$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。

注意: 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。

注意: 这个变量仅在 register_argc_argv 打开时可用。

范例

示例 #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)

注释

注意:

也可以在 $_SERVER['argc'] 中获取。

参见

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User Contributed Notes 4 notes

up
38
Tejesember
11 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
-6
karsten at typo3 dot org
13 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
-14
elm at r3m0ve dot gmx dot ch
8 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
-21
anonymous
3 years ago
int main(int argc, char *argv[])
{
  fprintf(stdout,"argumen count : %d\n",argc);
  fprintf(stdout,"argumen vector : %s\n",argv);
  return 0;
}
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备份地址：http://www.lvesu.com/blog/php/reserved.variables.argc.php