>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 <
<?php exec("mysql -u myuserid --password=mypassword myddbname ...", $result, $errorcode); echo implode('', $result); if ($errorcode){ echo "OS Error $errorcode Use 'man errno' to find meaning, but it's almost always path/permissions"; } ?> Oh yeah. You'll need the full path to mysql, almost for sure. -- Like Music? http://l-i-e.com/artists.htm Off-Topic: What is the moral equivalent of 'cat' in Windows? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php