On Tuesday 09 November 2004 19:12, Aaron Todd wrote: Please do not top post (google > top post). > I tried the double quotes but that didn't seem to matter. It does matter. > I also put the > command string in a variable and then am running the variable with exec. > Still it doesnt work. Splitting it up into 2 steps is to aid in debugging. > $filename = "/home/virtual/site341/fst/var/www/html/db_backup_" . > date("n-j-y").".sql"; > $command = "/home/virtual/site341/fst/usr/bin/mysqldump -u ***** -p***** > database > $filename"; > echo $command; If what is being echoed looks OK then > exec($command); make use of the extra arguments to exec() (see manual) so that you can see what's being output. -- Jason Wong -> Gremlins Associates -> www.gremlins.biz Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * ------------------------------------------ Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db ------------------------------------------ /* If you can survive death, you can probably survive anything. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php