Writing a PHP command line script that takes options

First of all, we make our file start like this.

#!/usr/bin/env php

This allows us to run the script without prefixing it with the “php” command, and instead we can run it like this:

chmod +x myscript  # This gives execution permissions to the script, do it only once
./myscript --first=option --second --third=option

If we want our script to take options like in the example above, we can use this snippet:

$options = array();
 
foreach ($argv as $arg){
	preg_match('/\-\-(\w*)\=?(.+)?/', $arg, $value);
	if ($value && isset($value[1]) && $value[1]) 
		$options[$value[1]] = isset($value[2]) ? $value[2] : null;
}

The $options array will hold the supplied options. You can then use them like this:

if (!isset($options['somevalue']))
	// show an error
 
if (isset($options['dosomething']))
	// do something
  1. Daemonic about 1 year ago
    Reply

    If you want to check against arguments without a value (i.e. –help) isset() will not work, you will need to use array_key_exists() as per the second example at http://uk2.php.net/manual/en/function.isset.php

    This is due to the default value of an empty argument being set to NULL

  2. AntonioCS about 1 year ago
    Reply

    Maybe you should have provided the link to the manual where it details php’s command line interface -> http://pt.php.net/features.commandline

    Nice post. I have made one or two scripts for the cli but very basic.

  3. Alan about 1 year ago
    Reply

    $argv[0] is always the filename so anything after the .php is your options

    • Guillermo Rauch about 1 year ago

      Right, but $argv is basically a explode(‘ ‘, $command). If you had read the code you’d know we’re parsing options preceded by — that can optionally include a value.

      $argv[1] in this case would only get you –something=something.