On Sunday 31 October 2004 18:32, Arne Essa Madsen wrote:
> I have the following problem:
>
> This one does not work:
>  $query = "SELECT * from user where name='$userid' and
> pass=password('$password')";

print $query;

>   $result = mysql_query($query, $link);

Use mysql_error().

-- 
Jason Wong -> Gremlins Associates -> www.gremlins.biz
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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