How to display default Apache 404 from PHP

Question

I have looked around, and haven't found any questions regarding what I need.

I would like to make something like an "admin.php", but I would like to make it where it returns a 404 if the conditions aren't right. However, a 404 that looks different from all the rest of the 404 pages will hint to the attacker that "admin.php" is a real file.

I don't want to make my own custom 404 page, I think Apache's default 404 is good enough for me. How do I send a 404 that triggers apache to send the already ErrorDocument. I assume this error document, after searching for it, is located (in Ubuntu): "/usr/share/apache2/error/HTTP_NOT_FOUND.html.var"

When I use things like:

<?php
    http_response_code(404);
    die();
?>

it looks like this in the web browser:

When I request a URL that truly isn't on the web server it looks like this:

I would like to know how I could make the "/admin.php" look like the second image, using Apache's default 404 page. Is there a way to do this without using .htaccess files, just from PHP? If not, I'll just have to figure out how to mess with Apache settings then.

sjdaws · Accepted Answer · 2017-06-19 02:55:15Z

9

There isn't a way to do this just by returning a 404 as Apache will find the file and therefore will no longer throw an error page. The best thing you can do is emulate the Apache error page:

<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<html><head>
<title>404 Not Found</title>
</head><body>
<h1>Not Found</h1>
<p>The requested URL /<?php echo htmlspecialchars($_SERVER['REQUEST_URI']); ?> was not found on this server.</p>
</body></html>

answered Jun 19, 2017 at 2:55

sjdaws

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kingunits · Accepted Answer · 2019-01-16 16:20:44Z

1

full example php embed (tested) is as follows :

if ( $_SESSION['login_user'] !== "admin" ) {

    echo("<!DOCTYPE HTML PUBLIC \"-//IETF//DTD HTML 2.0//EN\">");
    echo("<html><head>");
    echo("<title>404 Not Found</title>");
    echo("</head><body>");
    echo("<h1>Not Found</h1>");
    echo("<p>The requested URL ".htmlspecialchars($_SERVER['REQUEST_URI'])." was not found on this server.</p>"); 
    echo("<hr>");
    echo("<address>".apache_get_version()." Server at ".$_SERVER['SERVER_ADDR']." Port ".$_SERVER['SERVER_PORT']."</address>");
    echo("</body></html>");
    exit;
        }

answered Jan 16, 2019 at 16:20

kingunits

1076 bronze badges

1 Comment

Matix Over a year ago

This is bad practice. You should not echo each individual line. Also echo does not require the brackets.

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