Pass a variable to a PHP script running from the command line

Question

I have a PHP file that is needed to be run from the command line (via crontab). I need to pass type=daily to the file, but I don't know how. I tried:

php myfile.php?type=daily 

but this error was returned:

Could not open input file: myfile.php?type=daily

What can I do?

Answer 1

The ?type=daily argument (ending up in the $_GET array) is only valid for web-accessed pages.

You'll need to call it like php myfile.php daily and retrieve that argument from the $argv array (which would be $argv[1], since $argv[0] would be myfile.php).

If the page is used as a webpage as well, there are two options you could consider. Either accessing it with a shell script and Wget, and call that from cron:

#!/bin/sh wget http://location.to/myfile.php?type=daily 

Or check in the PHP file whether it's called from the command line or not:

if (defined('STDIN')) {   $type = $argv[1]; } else {   $type = $_GET['type']; } 

(Note: You'll probably need/want to check if $argv actually contains enough variables and such)

Answer 2

Just pass it as normal parameters and access it in PHP using the $argv array.

php myfile.php daily 

and in myfile.php

$type = $argv[1];