Shell运行/执行带参数的php脚本 [英] Shell run/execute php script with parameters

问题描述

我需要通过shell通过参数执行一个php文件。

I need to execute a php file with parameters through shell.

这里是我将运行php文件：

here is how I would run the php file:

php -q htdocs / file.php

php -q htdocs/file.php

我需要参数'显示通过和

I need to have the parameter 'show' be passed through and

php -q htdocs / file.php？show = show_name

php -q htdocs/file.php?show=show_name

不起作用

如果有人能告诉我什么命令执行以获取php文件与设置参数，这将是非常感激。

If someone could spell out to me what command to execute to get the php file to execute with set parameters, it would be much appreciated. If not, try to lead me the right direction.

推荐答案

test.php：

print_r($argv);

Shell：

$ php -q test.php foo bar
Array
(
    [0] => test.php
    [1] => foo
    [2] => bar
)

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