如何将参数从命令行传递到 php-script 中的 $_POST？
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How to pass parameters from command line to $_POST in php-script?
提问by hades
I know this could sound a little weird but I need to pass some parameters to a $_POST array. Similar to the way apache does it, or any other web server.
我知道这听起来有点奇怪，但我需要将一些参数传递给 $_POST 数组。与 apache 或任何其他 Web 服务器的工作方式类似。
Unfortunately I couldn't find libapache2-mod-php5anywhere for my Ubuntu.
不幸的是，我找不到libapache2-mod-php5适合我的 Ubuntu 的任何地方。
采纳答案by mario
That's not easily doable. You can invoke the php-cgibinary and pipe a fake POST request in. But you'll need to set up a whole lot of CGI environment variables:
这并不容易做到。您可以调用php-cgi二进制文件并通过管道发送一个假的 POST 请求。但是您需要设置大量的 CGI 环境变量：
echo 'var1=123&var2=abc' | REQUEST_METHOD=POST SCRIPT_FILENAME=script.php REDIRECT_STATUS=CGI CONTENT_TYPE=application/www-form-urlencoded php-cgi
Note: Insufficient, doesn't work like that. But something like that...
注意：不足，不能那样工作。但类似的事情...
It's certainly easier if you just patch the script, and let it load the $_POST array from a predefined environment variable.
如果您只是修补脚本，并让它从预定义的环境变量中加载 $_POST 数组，那肯定会更容易。
$_POST = parse_url($_SERVER["_POST"]);
Then you can invoke it like _POST=var=123 php script.phpfor simplicity.
然后，_POST=var=123 php script.php为了简单起见，您可以像调用它一样调用它。
回答by Schlangi
I was searching for a solution for this and came by, because it was the first hit at Google. The second onewas somehow mor useful for me, because it has a really easy solution, if you have access to the PHP script and can change it.
我正在为此寻找解决方案并来了，因为它是谷歌的第一个热门。第二个对我来说更有用，因为它有一个非常简单的解决方案，如果您可以访问 PHP 脚本并且可以更改它。
Just insert the following lines at the beginning of your script:
只需在脚本的开头插入以下几行：
/* if started from commandline, wrap parameters to $_POST and $_GET */
if (!isset($_SERVER["HTTP_HOST"])) {
parse_str($argv[1], $_GET);
parse_str($argv[1], $_POST);
}
This small piece of code does the trick (you may decide if you want to use $_GET or $_POST or, like I needed it, both.
After changing your script you can call it from commandline passing your args:
这一小段代码可以解决问题（您可以决定是否要使用 $_GET 或 $_POST ，或者，就像我需要的那样，两者都使用。
更改脚本后，您可以从传递参数的命令行调用它：
php yourscript.php 'arg1=x&arg2=y'
Have fun!
玩得开心！
回答by user926400
curl --data "name=ii" "param1=value1¶m2=value2" http://test.com/sample.php
