Using parameters via EXEC()?

[Jujubee]

I've got an update script that runs 3 times a day. The one php file call others with exec():


exec("/usr/bin/php -q home/public_html/update1.php");
exec("/usr/bin/php -q home/public_html/update2.php");


but if I try running one with a parameter:

exec("/usr/bin/php -q home/public_html/update1.php?action=dodo");

It doesn't work. I'm not too familar with UNIX command line use (or Unix at all), but I'm assuming the ? is causing problems.

Any of you Unix gurus know a work around for this? Thanks.

[freddydoesphp]

Is this run through a cron job? I am just wondering why you opted for the exec() command. You can use a cron job for this and then you can just specify the command to run


/usr/bin/php -q home/public_html/update1.php?action=dodo > /dev/null


Try reading up on how to set up cron jobs on your server, and you should be all set.

If you still insist on a php script try(not tested)


EscapeShellCmd("/usr/bin/php -q home/public_html/update1.php?action=dodo");

[Jujubee]

Yes, this is a CRON job, but I've got several scripts to run and adding them all to CRON is a hassle to keep track of.

I figured I could just set one PHP file that calls others. So if I need to change the times for the updates, I'd just update the one line in CRON.

In retrospect though, I rarely change the times, so I can't get it to work any other way, I'll do it directly as you mention. Thanks for the help.

I'll test out the Escape command.

[freddydoesphp]

Better yet stick them in a shell script

#!/bin/sh
/usr/bin/php -q home/public_html/update1.php?action=dodo > /dev/null 
/usr/bin/php -q home/public_html/update2.php?action=dodo > /dev/null 
/usr/bin/php -q home/public_html/update3.php?action=dodo > /dev/null