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I am trying to serve all requests with the backend (uwsgi for flask in this case) and if the backend has no such path, nginx should return 404.

Somehow I only can get the backend to take over when I use 

try_files $uri @uwsgi;

But that way, you can access all files in the root directory. I don't want to serve all files in the root directory.

I also tried this:

try_files @uwsgi =404;

But then I get an 404 even if the path is defined in my flask app.

More context: Ubuntu 14.04, nginx 1.4.6

    location @uwsgi {
            include uwsgi_params;
            uwsgi_pass unix:$document_root/myapp.sock;
    }

    location / {
            try_files $uri @uwsgi;
    }

    location /static/ {
            try_files $uri =404;
    }

The flask app is like this:

@app.route('/test')
def hello_world():
    return 'Hello World!'

So when I visit example.com/test I want to have the flask backend handle this. But I don't want nginx to serve all files in the root directory.

Sebastian
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1 Answers1

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Why use try_files if you never have any files you want to try and you always want to pass to your WSGI handler? Did you try simply making uwsgi_pass the root location?

location / {
    include uwsgi_params;
    uwsgi_pass unix:$document_root/myapp.sock;
}

You can combine it with a static location/static server name that serves files out of a regular folder but only for that static URL.

Alexander Ljungberg
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  • Ah yes, you are right, that is a solution to the problem I posted. Sorry I tried to make the problem as minimal as possible but got a bit confused by all the variations I tried. So why does `try_files @uwsgi =404;` is not working? The 404 returned by your solution is showing a different page than the `=404` page by the way. How would I go about returning another status code than 404? – Sebastian Aug 05 '15 at 09:41
  • When you proxy a request, nginx always returns the result just the way the proxied server returns them. So you're getting the 404 page of your flask app. If you want to use a custom 404 page (and not specify it within flask), you can use the "proxy_intercept_errors" directive and an "error_page" directive. – Alexander Ljungberg Aug 06 '15 at 11:27