RE: [PHP] passing mysql_error() out put to a different page?

> Well, the die() function takes a limited number of parameters, which as > you > noted is why it doesn't like the assignment of the mysql_error() and the > header > statements both inside the die() function. So, why not something like > this: > > $res = mysql_query(.....); > if ($res == FALSE) //which is what I _believe_ die() looks at > /* or at least this is what the manual says the function returns on a > query > error */ > { > $error = mysql_error(); > Header("Location: logerror.php?op=insert_error&err_value=$error"); If you do it this way, you'll want to urlencode(), rawurlencode(), or base64_encode() the $error. You only have to decode on the receiving end if you use the base64_encode. ---John Holmes...