on 5/26/02 12:47 PM, Dave Carrera at [EMAIL PROTECTED] appended the
following bits to my mbox:
All you need to do is create an array from your sql like this:
1st place your dumped sql statement into a varibale.
2nd create an array from your sql like this
if(isset($your_execution_var))
Well, where is your ';'? At the end of the statement? You don't
terminate queries with semi colons when talking directly to the database.
Just remove it.
-Rasmus
On Sun, 26 May 2002, Dave Carrera wrote:
Hi All
I have found the problem from my last post and need to know how to solve it.
- Original Message -
From: Dave Carrera [EMAIL PROTECTED]
To: php List [EMAIL PROTECTED]
Sent: Sunday, May 26, 2002 8:01 AM
Subject: [PHP-DB] db sql issue from var
Hi All
I have found the problem from my last post and need to know how to solve it.
Example sql var:
$sql =DROP
$sql= .
you forgot the :)
Thanks,
- Original Message -
From: Dave Carrera [EMAIL PROTECTED]
To: php List [EMAIL PROTECTED]
Sent: Sunday, May 26, 2002 11:01 AM
Subject: [PHP-DB] db sql issue from var
Hi All
I have found the problem from my last post and need to know how to solve
]
db sql issue from var
Well, where is your ';'? At the end of the statement? You
don'tterminate queries with semi colons when talking directly to the
database.Just remove it.-RasmusOn Sun, 26 May 2002,
Dave Carrera wrote: Hi All I have found the
problem from
I think it an issue with including multiple sql commands in one var...
Yup, you can't do that. Simply separate them and do them one at a time.
-Rasmus
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