pass crontab a variable and read it from PHP?

Question

I have created a crontab rule:

* * * * * php /my/directory/file.php

I want to pass a variable to be used in the file.php from this crontab.

Answer 1

Neither of the above methods worked for me. I run cron on my hoster's shared server. The cron task is created with the cPanel-like interface. The command line calls PHP passing it the script name and a couple arguments.

That is how the command line for cron looks:

php7.2 /server/path/to/my/script/my-script.php "test.tst" "folder=0"

Neither of the answers above with $argv worked for my case.

The issue was noone told you have to declare $argv as global before you get the access to the CLI arguments. This is neither mentioned in the official PHP manual.

Well, probably one has to declare $argv global ony for scripts run with server. Maybe in a local environment running script in CLI $argv does not require being declared global. When I test it I post here.

But nevertherless for my case the working configuration is:

global $argv;
echo "argv0: $argv[0]\n\r"; // echoes: argv0: /server/path/to/my/script/my-script.php
echo "argv1: $argv[1]\n\r"; // echoes: argv1: test.tst
echo "argv2: $argv[2]\n\r"; // echoes: argv2: folder=0

I got the same results with $_SERVER['argv'] superglobal array. One can make use of it like this:

 $myargv = $_SERVER['argv'];
 echo $myargv[1]; // echoes: test.tst

Hope that helps somebody.

Answer 2

Bear in mind that running PHP from the shell is completely different from running it in a web server environment. If you haven't done command-line programming before, you may run into some surprises.

That said, the usual way to pass information into a command is by putting it on the command line. If you do this:

 php /my/directory/file.php "some value" "some other value"

Then inside your script, $argv[1] will be set to "some value" and $argv[2] will be set to "some other value". ($argv[0] will be set to "/my/directory/file.php").

Answer 3

May I add to the $argv answers that for more sophisticated command-line parameters handling you might want to use getopt() : http://www.php.net/manual/en/function.getopt.php

Answer 4

When you execute a PHP script from command line, you can access the variable count from $argc and the actual values in the array $argv. A simple example.

Consider test.php

<?php
printf("%d arguments given:\n", $argc);
print_r($argv);

Executing this using php test.php a b c:

4 arguments given:
Array
(
    [0] => test.php
    [1] => a
    [2] => b
    [3] => c
)