could you post backtrace and CC the meesage to me? I'm almost sure pgsql module will never cause segfault like this, I can tell more if you send backtrace PS: If you don't know how to get backtrace, visit http://bugsphpnet/ -- Yasuo Ohgaki Kevin Traas wrote: Greetings everyone, I've been having a very frustrating problem over the past few months that I've ***finally*** tracked down the cause of Any questions/comments/thoughts/ideas you can provide would be *greatly* appreciated Symptoms: - IE, Opera, Lynx all report page cannot be displayed and/or server error about half the time - NS has no problems at all - Apache errorlog shows child (PID) seg faulted Problem post analysis: - The Apache processes can accept and successfully serve one connection from a client The next connection results in the Apache child process seg faulting, the error getting logged in errorlog, nothing gets reported in accesslog, and the Apache parent process respawns a new child - The newly spawned child will accept and serve one connection from a client before also segfaulting I've eliminated every possibility I could think of while troubleshooting this ie it's not SSL, it's not PHP itself, it's not Apache itself It's the pg_pconnect() function Throughout my troubleshooting, I came upon it by accident, really, that I discovered all my problems went away when I replaced the pg_pconnect() function with the (almost deprecated according to the docs) pg_connect() function In troubleshooting this, I created a shell script that continuously loads various pages from the site (using Lynx -dump -mime_header url), counts the iterations, and bails as soon as a page fails to load Running this against plain HTML files, PHP scripts (ie that simply echo something or provide phpinfo(), etc), over SSL or not all work just fine (15,000 loads in each case without fault) As soon as I try accessing a PHP script that includes the pg_pconnect() function, it dies almost immediately If I stop/start Apache right before trying this, I'll get X successful loads (where X = Apache's 'StartServers' property) After that, I'll get a continuous mixture of successful and failed loads from then on So, let me repeat here As soon as I switched from pg_pconnect() to pg_connect(), everything worked perfectly after that (ie 15000 loads without an error) More info: - My platform is up-to-date Debian GNU/Linux (sid) (Apache 1323, PHP 411, PostgreSQL 713, etc) - As above, this is *not* browser specific (Found a *lot* of google hits with people having problems with IE and SSL and seg faults, etc Completely unrelated) - Note from above, though, that NS is unaffected by this problem It *never* has a problem My assumption is that NS will simply silently retry (several times) if the connection fails for some reason, until it does get a valid response from the server (Kudos!!! ;-) So, any thoughts? Comments? Suggestions? Have I found a bug in pg_pconnect()? How can I provide more info to track this down further? Best regards, Kevin Traas Chief Information Officer, Co-founder Merilus, Inc ---=== wwwmeriluscom ===--- -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Re: Postgress+ODBC+BLOB
Hello as I wrote before I use Unified ODBC in order to make scripts database independent, but Unified ODBC does not support LO If you are using PostgreSQL and PHP, just use pgsql extension sorry but I do not understand : ( I didn't find any information in user guide about data type 'BYTEA' does anybody know what is it? its limit? ability to store binary data? -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Building secure authentification with sessions
Hi there, I did recently read an article about security Now I absolutly see the need of recoding my authentification procedure on a community site There are questions I hoped some of you guys can answer 1 Is storing sensitive data like permission level secure in session variables? 2 What could be a good way to session register a user and know which users are online, know their permission level in congungtion with a MySQL db? 3 Is it better to store the needed info about the user in a db table holding all current sessions, or to store it in more than 1 session variable 4 Maybe someone can relate to a good site dealing with security issuses on this topic I would really like to avoid that some hacker gets admin access on my website :-) Thanx for any hints, Cheers Andy -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] FW: PHP Question on MySQL grouping
I need to loop through a grouped recordset adding a hr at the end of each group, how can groups in recordsets be identified in PHP for specific formatting. SELECT name AS name, dishtype AS dishtype, price AS price FROM dishes WHERE used = 1 GROUP BY dishtype, name ORDER BY dishtype, price DESC The use is in the display of a menu for a restaurant, where there is a table containing all the dishes on the menu in any given instance, but they musst be displayed in a specific order, grouped by type. In other words, all Starters come first, followed by Main courses, followed by Desserts. Between each type there needs to be a seperator, the horizontal line. Thanks for you help Keiran -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: I'm really stuck!
First, could you be more specific? What isn't working? What error messages
are you getting? (I assume you are echoing them, because if not, you should
be... mysql_query($query) or die (mysql_error());
Second, this could be a problem. I think you need single quotes around the
variable here:
$query = select bin_data, filetype from image_data where
id=---'$id'; -
Beyond that, I can't really help much. Never stored image data in a db
before, so I can't tell you if anything else might be amiss. Good luck.
Jennifer Downey [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED];
Hi all,
I'm really stuck and I'm not asking anyone to re-write this just show me
what is wrong, or explain why it wont work. It just seems logical that this
should work.
The first query will print the pet id in the browser.
$query=SELECT id FROM wt_users WHERE uid={$session[uid]};
$ret = mysql_query($query);
while(list($pet)=
mysql_fetch_row($ret))
print(BRyour pet id is $pet);
So if $pet will print the id from the wt_users (has a value of 3) and I
assign $id = $pet (id also has a value of 3 in image_data) why doesn't it
show the image?
if($id) {
$id = $pet;
$query = select bin_data,filetype from image_data where id=$id;
$result = mysql_query($query);
$data = mysql_query($result,0,bin_data);
$type = mysql_query($result,0,filetype);
Header( Content-type: $type);
echo $data;
};
echo img src=\petdata.php?id=$id\;
Thanks
Jennifer Downey
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[PHP-DB] Re: PHP Question on MySQL grouping
depending on how you print the result set to the browser, you just need to add an echo hr at the end of each line. So, if you've fed your result set into an array called $row, then something like: print $row[0] . $row[1] . $row[2] . hr; That should print out each result followed by a horizontal rule. Something like that anyway. Hope that helps. Keiran Wynyard [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED]; I need to loop through a grouped recordset adding a hr at the end of each group, how can groups in recordsets be identified in PHP for specific formatting. SELECT name AS name, dishtype AS dishtype, price AS price FROM dishes WHERE used = 1 GROUP BY dishtype, name ORDER BY dishtype, price DESC The use is in the display of a menu for a restaurant, where there is a table containing all the dishes on the menu in any given instance, but they musst be displayed in a specific order, grouped by type. In other words, all Starters come first, followed by Main courses, followed by Desserts. Between each type there needs to be a seperator, the horizontal line. Thanks for you help Keiran -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] I'm really stuck!
Try doing echo stripslashes($data);
Or try setting the content type to text/plain - This should echo back
gobblygook (Just so you know that there IS image data there.
Also.. you have img src=\petdata.php?id=$id\ in another file right? If
you call petdata.php by itself, you should see just the image.
JD
-Original Message-
From: Jennifer Downey [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 28, 2002 4:41 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] I'm really stuck!
Hi all,
I'm really stuck and I'm not asking anyone to re-write this just show me
what is wrong, or explain why it wont work. It just seems logical that this
should work.
The first query will print the pet id in the browser.
$query=SELECT id FROM wt_users WHERE uid={$session[uid]};
$ret = mysql_query($query);
while(list($pet)=
mysql_fetch_row($ret))
print(BRyour pet id is $pet);
So if $pet will print the id from the wt_users (has a value of 3) and I
assign $id = $pet (id also has a value of 3 in image_data) why doesn't it
show the image?
if($id) {
$id = $pet;
$query = select bin_data,filetype from image_data where id=$id;
$result = mysql_query($query);
$data = mysql_query($result,0,bin_data);
$type = mysql_query($result,0,filetype);
Header( Content-type: $type);
echo $data;
};
echo img src=\petdata.php?id=$id\;
Thanks
Jennifer Downey
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[PHP-DB] PHP + Oracle + Redhat Linux Stupid Question Maybe? Sorry
This may be a stupid question but I am stuck Do I have to have the Oracle Linux Client to connet to a remote Oracle Database or can I just have the php --with-oci8 turned on and that be it? And I can't seem to find any how-tos on just installing the client and I am not sure what prerequisites I need on my computer to install just the client Any help would be appreciated Thanks -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] similar_text() equivalents
I do a lot of data cleaning, and I have really grown enamored of the similar_text() function Just curious, is there anything directly equivalent in perl and/or python? Thanks in advance aron -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Display in drop-down box
Hi there, I was wondering if anyone would be kind enough to show me how to select a field from a table in mysql and display it in a drop-down combo box? I am trying to show my users a list of all the parts from a table so that they can edit or delete it. Thanks! Alvin -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Display in drop-down box
Try doing a select distinct on that field and then loop through the results, adding each one to your drop-down box as you go. I don't have the exact code anymore because we stopped doing it that way, but that's the basic idea of how to go about it. I hope it helps! -Natalie -Original Message- From: Alvin Ang [SMTP:[EMAIL PROTECTED]] Sent: Friday, March 01, 2002 2:07 PM To: PHP Subject: [PHP-DB] Display in drop-down box Hi there, I was wondering if anyone would be kind enough to show me how to select a field from a table in mysql and display it in a drop-down combo box? I am trying to show my users a list of all the parts from a table so that they can edit or delete it. Thanks! Alvin -- 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] Display in drop-down box
Let me give this a shot :P
?
Mysql_connect(sqlserver,user,pass);
Mysql_select_db(db);
$resultset = mysql_query(select DISTINCT identifier, displayname from
table);
for($i=1;$i=mysql_num_rows($resultset);$i++)
{
$object = mysql_fetch_object($resultset);
$identifier = $object - identifier;
$displayname= $object - displayname;
print option value=\$identifier\$displayname/option;
}
?
I do believe that should do it. Correct me if I missed anything.
Ryan
-Original Message-
From: Leotta, Natalie (NCI/IMS) [mailto:[EMAIL PROTECTED]]
Sent: Friday, March 01, 2002 2:10 PM
To: PHP
Subject: RE: [PHP-DB] Display in drop-down box
Try doing a select distinct on that field and then loop through the results,
adding each one to your drop-down box as you go.
I don't have the exact code anymore because we stopped doing it that way,
but that's the basic idea of how to go about it.
I hope it helps!
-Natalie
-Original Message-
From: Alvin Ang [SMTP:[EMAIL PROTECTED]]
Sent: Friday, March 01, 2002 2:07 PM
To: PHP
Subject: [PHP-DB] Display in drop-down box
Hi there,
I was wondering if anyone would be kind enough to show me how to select a
field from a table in mysql and display it in a drop-down combo box?
I am trying to show my users a list of all the parts from a table so that
they can edit or delete it.
Thanks!
Alvin
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[PHP-DB] easy date format question
should be simple as pie, of course, but i can't find a straightforward
syntax example in the documentation i'm trying to change the PHP date
format to mysql's
my code:
$today = date(MMDD);
$sql = INSERT into table (date) values ('$today');
and so on
is filling the date field with zeros, instead of 2002-02-28
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[PHP-DB] Re: FW: PHP Question on MySQL grouping
Another way is to set a testing variable and then print an hr if it
changes
(this off the top of my head, probably bugs in it but you get the idea)
while $myrow=mysql_fetch_row($result) {
if $myrow[dishtype]$test {
echo hr;
//here you would print the rest of the info about the dish from the row
}
$test=$myrow[dishtype];
}
Keiran Wynyard wrote:
I need to loop through a grouped recordset adding a hr at the
end of each group, how can groups in recordsets be identified in
PHP for specific formatting
SELECT name AS name, dishtype AS dishtype, price AS price
FROM dishes
WHERE used = 1
GROUP BY dishtype, name
ORDER BY dishtype, price DESC
The use is in the display of a menu for a restaurant, where there
is a table containing all the dishes on the menu in any given
instance, but they musst be displayed in a specific order,
grouped by type In other words, all Starters come first,
followed by Main courses, followed by Desserts Between each type
there needs to be a seperator, the horizontal line
Thanks for you help
Keiran
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[PHP-DB] Retreiving data from a table
Hi, I got the following insert statement in my PHP code: $qid = db_query( INSERT INTO table1 ( Name ) VALUES ( '$frm[Name]' )); This is working like I want it to But in the table there is a column, which is AUTO_INCREMENT I would like to retreive this number right after I added the data Can anybody help me? Thanks, Morten -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
[PHP-DB] Rounding numbers
Hi again,
Would someone show me how to round numbers:
$bank_points= $ret;
$interest = 095;
$months = 12;
$weeks = 52;
$days=365;
$query=SELECT bank_points FROM wt_users WHERE uid={$session[uid]};
$ret = mysql_query($query);
while(list($bank_points)=
mysql_fetch_row($ret))
$yearly_interest=$bank_points * $interest;
print(BRYour yearly interest is : $yearly_interest);
$daily_interest=$yearly_interest / $days;
$round_daily_interest = round($daily_interest);
print(BRYour daily interest is : $daily_interest);
Thanks
Jennifer Downey
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[PHP-DB] answered my own question
for anyone else who needs it:
http://wwwphpbuildercom/snippet/downloadphp?type=snippetid=356
Matthew Crouch wrote:
should be simple as pie, of course, but i can't find a straightforward
syntax example in the documentation i'm trying to change the PHP date
format to mysql's
my code:
$today = date(MMDD);
$sql = INSERT into table (date) values ('$today');
and so on
is filling the date field with zeros, instead of 2002-02-28
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[PHP-DB] Re: Retreiving data from a table
i'm guessing you use mysql, so you can use the mysql_insert_id to retrieve the most recent addition, then select on it $newid=mysql_insert_id(); $sql=select [the auto_increment field] from table1 where id='$newid'; Does that help? Morten Nielsen wrote: Hi, I got the following insert statement in my PHP code: $qid = db_query( INSERT INTO table1 ( Name ) VALUES ( '$frm[Name]' )); This is working like I want it to But in the table there is a column, which is AUTO_INCREMENT I would like to retreive this number right after I added the data Can anybody help me? Thanks, Morten -- PHP Database Mailing List (http://wwwphpnet/) To unsubscribe, visit: http://wwwphpnet/unsubphp
Re: [PHP-DB] easy date format question
$today = date(Ymd);
$sql = INSERT into table (date) values ('$today');
when the $today variable is inserted into a mysql date column it will fit
just right.
but, if you want to insert it into a timestamp column, you will have to padd
it with 6 zero's
Jim Lucas
www.bend.com
- Original Message -
From: Matthew Crouch [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Friday, March 01, 2002 2:50 PM
Subject: [PHP-DB] easy date format question
should be simple as pie, of course, but i can't find a straightforward
syntax example in the documentation. i'm trying to change the PHP date
format to mysql's.
my code:
$today = date(MMDD);
$sql = INSERT into table (date) values ('$today');
and so on...
is filling the date field with zeros, instead of 2002-02-28
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[PHP-DB] Re: Add and Subtract
my guess is you've got a double-quote problem like that other guy
? $query=SELECT bank_points FROM wt_users WHERE uid={$session[uid]};
ends after the [ because of the double quote
Jennifer Downey wrote:
Hi All,
Would someone please help me with the following code?
I am trying to make this work like if I have 15 gold pieces and I deposit 5
it will subtract 5 from the gold pieces I have and deposit the 5 into my
account
showing that I now have 10 gold pieces I can deposit and 5 in the bank
I would really appreciate any help
Thanks in advance
Jen Downey
BRBRBR
FORM ACTION=bankphp METHOD=POST TITLE=bank
TABLE BORDER=0 WIDTH=25% CELLPADDING=0 CELLSPACING=0
TR
TDfont size=2Deposit/font/TD
TD INPUT TYPE=text VALUE= NAME=user_deposit/TD
/TR
TR
TD/TD
TDINPUT TYPE=submit VALUE=Make My Deposit NAME=deposit/TD
/TR
/TABLE
BRBRBRBRBRBR
? $query=SELECT bank_points FROM wt_users WHERE uid={$session[uid]};
$ret = mysql_query($query);
while(list($points)=
mysql_fetch_row($ret))
print(BRBRBRYou have $points Gold Pieces you may withdraw!); ?
BRBRBR
TABLE BORDER=0 WIDTH=25% CELLPADDING=0 CELLSPACING=0
TR
TDfont size=2Withdraw/font/TD
TD INPUT TYPE=text VALUE= NAME=user_withdraw/TD
/TR
TR
TD/TD
TDINPUT TYPE=submit VALUE=Make My Withdraw NAME=withdraw/TD
/TR
/TABLE
/FORM
?
$db[bank_points]=(UPDATE wt_users set bank_points = bank_points +
$user_deposit);
$result=mysql_query($db[bank_points]);
$db[bank_points]=(UPDATE wt_users set bank_points = bank_points -
$user_withdraw);
$result=mysql_query($db[bank_points]);
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[PHP-DB] Security concern with web forms (update of MySQL data base)
A php page, which includes an update statement for a MySQL data base: I am trying to figure out, how I can make sure that an update form on the web cannot include codes, that would update other parts of the database (or worse destroy a database). bye Ronald Ronald Wiplinger (ÃQ¤¯¯Ç), CEO, ELMIT - The Solution Provider Tel. +886 2 8809-7680, Fax. +886 2 2809-0183, Mobile: +886 915 653-452 Net2Phone:8869550066, ICQ: 111651169 http://www.elmit.comhttp://www.wiplinger.org
[PHP-DB] Rounding numbers again.
Thanks for answering my question I understand rounding now.
My next question is why the script will work like this:
$bank_points= $ret;
$interest = .095;
$months = 12;
$weeks = 52;
$days=365;
$query=SELECT bank_points FROM wt_users WHERE uid={$session[uid]};
$ret = mysql_query($query);
while(list($bank_points)=
mysql_fetch_row($ret))
$yearly_interest=$bank_points * $interest;
$round_yearly= round($yearly_interest);
print(BRYour yearly interest is : $round_yearly Gold Pieces);
$daily_interest=$round_yearly / $days;
$round_daily = round($daily_interest);
print(BRYour daily interest is : $round_daily Gold Pieces);
$query = UPDATE wt_users set bank_points = bank_points + $round_daily WHERE
uid={$session[uid]};
$result = mysql_query($query);
and not like this.
$bank_points = $ret;
$interest = .095;
$months = 12;
$weeks = 52;
$days=365;
$yearly_interest=$bank_points * $interest;
$round_yearly= round($yearly_interest);
$daily_interest=$round_yearly / $days;
$round_daily = round($daily_interest);
$query = UPDATE wt_users set bank_points = bank_points + $round_daily;
$ret = mysql_query($query);
or this.
$query = UPDATE wt_users set bank_points = bank_points + $round_daily;
$ret = mysql_query($query);
$bank_points = $ret;
$interest = .095;
$months = 12;
$weeks = 52;
$days=365;
$yearly_interest=$bank_points * $interest;
$round_yearly= round($yearly_interest);
$daily_interest=$round_yearly / $days;
$round_daily = round($daily_interest);
$query = UPDATE wt_users set bank_points = bank_points + $round_daily;
$ret = mysql_query($query);
If I use the users session id then it works fine. With the query all I'm doing is
updating ALL users bank_points right? I want to use cron to update all users at
midnight.
Thanks again
Jennifer Downey
[PHP-DB] Retrieving a date.
I have a test table with the ususal customer information in it along with a
field for the date of birth. I'm trying to return records of any customers
whose birthdates are the same day and month as the current date.
Here's the query result:
mysql SELECT * FROM Cinfo WHERE BirthDate = date('m d');
Empty set (0.00 sec)
Here's what is returned with a global query:
mysql select * from Cinfo;
+---+--+---+-+---++--+--+--+++
| FirstName | LastName | Address | City| State | Zip|
HomePhone| WorkPhone| CellPhone| BirthDate | id |
+---+--+---+-+---++--+--+--+++
| Joe | Blow | 1200 High St. #12 | St.Looy | UT| 844110 |
801-111-2299 | 801-111-3456 | 801-213-8956 | 1959-03-01 | 1 |
| | | | | ||
| | || 2 |
| | | | | ||
| | || 3 |
| | | | | ||
| | || 4 |
+---+--+---+-+---++--+--+--+++
4 rows in set (0.00 sec)
I have a feeling it's the date format in the table. I tried making it a date
field but just got the current date. What am I doing wrong?
Off Topic, I saw something about resetting the auto-increment counter, but
now I can't find reference to it. Can some kind soul enlighten me???
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Ken Thompson, North West Antique Autos
Payette, Idaho
Email: [EMAIL PROTECTED]
http://www.nwaa.com
Sales and brokering of antique autos and parts.
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RE: [PHP-DB] Security concern with web forms (update of MySQL data base)
A quick suggestion would be to build your query normally and then don't run the query if it has a semicolon that isn't inside quotes. Also, use single quotes in the update to make your checks easier: UPDATE table_name SET field1='value1' -Original Message- From: Ronald Wiplinger To: [EMAIL PROTECTED] Sent: 3/1/02 6:00 PM Subject: [PHP-DB] Security concern with web forms (update of MySQL data base) A php page, which includes an update statement for a MySQL data base: I am trying to figure out, how I can make sure that an update form on the web cannot include codes, that would update other parts of the database (or worse destroy a database). bye Ronald Ronald Wiplinger (ÃQ¤¯¯Ç), CEO, ELMIT - The Solution Provider Tel. +886 2 8809-7680, Fax. +886 2 2809-0183, Mobile: +886 915 653-452 Net2Phone:8869550066, ICQ: 111651169 http://www.elmit.comhttp://www.wiplinger.org -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
