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

These lines will convert the arguments of a CLI call like php myfile.php "type=daily&foo=bar" into the well known $_GET-array:

if (!empty($argv[1])) {
    parse_str($argv[1], $_GET);
}

Though it is rather messy to overwrite the global $_GET-array, it converts all your scripts quickly to accept CLI arguments.

See parse_str for details.

---

If you want the more traditional CLI style like php myfile.php type=daily foo=bar a small function can convert this into an associative array compatible with a $_GET-array:

// Convert $argv into associative array
function parse_argv(array $argv): array
{
    $request = [];
    foreach ($argv as $i => $a) {
        if (!$i) {
            continue;
        }

        if (preg_match('/^-*(.+?)=(.+)$/', $a, $matches)) {
            $request[$matches[1]] = $matches[2];
        } else {
            $request[$a] = true;
        }
    }

    return $request;
}

if (!empty($argv[1])) {
    $_GET = parse_argv($argv);
}

Using the getopt() function, we can also read a parameter from the command line. Just pass a value with the php running command:

php abc.php --name=xyz

File abc.php

$val = getopt(null, ["name:"]);
print_r($val); // Output: ['name' => 'xyz'];

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];

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)