: RE: [PHP-DB] A real question this time...
You are correct that an assignment between two variable is a copy, and that
is exactly what is happening, however, you need to know what $SA is
originally--it is a resource pointer in ways...it does not hold any data, it
is just an internal pointer to wh
On Fri, 2002-07-05 at 14:43, NIPP, SCOTT V (SBCSI) wrote:
>
> mysql_select_db($database, $Test);
> $query_SA = "SELECT sbcuid FROM contacts_sa";
> $SA = mysql_query($query_SA, $Test) or die(mysql_error());
> $PASS = $SA;
I think here is your problem you have the select element name pass just
You are correct that an assignment between two variable is a copy, and that
is exactly what is happening, however, you need to know what $SA is
originally--it is a resource pointer in ways...it does not hold any data, it
is just an internal pointer to where php is storing the data. So, when you
d
