I thought I had it but I didn't! I can only get LIKE to perform queries that
allow me use the result for a subsequent query but for accuracy, I need to
use actual values. The first statement below works OK (at least I get a
Resource id #) - courtesy of help received here - but will not return
$resultp = mysql_query("select Primaryid from primarycodes where Code =
'".$row['Primaryexpertise']."'") or die (mysql_error());
$resultp = mysql_query("select Primaryid from primarycodes where Code
like
'%$row[Primaryexpertise]%'") or die (mysql_error());
$pcodeid =
