Question

I have a server (nginx) which is routing some urls to static directories, and others to a web.py wsgi app. The server’s config also defines a 404 page, which is displayed when somebody tries to visit a nonexistant file within a static directory.

However, sometimes a user will visit a url which nginx passes to python, and only then do I realize it should 404. At this point web.py has no problem returning a custom error page along with the correct status code; the problem is that I want it to return the same 404 page as the server does for pages that aren’t sent to python. Currently python is manually reading and printing the 404.html file, but that seems very inefficient.

So, is there any way for python to tell the server to display its own 404 page?

Here is my current nginx.conf, with commented lines removed.

worker_processes  1;    

events {
    worker_connections  1024;
}

http {
    include            mime.types;
    default_type       application/octet-stream;
    sendfile           on;
    keepalive_timeout  65;

    server {
        listen       80;
        server_name  light.info;
        root         /www/example.com/html;
        index        index.html;

        location /s/     {}
        location /error/ {}
        location /       {
            fastcgi_pass   127.0.0.1:9001;
            fastcgi_param  REQUEST_URI     $request_uri;
            fastcgi_param  QUERY_STRING    $query_string;
            fastcgi_param  REQUEST_METHOD  $request_method;
            fastcgi_param  REMOTE_ADDR     $remote_addr;
        }

        error_page  404              /error/404.html;
        error_page  403              /error/403.html;
        error_page  500 502 503 504  /error/50x.html;
    }
}

Answer 1

You aren’t going to have python tell nginx what to do per se, you’re going to have nginx intercept the 404’s.

uwsgi_intercept_errors on; is what you are looking for I think assuming you’re using uwsgi. It would be helpful if you could post your current configuration…