Get name of file that is including a PHP script

Here is an examp开发者_如何学Gole of what I am trying to do:

index.php

<ul><?php include("list.php") ?></ul>

list.php

<?php
    if (PAGE_NAME is index.php) {
        //Do something
    }
    else {
        //Do something
    }
?>

How can I get the name of the file that is including the list.php script (PAGE_NAME)? I have tried basename(__FILE__), but that gives me list.php.

$_SERVER["PHP_SELF"]; returns what you want

If you really need to know what file the current one has been included from - this is the solution:

$trace = debug_backtrace();

$from_index = false;
if (isset($trace[0])) {
    $file = basename($trace[0]['file']);

    if ($file == 'index.php') {
        $from_index = true;
    }
}

if ($from_index) {
    // Do something
} else {
    // Do something else
}

In case someone got here from search engine, the accepted answer will work only if the script is in server root directory, as PHP_SELF is filename with path relative to the server root. So the universal solution is

basename($_SERVER['PHP_SELF'])

Also keep in mind, that this returns the top script, for example if you have a script and include a file, and then in included file include another file and try this, you will get the name of the first script, not the second.

In the code including list.php, before you include, you can set a variable called $this_page and then list.php can see the test for the value of $this_page and act accordingly.

Perhaps you can do something like the following:

<ul>
    <?php
        $page_name = 'index';
        include("list.php")
    ?>
</ul>

list.php

<?php
    if ($pagename == 'index') {
        //Do something
    }
    else {
        //Do something
    }
?>

The solution basename($_SERVER['PHP_SELF']) works but I recommend to put a strtolower(basename($_SERVER['PHP_SELF'])) to check 'Index.php' or 'index.php' mistakes.

But if you want an alternative you can do:
<?php if (strtolower(basename($_SERVER['SCRIPT_FILENAME'], '.php')) === 'index'): ?>.