Thanks for your suggestion ... Yes, I tried to print $sessionID (using
echo()) and I get the right value (the one I am looking for) ... Though the
query does not work. Any other idea ??? Thanks again !

    Jean-Marc

----- Original Message -----
From: "Jason Wong" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, August 18, 2002 11:53 AM
Subject: Re: [PHP] Help needed about queries with MySQL, thanks.


> On Sunday 18 August 2002 17:45, Jean-Marc Godart wrote:
>
> > I have a MySQL database (called "sessions") with a auto-increment INT
field
> > called "sessionID". I try to check if there is already a record in this
> > database with a given sessionID, which is called $sessionID in PHP. I
use
> > this query, which must be wrong (but I cannot find out why !!!) :
> >
> >    SELECT * FROM sessions WHERE sessionID = $sessionID
> >
> > Even though there is already a record with the right sessionID, my
program
> > does not find it (I always get 0 when using mysql_num_rows() after the
> > query). Though, if I replace "$sessionID" by, for example 4, then that
> > works and it says there is already a record with that ID.
> >
> > I guess the problem comes from comparing two different kinds of
variables,
> > but I can't fix it ... Could anyone please help ? By advance, thank you
!!!
>
> Have you tried printing out $sessionID to see whether it contains what you
> think it contains?
>
> --
> Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
> Open Source Software Systems Integrators
> * Web Design & Hosting * Internet & Intranet Applications Development *
>
> /*
> Under capitalism, man exploits man.  Under communism, it's just the
opposite.
> -- J.K. Galbraith
> */
>
>
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