从 include() 获取调用文件名

Get calling file name from include()(从 include() 获取调用文件名)

问题描述

I want to get the name of the file that includes another file from inside the included file.

I know there is the __FILE__ magic constant, but that doesn't help, since it returns the name of the included file, not the including one.

Is there any way to do this? Or is it impossible due to the way PHP is interpreted?

This is actually just a special case of what PHP templating engines do. Consider having this function:

function ScopedInclude($file, $params = array())
{
    extract($params);
    include $file;
}

Then A.php can include C.php like this:

<?php
// A.php
ScopedInclude('C.php', array('includerFile' => __FILE__));

Additionally, B.php can include C.php the same way without trouble.

<?php
// B.php
ScopedInclude('C.php', array('includerFile' => __FILE__));

C.php can know its includer by looking in the $params array.

<?php
// C.php
echo $includerFile;

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