RE: [PHP] Capturing System output

Hello, If you are working on a Linux system, you can try appending: 2>&1 To the end of your command, so that you end up with: Mysql -u $user -p{$pass} 2>&1 What this does is tell the shell to redirect anything from stderr to go through stdout. If you want to get -really- fancy you can use proc_open, which will give you handles for the process' stdout, stderr, and stdin. HTH, K. Bear > -----Original Message----- > From: Brad Bonkoski [mailto:bbonkoski@mediaguide.com] > Sent: Tuesday, August 15, 2006 10:38 AM > To: Stut > Cc: PHP List > Subject: Re: [PHP] Capturing System output > > > > Stut wrote: > > Brad Bonkoski wrote: > >> Had this problem in the past, and always programmed around it, but > >> wondering if there is an easier way. > >> > >> Good Example: > >> Creating a setup for connecting to a mysql database. Want to do > >> something simple to make sure they have entered a valid > >> username/password for the database. > >> So, the idea is something like: > >> $rc = exec("mysql -u $user -p{$pass}", $output); The > problem is one > >> error, the stderr does not go to the output array, but > rather to the > >> screen. > >> > >> Previously I would redirect the stderr to a file, and then > evaluate > >> the contents of the file, but is there an easier way to > get this into > >> the PHP variable with no risk of having the output make it > through to > >> the screen? > > > > I may be missing something, but why in the name of all that is holy > > would you want to shell out to try connecting to mysql? Why not use > > mysql_connect and avoid the potentially massive security > hole you're > > building? > > > > -Stut > > > Perhaps poor illustration of the question...the question > being how to issue system like commands in PHP which would > allow you to trap not only stdout, but also stderr. > -Brad > > -- > PHP General Mailing List (http://www.php.net/) To > unsubscribe, visit: http://www.php.net/unsub.php > > >