Custom 404 page - PHP

You need to use $_SERVER['HTTP_REFERER'] instead - that will be the address they requested first.

This only works in the exact case described in the question - where the browser has actually been redirected to the 404 page. In that situation, $_SERVER['REQUEST_URI'] contains the URI of the 404 page rather than the originally requested page as described.

Using Apache's ErrorDocument 404 /handle404.php in the site config or .htaccess would mean that $_SERVER['REQUEST_URI'] would actually work, but a more robust solution is the option in the update below.

Update:

Apparently $_SERVER['REDIRECT_URL'] might be a better bet however, having searched around a bit.

For both cases, as mentioned by the commenters below, bear in mind that any headers are just as prone to malicious content as $_POST, $_GET and others, so process them before outputting anything.

Update 2:

Didn't see the post from @Janoz below - he correctly mentions REDIRECT_URL.

From the perspective of the php page, that really is the request uri. Showing the error page is done by the webserver. Apache for example will add some extra server variables. REDIRECT_URL is probably the one you are looking for.

I did not write this function but it is what I use to do the same thing:

    function selfURL() {    $s = empty($_SERVER["HTTPS"]) ? ''        : ($_SERVER["HTTPS"] == "on") ? "s"        : "";    $protocol = strleft(strtolower($_SERVER["SERVER_PROTOCOL"]), "/").$s;    $port = ($_SERVER["SERVER_PORT"] == "80") ? ""        : (":".$_SERVER["SERVER_PORT"]);    return $protocol."://".$_SERVER['SERVER_NAME'].$port.$_SERVER['REQUEST_URI'];}function strleft($s1, $s2) {    return substr($s1, 0, strpos($s1, $s2));}

then to print it:

<?php print(selfURL()); ?>