如何将PHP脚本**结果**包含在另一个PHP脚本中？ [英] How can I get a PHP script **results** included into another PHP script?

问题描述

我想将给定脚本的结果（比方说a.php）包含在另一个脚本中（比方说b.php）。

I'd like to include the results of a given script (let's say a.php) into another script (let's say b.php).

虽然有PHP include语句和其他对应语句（include_once，require和require_once），但似乎所有这些都包含文件内容而不是其处理结果我希望。

Although there is the PHP include statement and their other counterparts (include_once, require and require_once), it seems all of them include the file contents instead of its processing results as I would expect.

那么如何首先处理a.php然后将其结果（而不是文件内容本身）包含在b.php脚本中呢？

So how can I get a.php processed first and then include its results (not the file contents itself) into b.php script ?

另一方面，如果我的包含文件（a.php）有一个return语句，那么整个文件应该首先处理，或者它的内容也包含在b.php中是吗？

On the other hand, if my included file (a.php) has a return statement, is the entire file expected to be processed first or its contents are also included into b.php as is ?

提前感谢您的回复。

推荐答案

a .php：

<?php
echo "world";

b.php：

<?php
ob_start();
include("a.php");
$a= ob_get_clean();
echo "Hello, $a!";  

也许？

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