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Custom 404 page - PHP

I have a custom 404 page which works fine except for the message I want to display on this page.

I would like it to say the url of the page which can't be found but instead it displays the url of the 404 page.

Here's what I have...

You were looking for <?php echo $_SERVER['REQUEST_URI'] ?>.

The htaccess file contains the line: ErrorDocument 404 /404/

like image 596
Tom Avatar asked Apr 14 '11 12:04

Tom


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3 Answers

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.

like image 196
Simon Avatar answered Oct 14 '22 19:10

Simon


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.

like image 29
Janoz Avatar answered Oct 14 '22 19:10

Janoz


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()); ?>
like image 1
George Reith Avatar answered Oct 14 '22 21:10

George Reith