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.

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

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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
}

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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.

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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.

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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'): ?>.