Return file with status code from remote server in Nginx

Question

I am trying to set up my SPA to return 404 for invalid resources by having nginx do a backcheck for the resource before responding. All this is to avoid invalid pages coming with 200 and thus having spiders crawling soft 404 pages.

My desired behaviour

Browser Nginx API | -- GET myapp.com/users/12 --> | | | | -- GET myapi:8000/users/12 | | | <----------- 404 -------- | | <-- 404 index.html ------------ | | | 

I expect my configuration to be something like below

server { listen 80; server_name localhost; root /usr/share/nginx/html; index index.html; location /users/ { # 1. Fetch from myapi:8000 # 2. Return index.html with status from fetched URI } } 

I've been digging around for a while now but I'm a bit lost on how to achieve this. I've seen many simple examples out there with try_files etc. but nothing that seems to fit my case, since it seems most examples do very simplistic forwarding.

How can I achieve the above behaviour?

Answer 1

I managed to solve this by having my backend API to always respond with index.html for Accept: text/html. I did try the Lua way with OpenResty but it felt too hacky and quirky and too complex to maintain.

Then my config looks like below:

 location / { try_files $uri $uri/ @forward_to_api; } location @forward_to_api { rewrite ^(.*) /api$1 break; proxy_set_header "Accept" "text/html"; proxy_pass http://localhost:8000; # Follow any redirects proxy_intercept_errors on; error_page 301 302 307 = @follow_redirect; } location @follow_redirect { set $moved_location $upstream_http_location; proxy_pass http://localhost:8000$moved_location; } 

This will redirect all $uri to localhost:8000/api/$uri and follow redirects.