如何在不加载的情况下从另一个PHP视图调用函数呢？

Question

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.

Answer 1

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.