If you've typed your code in accurately then...
> $detailqry = "SELECT id, parentitemid, itemtypeid, itemstatusid,
[etc]
> $result = mysql_query($detailqry) or die("Failed finding
> task details");
somewhere in about here you want to have something like:
$sqldata = mysql_fetch_array($res
Hi,
Tuesday, September 10, 2002, 1:41:23 PM, you wrote:
PH> Hello everyone..tryin to run this qry against a mysql db, but after it runs,
PH> it doesn't assign anything to the variables as it should. If i return all
PH> rows, and spit out each record in the result in an array, i have the same
P
Hello everyone..tryin to run this qry against a mysql db, but after it runs,
it doesn't assign anything to the variables as it should. If i return all
rows, and spit out each record in the result in an array, i have the same
problem, but have 24 'blank' records instead of 1. Any ideas? Thanks f
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