$argc

$argc — 传递给脚本的参数数目

说明

包含当运行于命令行下时传递给当前脚本的参数的数目。

Note: 脚本的文件名总是作为参数传递给当前脚本，因此 $argc 的最小值为 1。

Note: 这个变量仅在 register_argc_argv 打开时可用。

范例

Example #1 $argc 范例


<?php
var_dump($argc);
?>

当使用这个命令执行: php script.php arg1 arg2 arg3

以上例程的输出类似于：

int(4)

User Contributed Notes

elm at r3m0ve dot gmx dot ch 04-Apr-2014 01:57


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.

hvuaka at gmail dot com 04-Jan-2014 11:19


You can use php_sapi_name() to check whether it is cli or not

Tejesember 04-May-2011 06:50


To find out are you in CLI or not, this is much better in my opinion:

<?php

if (PHP_SAPI != "cli") {

    exit;

}

?>

karsten at typo3 dot org 18-Feb-2009 09:47


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