Re: mysql_fetch_array warnings.

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" <Martin.Andrew@kpmg-ct.com> wrote in message news:2670E10E092AD943BAAE845456A017245DD218@exchange-dh.kpmg-ct.com... > 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? > > > > > > > >