> $resultp = mysql_query("select Primaryid from primarycodes where Code =
> '".$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 [EMAIL PROTECTED]
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"
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