$argc

(PHP 4, PHP 5, PHP 7, PHP 8)

$argcThe number of arguments passed to script

Description

Contains the number of arguments passed to the current script when running from the command line.

Note: The script's filename is always passed as an argument to the script, therefore the minimum value of $argc is 1.

Note: This variable is not available when register_argc_argv is disabled.

Examples

Example #1 $argc example

<?php
var_dump
($argc);
?>

When executing the example with: php script.php arg1 arg2 arg3

The above example will output something similar to:

int(4)

Notes

Note:

This is also available as $_SERVER['argc'].

See Also

  • getopt() - Gets options from the command line argument list
  • $argv

add a note

User Contributed Notes 4 notes

up
45
Tejesember
13 years ago
To find out are you in CLI or not, this is much better in my opinion:
<?php
if (PHP_SAPI != "cli") {
exit;
}
?>
up
-9
karsten at typo3 dot org
15 years ago
Note: when using CLI $argc (as well as $argv) are always available, regardless of register_argc_argv, as explained at http://docs.php.net/manual/en/features.commandline.php
up
-15
elm at r3m0ve dot gmx dot ch
10 years ago
To decide whether my script is run from CLI I simply create a PHP script that handles only CLI invocations.

File cron.php:
<?php

// Set environment variables your application depends on
$_SERVER[ 'HTTP_HOST' ] = 'domain.tld';
// $_SERVER[ 'REQUEST_URI' ] = '/some/URI/if/needed';

// Use the environment to read out required values
$task = $_SERVER[ 'argv' ][ 1 ];

// Instanciate the dispatcher or whatever you use
$dispatcher = new Dispatcher();
$dispatcher->handle( $task );

?>

This way my application doesn't have to know about CLI at all.
up
-36
anonymous
5 years ago
int main(int argc, char *argv[])
{
fprintf(stdout,"argumen count : %d\n",argc);
fprintf(stdout,"argumen vector : %s\n",argv);
return 0;
}
To Top