PHP Command-Line PHP Parsing Program Arguments

PHP Command-Line PHP

Parsing Program Arguments

Problem

You want to process arguments passed on the command line.

Solution

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

       if ($argc != 2) {

            die("Wrong number of arguments: I expect only 1.");

       }

       $size = filesize($argv[1]);

       print "I am $argv[0] and report that the size of ";

       print "$argv[1] is $size bytes.";

Discussion

Example  Parsing commmand-line arguments

       for ($i = 1; $i < $argc; $i++) {

              switch ($argv[$i]) {

              case '-v':

                   // set a flag

                   $verbose = true;

                   break;

              case '-c':

                   // advance to the next argument

                   $i++;

                   // if it's set, save the value

                   if (isset($argv[$i])) {

                        $config_file = $argv[$i];

                   } else {

                        // quit if no filename specified

                        die("Must specify a filename after -c");

                   }

                   break;

              case '-q':

                   $quiet = true;

                   break;

              default:

                   die('Unknown argument: '.$argv[$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.

The $argc and $argv variables are concise, but they are not populated if the register_argc_argv config directive is turned off. However, $_SERVER['argc'] and $_SERVER['argv'] always contain the argument count and argument values. Those are good places to look for argument information if you want maximally portable code.