Use $argv, that failing it definately works with $_SERVER['argv'] Cheers, Rob.
On Wed, 2003-10-22 at 13:59, Dan Joseph wrote: > Hi, > > I'm having trouble remember something.... > > I have a php script that I want to run from a command line: > > % /usr/local/bin/php script.php 47264 > > 47264 is the argument, however, I can't figure out how to access it inside > the script. I thought it was the $ARG array or $ARGV, but I am mistaken... > can only refresh me? searching did no good. > > -Dan Joseph > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > -- .------------------------------------------------------------. | InterJinn Application Framework - http://www.interjinn.com | :------------------------------------------------------------: | An application and templating framework for PHP. Boasting | | a powerful, scalable system for accessing system services | | such as forms, properties, sessions, and caches. InterJinn | | also provides an extremely flexible architecture for | | creating re-usable components quickly and easily. | `------------------------------------------------------------' -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
