Your code is possibly failing at this line:
> $result = mysql_query($query);
Make sure that the value in $result is != 0
so, try this:
<snip>
##assuming you've already connected to the database, etc...
$query = "select * from users where UserName LIKE '$username'";
if ( $result = mysql_query($query) )
{
if ($row = mysql_fetch_array($result)) {
## ...
## ...
}
}
else
{
echo "Your query failed. Reason: " . mysql_error_msg();
}
</snip>
--
Nicole Amashta
Web Application Developer
www.aeontrek.com
"Martin.Andrew" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> testing locally I use the following code
>
> $query = "select * from users where UserName LIKE '$username'";
> $result = mysql_query($query);
> if ($row = mysql_fetch_array($result)) {
> ...
> ...
> }
>
> works fine.
>
> On the ISP machine I get the following warnings?
>
> Supplied argument is not a valid MySQL result resource !!
>
> Do I need to change a setting?
>
>
>
>
>
>
>
>
--
PHP Windows Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
