PHP Cookbook

20.1. Parsing Program Arguments

Problem

You want to process arguments passed on the command line.

Solution

Look in $_SERVER['argc'] for the number of arguments and $_SERVER['argv'] for their values. The first argument, $_SERVER['argv'][0], is the name of script that is being run:

if ($_SERVER['argc'] != 2) {
    die("Wrong number of arguments: I expect only 1.");
}

$size = filesize($_SERVER['argv'][1]);

print "I am $_SERVER[argv][0] and report that the size of ";
print "$_SERVER[argv][1] is $size bytes.";

Discussion

In order to set options based on flags passed from the command line, loop through $_SERVER['argv'] from 1 to $_SERVER['argc']:

for ($i = 1; $i < $_SERVER['argc']; $i++) {
    switch ($_SERVER['argv'][$i]) {
    case '-v':
        // set a flag
        $verbose = 1;
        break;
    case '-c':
        // advance to the next argument
        $i++;
        // if it's set, save the value
        if (isset($_SERVER['argv'][$i])) {
            $config_file = $_SERVER['argv'][$i];
        } else {
            // quit if no filename specified
            die("Must specify a filename after -c");
        }
        break;
    case '-q':
        $quiet = 1;
        break;
    default:
        die('Unknown argument: '.$_SERVER['argv'][$i]);
        break;
    }
}

In this example, the -v and -q arguments are flags that set $verbose and $quiet, but the -c argument is expected to be followed by a string. This string is assigned to $config_file.

See Also

Recipe 20.3 for more parsing arguments with getopt; documentation on $_SERVER['argc'] and $_SERVER['argv'] at http://www.php.net/reserved.variables.