Re: More on strings

> $resultp = mysql_query("select Primaryid from primarycodes where Code = > '".$row['Primaryexpertise']."'") or die (mysql_error()); > > $resultp = mysql_query("select Primaryid from primarycodes where Code like > '%$row[Primaryexpertise]%'") or die (mysql_error()); > > $pcodeid = mysql_fetch_array($resultp) or die("No pcodeid"); Two suggestions.: First, stick the query into a variable instead of using it directly; that way you can do some simple debugging (i.e. print the query to the page instead of posting this email). Second, take the array variable out of the double quotes, like you did in the first line. $sql = "select Primaryid from primarycodes where Code like'%" . $row[Primaryexpertise] . "%'"; print $sql . "<BR>\n"; $resultp = mysql_query($sql) or die (mysql_error()); - Darryl ---------------------------------------------------------------------- Darryl Friesen, B.Sc., Programmer/Analyst Darryl.Friesen@usask.ca Education & Research Technology Services, http://gollum.usask.ca/ Department of Computing Services, University of Saskatchewan ---------------------------------------------------------------------- "Go not to the Elves for counsel, for they will say both no and yes"