在PHP中包含后获取当前文件名

Let's pretend I have the following class.wrapper.php

// loop through events/ directory ($eventFile = filename of parsed file)
include $eventFile;

myevent.php

echo __FILE__;

myotherevent.php

echo __FILE__;

Both of the echo return the same thing: class.wrapper.php, is there a way to get the real file name ? not the one which included the code ?

The class.wrapper is part of an encrypted API which allows user to define custom php files in a given directory to invoke user defined code when a given event is fired in the main application, so I cannot edit the main class, I have to do this only by editing "events" files

If you want to get the real name, which is for your example the complete path to access this file, use the function realpath("yourfile.extension").

I let you with this short example :

Let's say I have a file named index.php. I want to display its complete path. Here is the code to do so :

<?php
    echo realpath("index.php");
?>

In my example, I saved my file in C:\xampp\htdocs\test\index.php. The result will be a display of this exact path.

Hope it helps.

Try the below function

function GetInclude($file, $params = array())
    {
        extract($params);
        include $file;
    }
    myevent.php
    <?php GetInclude('class.wrapper.php', array('includerFile' => __FILE__)); ?>
    myotherevent.php
    <?php GetInclude('class.wrapper.php', array('includerFile' => __FILE__)); ?>
    class.wrapper.php
    echo $includerFile;