I suggest creating a shell wrapper for PHP that will write the command to a file for you and then call PHP with the appropriate arguments. PHP won't even see most of the command that you originally posted. Thank you, Micah Gersten onShore Networks Internal Developer http://www.onshore.com Jochem Maas wrote: > Robert Cummings schreef: >> On Thu, 2008-08-28 at 10:22 -0500, Micah Gersten wrote: >>> Does this work? >>> $command = implode(' ', $argv); > > no syntax errors, so in that sense it works. > but it doesn't answer my question (check the body of the post as > well as the subject and that might become clear). > >> Only do it that way if it's for a log of general output. If you're going >> to punt any of those parts back to the OS in another command you'll need >> to properly escape them. > > true, and yes it's for a log and no join(' ', $argv); doesn't cut it :-) > >> >> Cheers, >> Rob. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
