Since it works for you then it MUST be a version thing. How odd that it works
from the command line.
It would have been a pain to do an upgrade just to find out the problem didnt
go away. I appreciate your taking the time to test it for me.
Thanks again!!!!
Johannes Janson wrote:
> I'm not sure but couldn't it really be that this is
> due to the MySQL version? PHP is just passing
> a string which is then executed by MySQL. I think
> I recall having no problems with aliases. I don't quite
> remember which version it was but I think it was 3.23.*.
>
> > Now I am using a shareware version which, perhaps, is buggy:
> > 3.22.24-shareware-debug
>
> Probably is a problem of the version...
> I simply had to test it again. Here's very simple a script that worked fine.
>
> $sql = "SELECT a.id as ida, b.id as idb FROM alleuebungen a,
> allevorlesungen b WHERE a.id=b.id";
>
> $result = mysql_query($sql) or die(mysql_error());
>
> while (list($ida, $idb) = mysql_fetch_array($result)){
>
> echo "$ida, $idb<br>";
> }
>
> I'm using 3.23.36.
>
> hope it helps
> Johannes
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
> To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
