Hello,
3 tables hold money value but they are inputed into
the database without the dollar symbol (without $).
So the value could be 3.00 etc...
I want to calculate by adding (+) them together. A
username is in each table. So only the username
matching in each table must calculate. So if in
> What is the best PHP code to use for this inside a
> PHP
> code?
I meant ...
What is the best PHP code to use for this inside a PHP
page/file.
(e.g. calculate.php)
Also the total of the 3 tables should than be updated
to a 4th table which I did not mention.
So the total of table 1, 2 and
First of all, check the SELECT statement in your database documentation,
because you do this within your select. It is important you that you inform
yourself about this, because you will use it a lot. So this is not really a
PHP question. The statement is written something like this:
SELECT a.c
I don't see many SQL Server posts here but I will give it a shot. I am
using PHP 4.3.10, SQL Server 2000, and W2000.
I have had no problems with scripting against PHP 5.x but I have had
real problems when running these same scripts against 4.3.10.
For example, all my select and insert statements
Does the mssql account user have the correct priviledges to be able to run
those queries? Are you using proper error reporting to see if there are any
error messages returned by mssql wrt those queries? (are you using
http://ca3.php.net/manual/en/function.mssql-get-last-message.php to get the
- Original Message -
From: "Bastien Koert" <[EMAIL PROTECTED]>
Newsgroups: php.db
To: <[EMAIL PROTECTED]>;
Sent: Sunday, March 13, 2005 6:19 PM
Subject: RE: [PHP-DB] Problems with 4.3.10 and SQL Server
> Does the mssql account user have the correct priviledges to be able
to run
> those
I have created a stored procedure wich is accessing an remote linked MSSQL
2000 server and works perfectly from Querry Analyzer. The problem occurs
when I try to execute the stored procedure from php 5.0.3 (by using mssql_bind
or mssql_query); I've got no error but no result either. It seems that p
