On Wednesday 10 July 2002 13:27, Fargo Lee wrote: > Thanks for pointing out the syntax error. I added the space after the -u > but it did not make any difference. It still gives the same result, that is > "Success", when it actually fails. What I am trying to figure out is how I > can tell if it failed (did not create the tables)? The $status variable > does not appear to hold the output of the system() function. Anyone know > how to get the output of system, passthru or exec into the $status variable > to check for success or failure?
Your test for failure should be: if ($status === FALSE) -- Jason Wong -> Gremlins Associates -> www.gremlins.com.hk Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * /* Extreme fear can neither fight nor fly. -- William Shakespeare, "The Rape of Lucrece" */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php