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Accepted
alexus
Asked: 2014-11-22 10:03:54 +0800 CST

return error 410 for location regex? in nginx

  • 772

I'm using nginx-1.6.2,2 and I'd like return error 410 for URL that matches /browse.php?u=http, so requests such as this will get 410:

162.218.208.16 - - [21/Nov/2014:12:35:40 -0500] "GET /browse.php?u=http%3A%2F%2Fwww.bing.com%2Fsearch%3Fq%3DBroke%2C%2BUSA%2Bfiletype%3Apdf%26first%3D0%26count%3D20%26format%3Drss&b=156&f=norefer HTTP/1.1" 403 570 "http://ww.thespacesurf.com/browse.php?u=http%3A%2F%2Fwww.bing.com%2Fsearch%3Fq%3DBroke%2C%2BUSA%2Bfiletype%3Apdf%26first%3D0%26count%3D20%26format%3Drss&b=156&f=norefer" "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.0.3705; .NET CLR 1.1.4322)"

I don't really need regex, so something like this, but it's not working:

location = /browse.php?u=http {
        return 410;
}

couple of days later, I grep -c for 410 in /var/log/nginx-access.log

$ bzip2 -cd /var/log/nginx-access.log.0.bz2 | grep -c ' 410 '
5665
$

and that made me feel warm and fuzzy) thanks again!

nginx

2 Answers

  1. You can't match the query string in the declaration of a location statement so you need something like this :

    location = /browse.php {
    if ($arg_u ~ "^http") { 
        return 410;
    }
}
    • 4
  2. Best Answer

    That's because location is /browse.php, not the one you specified.

    There are several ways to do it. For example, something like this should do:

    location = /browse.php {
        if ($query_string ~ ^u=http) {
                return 410;
        }
}

    P.S.: or use ~* instead of ~ if u=http is case-insensitive.

    • 0