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
$pcodeid from the third statement - it dies giving me "No pcodeid". Using
the LIKE version (second one below) works fine and I can use the $pcodeid
array elements as I like. Any thoughts gratefully received.
$resultp = mysql_query("select Primaryid from primarycodes where Code =
'".$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");
Regards
Mick Lloyd
[EMAIL PROTECTED]
Tel: +44 (0)1684 560224
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