On Fri, Feb 6, 2009 at 12:44, Terion Miller <webdev.ter...@gmail.com> wrote:
>
> ---------------------
> $sql = "INSERT INTO admin (UserName, Password, Name, Email, Property,
> Department, AddWorkOrder, ";
> $sql .= "ViewAllWorkOrders, ViewNewOrders, ViewNewArt,
> ViewPendingWorkOrders, ViewPendingArtwork, ViewCompletedArt, ";
> $sql .= "ViewCompletedWorkOrders, SearchWorkOrder, EditWorkOrder,
> DelWorkOrder, ChangeStatus, AddEditAdmin";
> $sql .= ") VALUES( '$UserName', '$Password', '$Name', '$Email',
> '$Property', '$Department', '$AddWorkOrder', ";
> $sql .= "'$ViewAllWorkOrders', '$ViewNewOrders', '$ViewNewArt',
> '$ViewPendingWorkOrders', '$ViewPendingArtwork', ";
> $sql .= "'$ViewCompletedArt', '$ViewCompletedWorkOrders',
> '$SearchWorkOrder', '$EditWorkOrder', '$DelWorkOrder', ";
> $sql .= "'$ChangeStatus', '$AddEditAdmin', '$ViewMyOrders')";
> $result = mysql_query($sql);
1.) You have 19 columns named, but 20 values given.
2.) Make sure you use mysql_real_escape_string() or otherwise
sanitize the data.
3.) When you run into similar issues, use <?php mysql_query($sql)
or die(mysql_error()); ?>
--
</Daniel P. Brown>
daniel.br...@parasane.net || danbr...@php.net
http://www.parasane.net/ || http://www.pilotpig.net/
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