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?
"Analysis & Solutions" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > On Tue, Jul 09, 2002 at 06:09:36PM -0700, Fargo Lee wrote: > > > > $status = system("mysql -umyuserid -pmypassword mydbname < > > You need a space between "-u" and "myuserid" > > --Dan > > -- > PHP classes that make web design easier > SQL Solution | Layout Solution | Form Solution > sqlsolution.info | layoutsolution.info | formsolution.info > T H E A N A L Y S I S A N D S O L U T I O N S C O M P A N Y > 4015 7 Av #4AJ, Brooklyn NY v: 718-854-0335 f: 718-854-0409 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
