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

Reply via email to