PHP: What's the best way to check equality of $_SERVER['HTTP_REFERER'] ?

Question

I have a PHP script that checks the HTTP Referer.

if ($_SERVER['HTTP_REFERER'] == 'http://www.example.com/') {...}

However, this seems inherintly unsafe ... because what happens if the user goes to 'http://example.com/' or 'http://www.ExaMple.com' (both of which don't match the equality test).

Question: what's a better equality test to ensure that the HTTP Referer is coming from 'example.com' ?

Answer 1

Obligatory response: HTTP_REFERER can be spoofed so there is no way to be 100% sure anyone came from a specific website.

However if you do want to rely on it you can use a regex to look for "example.com" in the HTTP_REFERER. stristr() would also work, and probably would be recommended since it would be faster then a regex. It's also case insensitive so it would match "ExaMple.com" as well as 'example.com".

Answer 2

Hope you don't check it for anything significant.

you can use parse_url() function to get hostname part and to check if HTTP_REFERER contain actually an url.
you can also just substr www. from hostname.
character case is not important as it's always lowercase

or use a regexp.

Answer 3

if (strtolower($_SERVER['HTTP_REFERER']) == 'http://www.example.com/') {...}

How about...

$parts = explode('/',$_SERVER['HTTP_REFERER']);
if (in_array('example.com',$parts) || in_array('www.example.com',$parts)) {...}

Answer 4

parse_url() combined with a bit of string juggling should do what you want. Try this:

$url = parse_url($_SERVER['HTTP_REFERER']);
//take the last two 'dot segments' of the host
$hostOnly = implode('.',array_slice(explode('.',$url['host']),-2));
if (strtolower($hostOnly) == 'example.com') {
    //stuff
}

Note that parse_url() can fail on badly formed URLs, so you might want to add some error checking to be safe. HTTP_REFERER could easily be filled with junk.

Answer 5

Using a regex, this would become

if(preg_match("%(http://)?(www\.)?example\.com(/)?%i",$_SERVER['HTTP_REFERER'])) { ... }