Jason, That's exactly what I'm trying to do, and it's not working:
My Script: #!/usr/bin/php -f <?php $test = $argv[1]; print $test; $demo = "This Works"; print $demo; ?> Running: ./test.php blah Yiels only "This Works", but not "blah". I'm using version 4.0.6 --- Jason Young <[EMAIL PROTECTED]> wrote: > I can't say I'm really too familiar with the php > commandline.. > > You're using /usr/bin/php (or equivalent) and > attempting to use your > exec() that way? > > If you do 'php -?' you'll get a list of commands > that you can use, and I > don't see a way to pass cmdline arguments as > variables.. > > Having said that, I Just went and looked further > into it.. if I make a > test script, and at the top I put: > $hi = $argv[1]; > > then $hi becomes whatever you've specified as the > first argument.. I'm > assuming this is what you want? > > To clarify: > phpfile.php contains: > <? > $hi = $argv[1]; > echo $hi; > ?> > > Running the command "php -f phpfile.php test" > returns "test" > > Does this help at all?? > > -Jason > > Daren Cotter wrote: > > Jason, > > > > I'm not using a web script any longer, I'm using > > command-line (I determined that it is installed on > the > > server). > > > > I read about $argc and $argv, but when I call the > > script passing two arguments, both $argc and $argv > are > > blank. Is this a php.ini setting I need to change > or > > somethign? > > > > --- Jason Young <[EMAIL PROTECTED]> wrote: > > > >>Sorry to butt in :) > >> > >>Arguments to web scripts are done in the format: > >>page.php?arg1=data1&arg2=data2 > >> > >>So you would use that full string as the lynx > path. > >> > >>Hope this helps :) > >>-Jason > >> > >>Daren Cotter wrote: > >> > >>>Thanks for the info Chris, it works! > >>> > >>>How do I pass arguments to the script? I'm > >> > >>assuming > >> > >>>it'd just be: > >>> > >>>test.php arg1 arg2 > >>> > >>>The stuff I've read says $argc should be the > count > >> > >>of > >> > >>>the # of arguments, and $argv should be an array > >>>holding them...but when I do a simple: > >>>print "# of Arguments: $argc\n"; > >>>It prints nothing, not even 0 > >>> > >>> > >>>--- Chris Hewitt <[EMAIL PROTECTED]> > >> > >>wrote: > >> > >>>>>On Wed, 25 Sep 2002, Daren Cotter wrote: > >>>>> > >>>>> > >>>>> > >>>>>>My problem, is that I absolutely NEED to run a > >>>>> > >>PHP > >> > >>>>>>script using crontab. The script needs to send > >>>>>>numerous queries to a database every hour. Is > >>>>> > >>>>there > >>>> > >>>> > >>>>>>any way I can accomplish this, directly or > >>>>> > >>>>indirectly? > >>>> > >>>>Are you sure its not already there? Commonly in > >>>>/usr/bin. Try a "which > >>>>php" and see if it finds anything? > >>>> > >>>>HTH > >>>>Chris > >>>> > >>> > >>> > >>> > >>>__________________________________________________ > >>>Do you Yahoo!? > >>>New DSL Internet Access from SBC & Yahoo! > >>>http://sbc.yahoo.com > >> > >> > >>-- > >>PHP General Mailing List (http://www.php.net/) > >>To unsubscribe, visit: > http://www.php.net/unsub.php > >> > > > > > > > > __________________________________________________ > > Do you Yahoo!? > > New DSL Internet Access from SBC & Yahoo! > > http://sbc.yahoo.com > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > __________________________________________________ Do you Yahoo!? New DSL Internet Access from SBC & Yahoo! http://sbc.yahoo.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
