Thanks for the information, I will try that in this situation, however, I
have done this in other select statements and it has worked fine. I will
follow-up in the PHP list if I still have problems.

Please note also that since I assumed this was a MySQL problem I posted it
to this group. Since these two programs are so closely interlinked, it is
hard for us newbies to tell where the problem may lie. I have in the past
posted a question to both groups as I wasn't sure of it either. These lists
are for those that know and those that don't and both have to be tolerant of
the other.

Beauford

----- Original Message -----
From: "Basil Hussain" <[EMAIL PROTECTED]>
To: "C. Reeve" <[EMAIL PROTECTED]>; "MySQL List"
<[EMAIL PROTECTED]>
Sent: Wednesday, December 18, 2002 11:57 AM
Subject: RE: $Variables and Select statements


> Hi,
>
> >         while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
> >
> >                 echo "<TR WIDTH=150><TD
> > ALIGN=left>".$line['name']."</TD><TD
> > ALIGN=right>".$line['$wk']."</TD>";
> >
> >         }
>
> Here is your problem. Your are indeed substituting $wk into the query
> correctly, but when you go to output the result rows you are not.
>
> You should simply remove the quote marks from:
>
> $line['$wk']
>
> so that it is:
>
> $line[$wk]
>
> PHP does not interpret variable names in single-quoted strings - only in
> double-quoted strings. With your current code, PHP is only looking for an
> array element called '$wk', which doesn't exist!
>
> In future, may I suggest you that post questions like this to the PHP
> mailing list, as this is nothing to do with MySQL.
>
> Regards,
>
> Basil Hussain
> ---------------------------------------
> Internet Developer, Kodak Weddings
> E-Mail: [EMAIL PROTECTED]
>
>
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