Question

I have a PHP script that takes to command line arguments. I want the user to type the name of the program with no arguments:

$ ./foo.php

and I want it to output something like:

usage: $ ./foo arg1 arg2 where arg1 is something and arg2 is something else

Is there a standard way of doing this?

Many thanks :).

Answer 1

Just output it.

if (count($_SERVER['argv']) <= 1) {
  echo 'Usage: $ ' . $_SERVER['argv'][0] . ' arg1 arg2 where arg1 is something and arg2 is something else' . PHP_EOL;
}

$_SERVER['argv'] contains all arguments from the command line and especially the first item is always the scriptname.

See “reserved variables: $argv” and “reserved variables: $_SERVER”. Note, that $argv is no available in any case, thus I recommend using $_SERVER['argv']