如何在不加载的情况下从另一个PHP视图调用函数呢？
问题描述：
How to call a function from another view in current PHP view? Can anyone help me on this? I only want it to display part of the include view, not the whole thing.
如何在当前PHP视图中从另一个视图调用函数？ 任何人都可以帮我吗？ 我只希望它显示包含视图的一部分，而不是整个事物。 p> div>
答
It's possible with output buffering.
If you have a file like this (we'll call it example2.php):
<?php
function example() {
echo 'For example';
}
?>
Here's some text that you don't want to see.
You can include it in example.php in an output buffer. Clear the buffer without printing its contents, and you'll still have access to the function.
<?php
ob_start();
include 'example2.php';
ob_clean();
example();
Keep in mind that all the PHP code in example2.php will be executed, and there may be undesirable side effects.