Passing variables to a script with Cron

icrwly

Hi...I'm using MAMP and I have a cron set up like this:

5 /usr/bin/curl --silent --compressed http://localhost/cron.php

But I would like to pass in a variable like this:

5 /usr/bin/curl --silent --compressed http://localhost/cron.php?q=mets

Is this possible?

devinemke

did you try it?

icrwly

if you don't want to help, don't respond.

devinemke

with all due respect, i think my suggestion was a rather good one and might have gone a long way in solving your problem.

indeed i do wish to help, and to that end i will answer your direct question:

Q: is this possible?
A: yes

dagon

icrwly,
"will this work" type questions are stupid, all you have to do to answer them is to try it. If you don't even want to attempt to do it yourself, hire some one. If you do want to do it yourself, then try it and if you then have a problem ask.

Weedpacket

There are two ways of running a PHP script locally.

One is to have cron start a web client, pass it the name of the php script as a command-line argument, have the client request the script from the web server, and throw away the result the response from the web server.

The other is to have cron start the PHP executable, pass it the name of the php script as a command line argument, and not have the script generate any output that needs throwing away.

Advantage of the first method is that things like $_SERVER and server functions are available. The advantage of the second is that it's a lot simpler and requires neither a server nor a client.

Running PHP from the command line (a crontab is just a list of scheduled shell commands) is covered in the manual's discussion of PHP's [man]cli[/man] SAPI.

Of course, this isn't an answer to the question that was asked, because that has already been answered.

NogDog

cron:

5 * * * *  /usr/bin/php /absolute/path/to/cron.php mets

cron.php

<?php
if(isset($argv[1]))
{
   $q = $argv[1];
}
else
{
   die('No "q" specified');
}
echo $q;
?>