PHP，将参数从命令行传递到PHP脚本 | 将参数从命令行传递到PHP脚本

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问题描述

我想从PHP命令行界面传递参数，然后使用PHP脚本读取值，如下所示:

I want to pass parameters from PHP Command Line Interface, and then read in the values using PHP script, something like this:

<?php
  $name1 = $argv[1];
  echo $name1;
?>

我这样从CLI传递变量:

I pass the variable from CLI like this:

C:\xampp\php\php.exe name.php Robby

上面的作品，我得到罗比作为输出.

The above works, I get Robby as the output.

但是我想做这样的事情:

But I want to do something like this:

C:\xampp\php\php.exe name.php -inputFirstName="Robby"

这样，用户就会被告知在正确的位置输入正确的参数.解析这些参数的合适方法是什么?

So that the user is well informed to enter the correct parameters in the correct places. What is the appropriate way to parse these parameters?

推荐答案

从命令行调用PHP脚本时，可以使用$ argc找出要传递的参数数量，并可以使用$ argv来访问它们.例如，运行以下脚本:

When calling a PHP script from the command line you can use $argc to find out how many parameters are passed and $argv to access them. For example running the following script:

<?php
    var_dump($argc); //number of arguments passed
    var_dump($argv); //the arguments passed
?>

像这样:-

php script.php arg1 arg2 arg3

将给出以下输出

int(4)
array(4) {
  [0]=>
  string(21) "d:\Scripts\script.php"
  [1]=>
  string(4) "arg1"
  [2]=>
  string(4) "arg2"
  [3]=>
  string(4) "arg3"
}

请参见 $ argv 和 $ argc 了解更多信息.

See $argv and $argc for further details.

要做你想做的事

php script.php arg1=4

您需要将等号上的参数展开:-

You would need to explode the argument on the equals sign:-

list($key, $val) = explode('=', $argv[1]);
var_dump(array($key=>$val));

这样，您可以在等号前面拥有所需的任何内容而不必解析它，只需检查key => value对是否正确即可.但是，这只是浪费，只是指导用户按照正确的顺序传递参数.

That way you can have whatever you want in front of the equals sign without having to parse it, just check the key=>value pairs are correct. However, that is all a bit of a waste, just instruct the user on the correct order to pass the arguments.

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