> Instead of checking if(!$id), perhaps you would be better off to check the
> result of your query (which in this example was successful, since you got
> a return from mysql() ).
I am. I just didn't include it in my previous message as it as I was trying
to
keep extraneous code down to a minimum. Plus, I mentioned in my initial
post that there were no errors being returned and mysql() was in fact
returning
a 1 (true) as opposed to a 0 (false) as it's result. Both indicate that the
query
was successful.
> Also, what is the last_insert_id() function that you are referencing in
your
> $insertIDQuery string?
Internal mySQL function.
> As for things getting messed up when two queries hit the system at the
same
> time, you shouldn't need to worry about that either because the two people
> coming in would have different values for the $link variable and since
> mysql_insert_id goes off of the link identifier of the connection (which
will be
> different for the two different people running queries), you should not
have a
> conflict.
My worry comes from the fact that I'm running an internal mySQL function
and I'm not certain that mySQL would know who was who and which insert
ID to return for 2 simultaneous queries. But now that I think about it, I
see
that worry is misplaced.
Chris
