包含PHP脚本时传递参数 [英] Passing parameter while including PHP script

问题描述

我想这样做，但它给出了错误:(为了更好地理解我的问题，我给出了一个例子：

I want to do this, but it gives error :( for better understanding my problem I'm giving an example:

<?php
include 'script.php?text=hiii';
?>

script.php的内容

content of the script.php

<?php
echo $_GET['text'];
?>

那么，我怎样才能在包含脚本页？

So, how can i pass an argument while including the script page?

推荐答案

你可以设置 $ _ GET ['text'] 在包含文件之前：

You could set $_GET['text'] before including the file:

$_GET['text'] = 'hiii';
include 'script.php';

但这显然不会影响其他变量，例如 $ _ SERVER ['REQUEST_URI'] ， $ _ SERVER ['QUERY_STRING'] 等。

But this obviously won’t affect other variables like $_SERVER['REQUEST_URI'], $_SERVER['QUERY_STRING'] etc.

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