Flask+nginx+uwsgi: only serve url with nginx if flask doesn't have a route for it










1















nginx config for the server (the main nginx one is the default one on debian 9):



server 
listen 80;

server_name subdomain.domain.com;
include /etc/nginx/mime.types;
location /galleries
autoindex on;
alias /srv/galleries/;

location /
include uwsgi_params;
uwsgi_pass unix:/tmp/scraper.sock;




uwsgi config:



[uwsgi]
module = wsgi:app

master = true
processes = 5

socket = /tmp/scraper.sock
chmod-socket = 777
uid = www-data
gid = www-data
vacuum = true

die-on-term = true
plugins = python3
py-autoreload = 1


If I try creating a route for /galleries/whatever, ie like this:



@app.route("/galleries/whatever")
def test():
return "Hello"


I'll just see the indexed files inside /galleries/whatever through nginx instead of going through flask.



Is there a way for me to force nginx to only handle requests if flask returns 404? Alternatively, is there a better way for me to serve files while still having them available under those urls? Keep in mind the /galleries folder is pretty big and generated by another program.



I run the server with "uwsgi --ini server.ini" and nothing else.










share|improve this question






















  • Maybe using a subdomain name is an idea to overcome this problem?

    – Peshmerge
    Nov 14 '18 at 15:00











  • You mean using a subdomain specifically for the files, having the rest of the path be identical to flask's routes? That's an option, but I was hoping on some simpler solution - I'd like to make my app portable rather than relying on configuring subdomains every time.

    – msaba92
    Nov 14 '18 at 15:44
















1















nginx config for the server (the main nginx one is the default one on debian 9):



server 
listen 80;

server_name subdomain.domain.com;
include /etc/nginx/mime.types;
location /galleries
autoindex on;
alias /srv/galleries/;

location /
include uwsgi_params;
uwsgi_pass unix:/tmp/scraper.sock;




uwsgi config:



[uwsgi]
module = wsgi:app

master = true
processes = 5

socket = /tmp/scraper.sock
chmod-socket = 777
uid = www-data
gid = www-data
vacuum = true

die-on-term = true
plugins = python3
py-autoreload = 1


If I try creating a route for /galleries/whatever, ie like this:



@app.route("/galleries/whatever")
def test():
return "Hello"


I'll just see the indexed files inside /galleries/whatever through nginx instead of going through flask.



Is there a way for me to force nginx to only handle requests if flask returns 404? Alternatively, is there a better way for me to serve files while still having them available under those urls? Keep in mind the /galleries folder is pretty big and generated by another program.



I run the server with "uwsgi --ini server.ini" and nothing else.










share|improve this question






















  • Maybe using a subdomain name is an idea to overcome this problem?

    – Peshmerge
    Nov 14 '18 at 15:00











  • You mean using a subdomain specifically for the files, having the rest of the path be identical to flask's routes? That's an option, but I was hoping on some simpler solution - I'd like to make my app portable rather than relying on configuring subdomains every time.

    – msaba92
    Nov 14 '18 at 15:44














1












1








1








nginx config for the server (the main nginx one is the default one on debian 9):



server 
listen 80;

server_name subdomain.domain.com;
include /etc/nginx/mime.types;
location /galleries
autoindex on;
alias /srv/galleries/;

location /
include uwsgi_params;
uwsgi_pass unix:/tmp/scraper.sock;




uwsgi config:



[uwsgi]
module = wsgi:app

master = true
processes = 5

socket = /tmp/scraper.sock
chmod-socket = 777
uid = www-data
gid = www-data
vacuum = true

die-on-term = true
plugins = python3
py-autoreload = 1


If I try creating a route for /galleries/whatever, ie like this:



@app.route("/galleries/whatever")
def test():
return "Hello"


I'll just see the indexed files inside /galleries/whatever through nginx instead of going through flask.



Is there a way for me to force nginx to only handle requests if flask returns 404? Alternatively, is there a better way for me to serve files while still having them available under those urls? Keep in mind the /galleries folder is pretty big and generated by another program.



I run the server with "uwsgi --ini server.ini" and nothing else.










share|improve this question














nginx config for the server (the main nginx one is the default one on debian 9):



server 
listen 80;

server_name subdomain.domain.com;
include /etc/nginx/mime.types;
location /galleries
autoindex on;
alias /srv/galleries/;

location /
include uwsgi_params;
uwsgi_pass unix:/tmp/scraper.sock;




uwsgi config:



[uwsgi]
module = wsgi:app

master = true
processes = 5

socket = /tmp/scraper.sock
chmod-socket = 777
uid = www-data
gid = www-data
vacuum = true

die-on-term = true
plugins = python3
py-autoreload = 1


If I try creating a route for /galleries/whatever, ie like this:



@app.route("/galleries/whatever")
def test():
return "Hello"


I'll just see the indexed files inside /galleries/whatever through nginx instead of going through flask.



Is there a way for me to force nginx to only handle requests if flask returns 404? Alternatively, is there a better way for me to serve files while still having them available under those urls? Keep in mind the /galleries folder is pretty big and generated by another program.



I run the server with "uwsgi --ini server.ini" and nothing else.







python nginx flask uwsgi






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 13 '18 at 21:52









msaba92msaba92

187110




187110












  • Maybe using a subdomain name is an idea to overcome this problem?

    – Peshmerge
    Nov 14 '18 at 15:00











  • You mean using a subdomain specifically for the files, having the rest of the path be identical to flask's routes? That's an option, but I was hoping on some simpler solution - I'd like to make my app portable rather than relying on configuring subdomains every time.

    – msaba92
    Nov 14 '18 at 15:44


















  • Maybe using a subdomain name is an idea to overcome this problem?

    – Peshmerge
    Nov 14 '18 at 15:00











  • You mean using a subdomain specifically for the files, having the rest of the path be identical to flask's routes? That's an option, but I was hoping on some simpler solution - I'd like to make my app portable rather than relying on configuring subdomains every time.

    – msaba92
    Nov 14 '18 at 15:44

















Maybe using a subdomain name is an idea to overcome this problem?

– Peshmerge
Nov 14 '18 at 15:00





Maybe using a subdomain name is an idea to overcome this problem?

– Peshmerge
Nov 14 '18 at 15:00













You mean using a subdomain specifically for the files, having the rest of the path be identical to flask's routes? That's an option, but I was hoping on some simpler solution - I'd like to make my app portable rather than relying on configuring subdomains every time.

– msaba92
Nov 14 '18 at 15:44






You mean using a subdomain specifically for the files, having the rest of the path be identical to flask's routes? That's an option, but I was hoping on some simpler solution - I'd like to make my app portable rather than relying on configuring subdomains every time.

– msaba92
Nov 14 '18 at 15:44













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