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    $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。

    示例

    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

    以上示例的输出类似于：

    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
    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
    -6
    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
    -14
    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
    -21
    anonymous
    6 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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