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Prepend a prefix to all results from url_for, without making the prefix required by the routes [duplicate]

I have a prefix that I want to add to every route. Right now I add a constant to the route at every definition. Is there a way to do this automatically?
PREFIX = "/abc/123"
#app.route(PREFIX + "/")
def index_page():
return "This is a website about burritos"
#app.route(PREFIX + "/about")
def about_page():
return "This is a website about burritos"

You can put your routes in a blueprint:
bp = Blueprint('burritos', __name__,
template_folder='templates')
#bp.route("/")
def index_page():
return "This is a website about burritos"
#bp.route("/about")
def about_page():
return "This is a website about burritos"
Then you register the blueprint with the application using a prefix:
app = Flask(__name__)
app.register_blueprint(bp, url_prefix='/abc/123')

The answer depends on how you are serving this application.
Sub-mounted inside of another WSGI container
Assuming that you are going to run this application inside of a WSGI container (mod_wsgi, uwsgi, gunicorn, etc); you need to actually mount, at that prefix the application as a sub-part of that WSGI container (anything that speaks WSGI will do) and to set your APPLICATION_ROOT config value to your prefix:
app.config["APPLICATION_ROOT"] = "/abc/123"
#app.route("/")
def index():
return "The URL for this page is {}".format(url_for("index"))
# Will return "The URL for this page is /abc/123/"
Setting the APPLICATION_ROOT config value simply limit Flask's session cookie to that URL prefix. Everything else will be automatically handled for you by Flask and Werkzeug's excellent WSGI handling capabilities.
An example of properly sub-mounting your app
If you are not sure what the first paragraph means, take a look at this example application with Flask mounted inside of it:
from flask import Flask, url_for
from werkzeug.serving import run_simple
from werkzeug.middleware.dispatcher import DispatcherMiddleware
app = Flask(__name__)
app.config['APPLICATION_ROOT'] = '/abc/123'
#app.route('/')
def index():
return 'The URL for this page is {}'.format(url_for('index'))
def simple(env, resp):
resp(b'200 OK', [(b'Content-Type', b'text/plain')])
return [b'Hello WSGI World']
app.wsgi_app = DispatcherMiddleware(simple, {'/abc/123': app.wsgi_app})
if __name__ == '__main__':
app.run('localhost', 5000)
Proxying requests to the app
If, on the other hand, you will be running your Flask application at the root of its WSGI container and proxying requests to it (for example, if it's being FastCGI'd to, or if nginx is proxy_pass-ing requests for a sub-endpoint to your stand-alone uwsgi / gevent server then you can either:
Use a Blueprint, as Miguel points out in his answer.
or use the DispatcherMiddleware from werkzeug (or the PrefixMiddleware from su27's answer) to sub-mount your application in the stand-alone WSGI server you're using. (See An example of properly sub-mounting your app above for the code to use).

You should note that the APPLICATION_ROOT is NOT for this purpose.
All you have to do is to write a middleware to make the following changes:
modify PATH_INFO to handle the prefixed url.
modify SCRIPT_NAME to generate the prefixed url.
Like this:
class PrefixMiddleware(object):
def __init__(self, app, prefix=''):
self.app = app
self.prefix = prefix
def __call__(self, environ, start_response):
if environ['PATH_INFO'].startswith(self.prefix):
environ['PATH_INFO'] = environ['PATH_INFO'][len(self.prefix):]
environ['SCRIPT_NAME'] = self.prefix
return self.app(environ, start_response)
else:
start_response('404', [('Content-Type', 'text/plain')])
return ["This url does not belong to the app.".encode()]
Wrap your app with the middleware, like this:
from flask import Flask, url_for
app = Flask(__name__)
app.debug = True
app.wsgi_app = PrefixMiddleware(app.wsgi_app, prefix='/foo')
#app.route('/bar')
def bar():
return "The URL for this page is {}".format(url_for('bar'))
if __name__ == '__main__':
app.run('0.0.0.0', 9010)
Visit http://localhost:9010/foo/bar,
You will get the right result: The URL for this page is /foo/bar
And don't forget to set the cookie domain if you need to.
This solution is given by Larivact's gist. The APPLICATION_ROOT is not for this job, although it looks like to be. It's really confusing.

This is more of a python answer than a Flask/werkzeug answer; but it's simple and works.
If, like me, you want your application settings (loaded from an .ini file) to also contain the prefix of your Flask application (thus, not to have the value set during deployment, but during runtime), you can opt for the following:
def prefix_route(route_function, prefix='', mask='{0}{1}'):
'''
Defines a new route function with a prefix.
The mask argument is a `format string` formatted with, in that order:
prefix, route
'''
def newroute(route, *args, **kwargs):
'''New function to prefix the route'''
return route_function(mask.format(prefix, route), *args, **kwargs)
return newroute
Arguably, this is somewhat hackish and relies on the fact that the Flask route function requires a route as a first positional argument.
You can use it like this:
app = Flask(__name__)
app.route = prefix_route(app.route, '/your_prefix')
NB: It is worth nothing that it is possible to use a variable in the prefix (for example by setting it to /<prefix>), and then process this prefix in the functions you decorate with your #app.route(...). If you do so, you obviously have to declare the prefix parameter in your decorated function(s). In addition, you might want to check the submitted prefix against some rules, and return a 404 if the check fails. In order to avoid a 404 custom re-implementation, please from werkzeug.exceptions import NotFound and then raise NotFound() if the check fails.

So, I believe that a valid answer to this is: the prefix should be configured in the actual server application that you use when development is completed. Apache, nginx, etc.
However, if you would like this to work during development while running the Flask app in debug, take a look at this gist.
Flask's DispatcherMiddleware to the rescue!
I'll copy the code here for posterity:
"Serve a Flask app on a sub-url during localhost development."
from flask import Flask
APPLICATION_ROOT = '/spam'
app = Flask(__name__)
app.config.from_object(__name__) # I think this adds APPLICATION_ROOT
# to the config - I'm not exactly sure how!
# alternatively:
# app.config['APPLICATION_ROOT'] = APPLICATION_ROOT
#app.route('/')
def index():
return 'Hello, world!'
if __name__ == '__main__':
# Relevant documents:
# http://werkzeug.pocoo.org/docs/middlewares/
# http://flask.pocoo.org/docs/patterns/appdispatch/
from werkzeug.serving import run_simple
from werkzeug.wsgi import DispatcherMiddleware
app.config['DEBUG'] = True
# Load a dummy app at the root URL to give 404 errors.
# Serve app at APPLICATION_ROOT for localhost development.
application = DispatcherMiddleware(Flask('dummy_app'), {
app.config['APPLICATION_ROOT']: app,
})
run_simple('localhost', 5000, application, use_reloader=True)
Now, when running the above code as a standalone Flask app, http://localhost:5000/spam/ will display Hello, world!.
In a comment on another answer, I expressed that I wished to do something like this:
from flask import Flask, Blueprint
# Let's pretend module_blueprint defines a route, '/record/<id>/'
from some_submodule.flask import module_blueprint
app = Flask(__name__)
app.config['APPLICATION_ROOT'] = '/api'
app.register_blueprint(module_blueprint, url_prefix='/some_submodule')
app.run()
# I now would like to be able to get to my route via this url:
# http://host:8080/api/some_submodule/record/1/
Applying DispatcherMiddleware to my contrived example:
from flask import Flask, Blueprint
from flask.serving import run_simple
from flask.wsgi import DispatcherMiddleware
# Let's pretend module_blueprint defines a route, '/record/<id>/'
from some_submodule.flask import module_blueprint
app = Flask(__name__)
app.config['APPLICATION_ROOT'] = '/api'
app.register_blueprint(module_blueprint, url_prefix='/some_submodule')
application = DispatcherMiddleware(Flask('dummy_app'), {
app.config['APPLICATION_ROOT']: app
})
run_simple('localhost', 5000, application, use_reloader=True)
# Now, this url works!
# http://host:8080/api/some_submodule/record/1/

Another completely different way is with mountpoints in uwsgi.
From the doc about Hosting multiple apps in the same process (permalink).
In your uwsgi.ini you add
[uwsgi]
mount = /foo=main.py
manage-script-name = true
# also stuff which is not relevant for this, but included for completeness sake:
module = main
callable = app
socket = /tmp/uwsgi.sock
If you don't call your file main.py, you need to change both the mount and the module
Your main.py could look like this:
from flask import Flask, url_for
app = Flask(__name__)
#app.route('/bar')
def bar():
return "The URL for this page is {}".format(url_for('bar'))
# end def
And a nginx config (again for completeness):
server {
listen 80;
server_name example.com
location /foo {
include uwsgi_params;
uwsgi_pass unix:///temp/uwsgi.sock;
}
}
Now calling example.com/foo/bar will display /foo/bar as returned by flask's url_for('bar'), as it adapts automatically. That way your links will work without prefix problems.

from flask import Flask
app = Flask(__name__)
app.register_blueprint(bp, url_prefix='/abc/123')
if __name__ == "__main__":
app.run(debug='True', port=4444)
bp = Blueprint('burritos', __name__,
template_folder='templates')
#bp.route('/')
def test():
return "success"

I needed similar so called "context-root". I did it in conf file under /etc/httpd/conf.d/ using WSGIScriptAlias :
myapp.conf:
<VirtualHost *:80>
WSGIScriptAlias /myapp /home/<myid>/myapp/wsgi.py
<Directory /home/<myid>/myapp>
Order deny,allow
Allow from all
</Directory>
</VirtualHost>
So now I can access my app as : http://localhost:5000/myapp
See the guide - http://modwsgi.readthedocs.io/en/develop/user-guides/quick-configuration-guide.html

My solution where flask and PHP apps coexist
nginx and PHP5.6
KEEP Flask in root and PHP in subdirectories
sudo vi /etc/php/5.6/fpm/php.ini
Add 1 line
cgi.fix_pathinfo=0
sudo vi /etc/php/5.6/fpm/pool.d/www.conf
listen = /run/php/php5.6-fpm.sock
uwsgi
sudo vi /etc/nginx/sites-available/default
USE NESTED LOCATIONS for PHP and let FLASK remain in root
server {
listen 80 default_server;
listen [::]:80 default_server;
# SSL configuration
#
# listen 443 ssl default_server;
# listen [::]:443 ssl default_server;
#
# Note: You should disable gzip for SSL traffic.
# See: https://bugs.debian.org/773332
#
# Read up on ssl_ciphers to ensure a secure configuration.
# See: https://bugs.debian.org/765782
#
# Self signed certs generated by the ssl-cert package
# Don't use them in a production server!
#
# include snippets/snakeoil.conf;
root /var/www/html;
# Add index.php to the list if you are using PHP
index index.html index.htm index.php index.nginx-debian.html;
server_name _;
# Serve a static file (ex. favico) outside static dir.
location = /favico.ico {
root /var/www/html/favico.ico;
}
# Proxying connections to application servers
location / {
include uwsgi_params;
uwsgi_pass 127.0.0.1:5000;
}
location /pcdp {
location ~* \.php$ {
try_files $uri =404;
fastcgi_split_path_info ^(.+\.php)(/.+)$;
fastcgi_pass unix:/var/run/php/php5.6-fpm.sock;
fastcgi_index index.php;
fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
include fastcgi_params;
}
}
location /phpmyadmin {
location ~* \.php$ {
try_files $uri =404;
fastcgi_split_path_info ^(.+\.php)(/.+)$;
fastcgi_pass unix:/var/run/php/php5.6-fpm.sock;
fastcgi_index index.php;
fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
include fastcgi_params;
}
}
# pass the PHP scripts to FastCGI server listening on 127.0.0.1:9000
#
#location ~ \.php$ {
# include snippets/fastcgi-php.conf;
#
# # With php7.0-cgi alone:
# fastcgi_pass 127.0.0.1:9000;
# # With php7.0-fpm:
# fastcgi_pass unix:/run/php/php7.0-fpm.sock;
#}
# deny access to .htaccess files, if Apache's document root
# concurs with nginx's one
#
#location ~ /\.ht {
# deny all;
#}
}
READ carefully
https://www.digitalocean.com/community/tutorials/understanding-nginx-server-and-location-block-selection-algorithms
We need to understand location matching
(none): If no modifiers are present, the location is interpreted as a prefix match. This means that the location given will be matched against the beginning of the request URI to determine a match.
=: If an equal sign is used, this block will be considered a match if the request URI exactly matches the location given.
~: If a tilde modifier is present, this location will be interpreted as a case-sensitive regular expression match.
~*: If a tilde and asterisk modifier is used, the location block will be interpreted as a case-insensitive regular expression match.
^~: If a carat and tilde modifier is present, and if this block is selected as the best non-regular expression match, regular expression matching will not take place.
Order is important, from nginx's "location" description:
To find location matching a given request, nginx first checks locations defined using the prefix strings (prefix locations). Among them, the location with the longest matching prefix is selected and remembered. Then regular expressions are checked, in the order of their appearance in the configuration file. The search of regular expressions terminates on the first match, and the corresponding configuration is used. If no match with a regular expression is found then the configuration of the prefix location remembered earlier is used.
It means:
First =. ("longest matching prefix" match)
Then implicit ones. ("longest matching prefix" match)
Then regex. (first match)

For people still struggling with this, the first example does work, but the full example is here if you have a Flask app that is not under your control:
from os import getenv
from werkzeug.middleware.dispatcher import DispatcherMiddleware
from werkzeug.serving import run_simple
from custom_app import app
application = DispatcherMiddleware(
app, {getenv("REBROW_BASEURL", "/rebrow"): app}
)
if __name__ == "__main__":
run_simple(
"0.0.0.0",
int(getenv("REBROW_PORT", "5001")),
application,
use_debugger=False,
threaded=True,
)

In flask blueprint, we can use -
app = Flask(__name__)
app.config['APPLICATION_ROOT'] = '/prefix-text'
Anyone looking to do in flask-restful can make use of -
doc link
app = Flask(__name__)
api = Api(app, prefix='/pefix-text')
Now, all your routes will be prefixed with /prefix-text. Just make sure you use url_for('link') in places where you might have simply used a /link.

I think su27's answer is right. And I am using gevent, here is my code and it works fine:
from gevent import pywsgi
# your flask code ...
# app = Flask(__name__)
if __name__ == "__main__":
class MyHandler(pywsgi.WSGIHandler):
def get_environ(self):
prefix = "/your_prefix"
env = super().get_environ()
if env['PATH_INFO'].startswith(prefix):
env['PATH_INFO'] = env['PATH_INFO'][len(prefix):]
env['SCRIPT_NAME'] = prefix
return env
server = pywsgi.WSGIServer(('', 8080), app, handler_class=MyHandler)
server.serve_forever()

From all the answers I have seen above, they are either too simplistic or over complicating.
That said, I like to accomplish it using nested blueprints:
from .blueprints import blueprint1, blueprint2, blueprint3, etc
app = Flask(__name__)
url_prefix = "/abc/123"
parent = Blueprint('index', __name__, url_prefix=url_prefix)
index.register_blueprint(blueprint1)
index.register_blueprint(blueprint2)
index.register_blueprint(blueprint3)
app.register_blueprint(index)
This way, you basically link your child blueprints to a parent blueprint, where you define the prefix. This is documented here.
With your example, you would simply rewrite it to:
blueprint1 = Blueprint('blueprint1', __name__)
#blueprint1.route("/")
def index_page():
return "Index page"
#blueprint1.route("/about")
def about_page():
return "About page"

If your purpose is to add the prefix in some way,
take a look at the answer https://stackoverflow.com/a/73883005/553095
and https://github.com/mskimm/prefixed-superset

If you want to handle the prefix when using Nginx as a reverse proxy. Werkzeug's ProxyFix middleware will be a simpler solution:
from werkzeug.middleware.proxy_fix import ProxyFix
from flask import Flask
app = Flask(__name__)
app.wsgi_app = ProxyFix(
app.wsgi_app, x_for=1, x_proto=1, x_host=1, x_prefix=1
)
Werkzeug will read the X-Forwarded-Prefix header and set it to SCRIPT_NAME. So be sure to set the X-Forwarded-Prefix header in Nginx config:
server {
listen 80;
server_name _;
location /api {
proxy_pass http://127.0.0.1:5000/;
proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;
proxy_set_header X-Forwarded-Proto $scheme;
proxy_set_header X-Forwarded-Host $host;
proxy_set_header X-Forwarded-Prefix /api;
}
}
See more details at https://stackoverflow.com/a/75123044/5511849

Unable to render create-react-app boilerplate views from Flask backend

I'm having a hard time integrating create-react-app single page application to my flask backend. I want to be able to make a fetch/axios call from my front end like so: axios.get('/getResults') and fetch('/getResults'). Some things I have tried but not limited to is specifying the Flask port as 3000 which is the same used by create-react-app. Also, used the proxy configuration feature on the "package.json" file of create-react-app but to no avail. I suspect my folder structure and Flask code implementation may likely be causing this. Below is my folder structure and "app.py" code. Any help I could get will be appreciated. I can provide additional information if necessary. Thanks
Project -build(contains static folder, index.html...Other meta files)-node_modules-public-srcapp.pypackage.jsonrequirements.txt
app.py:
from flask import Flask, Response, request, jsonify, make_response, send_from_directory,render_template
app = Flask(__name__, static_path='/build/static/')
app.debug=True
#app.route('/')
def root():
print('Inside root function')
return app.send_static_file('index.html')
#app.route('/getResults', methods=["GET"])
def results():
print('Inside getResults path')
return app.send_static_file('index.html')
#app.route('/postData', methods=["POST"])
def data_results():
print('Inside postData path')
data = request.get_json
return jsonify(data)
#app.route('/<path:path>')
def send_js(path):
print("inside send_js fxn")
return send_from_directory('./build/static',path)
if __name__ == "__main__":
print("inside main host call")
app.run(host='0.0.0.0', port=3000)
Errors I get when I run "python app.py" are:
On the terminal: Inside root function
127.0.0.1 - - [12/Jun/2017 09:42:24] "GET / HTTP/1.1" 404 -
On the browser:Not Found - The requested URL was not found on the server. If you entered the URL manually please check your spelling and try again.

I was having the exact same issue and I was able to solve it by appeasing Flask with symlinks.
Keep the templates and static directory paths at their defaults and in the directory with your Flask main file, add these symlinks
ln -s build/ templates
ln -s build/static static
In case you were curious, this was my specific problem, which just involved a few more nested directories but was in essence the same:
Running NPM Build from Flask
You can then use Nurzhan's root configuration:
#app.route('/')
def root():
print('Inside root function')
return render_template('index.html')
But you only require your app declaration to be: app = Flask(__name__)
The only thing that doesn't work for me is the favicon, and I will update this answer once I figure that out.

In development mode, you need to configure your create-react-app package.json to forward "ajax" request to the flask server.
Here is what my package.json looks like:
{
"name": "socialite",
"version": "0.1.0",
"private": true,
"proxy": "http://localhost:8080",
"devDependencies": {
"react-scripts": "1.0.10"
},
"dependencies": {
"react": "^15.6.1",
"react-dom": "^15.6.1"
},
"scripts": {
"start": "react-scripts start",
"build": "react-scripts build",
"test": "react-scripts test --env=jsdom",
"eject": "react-scripts eject"
}
}
See the proxy field? That's where the magic happens, replace its value with the flask server address. That way, you can take advantage of CRA hot reloading feature. This is documented at in create-react-app as "Proxying API Requests in Development"
Then do run your application, you go at localhost:3000 or whatever port yarn opens for you. And when you do an API call in javascript over the rainbow to the server for instance: fetch('/api/model/') or something nodejs' server will forward to the flask app. I think the nodejs server does look at the content-type field of the ajax request to know whether it should forward the request to the backend server or not.
I recommend you prefix all your backend routes with something like /api/v1/ or something so the nginx configuration is neat and easy to write.

I think you have a number of misunderstandings.
The create-react-app runs on its own server on port 3000 and if you try to run your flask app on the same port on the same machine it will complain that port 3000 is already in use. So from this we move to another question - the structure of your application.
Will it be a separate reactjs based client on the frontend and api based on flask in the backend which will be 2 separate applications communicating with each other over HTTP? In this case the frontend and backend will usually run on separate servers.
Or it will one flask application which will use reactjs in its template pages?
You can fix your current problem with not finding URL by changing to this in your code:
#app.route('/')
def root():
print('Inside root function')
return render_template('index.html')
And this:
template_dir = os.path.abspath('build/templates')
app = Flask(__name__, static_path='/build/static/',
template_folder=template_dir)
Since your templates folder is in the build directory.

python flask redirect to https from http

I have a website build using python3.4 and flask...I have generated my own self-signed certificate and I am currently testing my website through localhost.
I am using the python ssl module along with this flask extension: https://github.com/kennethreitz/flask-sslify
context = ('my-cert.pem', 'my-key.pem')
app = Flask(__name__)
sslify = SSLify(app)
...
if __name__ == '__main__':
app.debug = False
app.run(
host="127.0.0.1",
port=int("5000"),
ssl_context=context
)
This does not seem to be working however. I took a look in the sslify source code and this line does not seem to be working
def init_app(self, app):
"""Configures the configured Flask app to enforce SSL."""
app.before_request(self.redirect_to_ssl)
app.after_request(self.set_hsts_header)
Specifically the function call to redirect_to_ssl (I added my own print statement under the redirect_to_ssl function and my statement was never printed)
def redirect_to_ssl(self):
print("THIS IS WORKING")
"""Redirect incoming requests to HTTPS."""
Should we redirect?
criteria = [
request.is_secure,
current_app.debug,
request.headers.get('X-Forwarded-Proto', 'http') == 'https'
]
if not any(criteria) and not self.skip:
if request.url.startswith('http://'):
url = request.url.replace('http://', 'https://', 1)
code = 302
if self.permanent:
code = 301
r = redirect(url, code=code)
return r
I am pretty new to python. Any ideas?

To me, it appears you're making it more complicated than it needs to be. Here is the code I use in my views.py script to force user to HTTPS connections:
#app.before_request
def before_request():
if not request.is_secure:
url = request.url.replace('http://', 'https://', 1)
code = 301
return redirect(url, code=code)

According with the docs, after pip install Flask-SSLify you only need to insert the following code:
from flask import Flask
from flask_sslify import SSLify
app = Flask(__name__)
sslify = SSLify(app)
I have done it and it works very well. Am I missing something in the discussion ?

The Flask Security Guide recommends using Flask-Talisman.
$ pip install flask-talisman
Usage example:
from flask import Flask
from flask_talisman import Talisman
app = Flask(__name__)
Talisman(app)
It forces HTTPS by default (from the README):
force_https, default True, forces all non-debug connects to https.
Personally, I got some errors relating to CSP (Content Security Policy) which I disabled with:
Talisman(app, content_security_policy=None)
But use this at your own risk :)

Thanks to answer from Kelly Keller-Heikkila and comment by jaysqrd I ended up doing this in my Flask app:
from flask import request, redirect
...
#app.before_request
def before_request():
if app.env == "development":
return
if request.is_secure:
return
url = request.url.replace("http://", "https://", 1)
code = 301
return redirect(url, code=code)
I tried the flask_sslify solution suggested by Rodolfo Alvarez but ran into this issue and went with the above solution instead.
If the app is running in development mode or the request is already on https there's no need to redirect.

Here is a flask solution if you are on aws and behind a load balancer. Place it in your views.py
#app.before_request
def before_request():
scheme = request.headers.get('X-Forwarded-Proto')
if scheme and scheme == 'http' and request.url.startswith('http://'):
url = request.url.replace('http://', 'https://', 1)
code = 301
return redirect(url, code=code)

The standard solution is to wrap the request with an enforce_ssl decorator that after checking some flags in the app configuration (flags you can set depending on your debugging needs) modifies the request's url with request.url.
As it is written here.
You can modify the code to make it working with before_request as suggested by #kelly-keller-heikkila

I use a simple extra app that runs on port 80 and redirect people to https:
from flask import Flask,redirect
app = Flask(__name__)
#app.route('/')
def hello():
return redirect("https://example.com", code=302)
if __name__ == '__main__':
app.run(host='0.0.0.0', port=80)

An alternative to the other answers that I've been able to use with great success:
from http import HTTPStatus
from typing import Optional
from flask import Response, redirect, request, url_for
def https_redirect() -> Optional[Response]:
if request.scheme == 'http':
return redirect(url_for(request.endpoint,
_scheme='https',
_external=True),
HTTPStatus.PERMANENT_REDIRECT)
# ..
if app.env == 'production':
app.before_request(https_redirect)
# ..

On app engine flex, add:
from werkzeug.middleware.proxy_fix import ProxyFix
def create_app(config=None):
app = Flask(__name__)
app.wsgi_app = ProxyFix(app.wsgi_app)
In addition to the solution of:
#app.before_request
def before_request():
if not request.is_secure:
url = request.url.replace('http://', 'https://', 1)
code = 301
return redirect(url, code=code)
Otherwise it'll cause infinite redirects since SSL is unwrapped behind the proxy but is noted in the headers.

I ran into the same solution running a Flask application in AWS Elastic Beanstalk behind a load balancer. The following AWS docs provided two steps to configure the environment for http redirects: https://docs.aws.amazon.com/elasticbeanstalk/latest/dg/configuring-https-httpredirect.html Following both steps fixed my issue.
One thing to note is that you'll have to create the .ebextenions folder at the root level of your application source bundle and add the config file to that .ebextensions folder. The readme here: https://github.com/awsdocs/elastic-beanstalk-samples explains this in a bit more detail.

For some reason it seems, requests from a Private AWS API Gateway with a VPC endpoint don't include the "X-Forwarded-Proto" header. This can break some of the other solutions (either it doesn't work or it continuously redirects to the same url). The following middleware forces https on most flask generated internal redirects:
class ForceHttpsRedirects:
def __init__(self, app):
self.app = app
def __call__(self, environ, start_response):
environ["wsgi.url_scheme"] = "https"
return self.app(environ, start_response)
# Usage
app = flask.Flask(__name__)
app.wsgi_app = ForceHttpsRedirects(app.wsgi_app) # Add middleware to force all redirects to https

Use:
app.run(port="443")
All modern browsers automatically use HTTPS when the port is 443 or 8443.

I'm using cloud foundry python app which is behind a load balancer (like https://stackoverflow.com/users/5270172/kelly-keller-heikkila said) .
This resolution helped me by adding (_external and _Scheme to the url_for function). https://github.com/pallets/flask/issues/773

I had the same issue and mine is a brute-force solution, but it works.
Heroku in the past suggested flask_sslify, which is not maintained anymore. Nowadays the proper way in Flask should be flask-talisman, but I tried it and it has bad interactions with boostrap templates.
I tried the anulaibar solution but it did not always worked for me.
The following is what I came up with:
#app.before_request
def before_request():
# If the request is sicure it should already be https, so no need to redirect
if not request.is_secure:
currentUrl = request.url
if currentUrl.startswith('http://'):
# http://example.com -> https://example.com
# http://www.example.com -> https://www.example.com
redirectUrl = currentUrl.replace('http://', 'https://', 1)
elif currentUrl.startswith('www'):
# Here we redirect the case in which the user access the site without typing any http or https
# www.example.com -> https://www.example.com
redirectUrl = currentUrl.replace('www', 'https://www', 1)
else:
# I do not now when this may happen, just for safety
redirectUrl = 'https://www.example.com'
code = 301
return redirect(redirectUrl, code=code)
I have the domain registered in godaddy which is also redirecting to https://www.example.com.

In my case Flask app is sitting behind AWS API Gateway and solutions with #app.before_request were giving me permanent redirects.
The following simple solution finally worked:
#app.after_request
def adjust_response(response):
....
if response.location:
if app.env != "development":
response.location = response.location.replace("http://", "https://", 1)
return response

Unable to link Flask module to nginx through uwsgi

I'm sure I am missing something simple here. I can get PART of a python/flask script to be available through nginx, but the important bits are just not working. Here is my python:
#!/usr/bin/python
import flask
from flask import Flask, jsonify, render_template, request
import os
app = flask.Flask(__name__)
app.config['SERVER_NAME']='localhost'
#app.route("/stuff")
def index():
return render_template('index.html')
#app.errorhandler(404)
def page_not_found(e):
return render_template('404.html'), 404
def application(environ, start_response):
start_response("200 OK", [("Content-Type", "text/plain")])
return ["Hello World!"]
Here is my uwsgi start up:
sudo -u www-data uwsgi -s /tmp/myApp.sock --module MyApp
The socket is correctly linked and available to nginx.
And here is my nginx config snippet:
location /test {
uwsgi_pass unix:/tmp/myApp.sock;
include uwsgi_params;
}
When I go to myserver/test I get the "Hello World!" like I Would expect. But when I go to myserver/test/stuff, I ALSO get "Hello World!" rather than the contents of my index.html(Which is valid, I use it elsewhere). And if I enter myserver/test/garbage.html, I get a generic nginx 404, rather than my custom one.
Can anyone point me in a direction?
Thanks
--edit--
Thank you for the answer, it does help, but does not solve my entire issue.
Adding "--callable app" to my uwsgi startup line DOES link the uwsgi server to nginx. Yah! I can confirm this by the fact that my customized 404 file IS being returned by nginx.
But that 404 is ALL I can get. I cannot get to my index.html, which is present in the same directory as the 404.html, has the same owner, and the same rights. It is almost the same file really with slightly different text.
This MAY be a problem with expectations. I am expecting to find my index.html at http://(myserver)/test/stuff But I get a 404.
Am I looking in the wrong place? Or is there something off in my flask, uwsgi, or nginx?
Thanks

You're application function does not call your flask app, which is why every route returns "Hello World", 200. I'm pretty sure you have two easy options.
The first is to drop the application function and replace it with application = app.
The second would be to change the uwsgi line to
sudo -u www-data uwsgi -s /tmp/myApp.sock --module MyApp --callable app
which renders your application function irrelevant anyway.
You can read more about using uwsgi here.
edit
As far as my knowledge about nginx goes, your nginx config should look like this
location = /test { rewrite ^ /test/; }
location /test { try_files $uri #test; }
location #test {
include uwsgi_params;
uwsgi_param SCRIPT_NAME /test;
uwsgi_modifier1 30;
uwsgi_pass unix:/tmp/myApp.sock;
}
It is the same as the recommended one on the linked uwsgi page above. You seem to be running not on the root url, so the basic config will not work.

Creating a dynamic redirect by using any of Flask, Pyramid, or Bottle?

I want to create a webapp that dynamically redirects to a URL, based on address that user typed. When a user visit my site by a address like this:
http://mydomain1.com/a1b2c3d4
I want redirect this user to URL:
http://mydomain2.com/register.php?id=a1b2c3d4&from=mydomain1.com

Yay, I love a good fight!
from pyramid.config import Configurator
from pyramid.httpexceptions import HTTPFound
from paste.httpserver import serve
config = Configurator()
config.add_route('redirect', '/{arg}')
def redirect_view(request):
dst = 'http://mydomain2.com/register.php?id={id}&from={host}'
args = {
'id': request.matchdict['arg'],
'host': request.host,
}
return HTTPFound(dst.format(**args))
config.add_view(redirect_view, route_name='redirect')
serve(config.make_wsgi_app(), host='0.0.0.0', port=80)

Here goes my attempt, I'm almost newbie in flask, so it should have room to improve
from flask import Flask, redirect, request
app = Flask(__name__)
host = 'domain2.org'
#app.route('/<path>')
def redirection(path):
return redirect('http://'+host+'/register.php?id='+path+'&from='+request.host)
if __name__ == '__main__':
app.run()
Edited to add the host to the from parameter

My solution was to use a Werkzeug rule using the path type :
host = 'domain2.org'
#app.route('/<path:path>')
def redirection(path):
return redirect('http://%s/%s' % (host, path), code=301)
This can be useful if you move a site and want another site instead with redirection on others pages.

There's a pyramid_rewrite extension (https://pypi.python.org/pypi/pyramid_rewrite/) that looks unmaintained, but seems to work. I had a use case it didn't handle, though: using Configure.include() with the route_prefix parameter.
It occurred to me that the usual approach is to do URL rewrites in the server, and I was using a WSGI server from the Python standard library. How hard could it be?
Make a custom request handler class:
from wsgiref.simple_server import make_server, WSGIRequestHandler
class MyReqHandler(WSGIRequestHandler):
def get_environ(self):
env = WSGIRequestHandler.get_environ(self)
if env['PATH_INFO'].startswith('/foo'):
env['PATH_INFO'] = env['PATH_INFO'].replace('foo', 'bar', 1)
return env
Pass it to make_server() when creating your server:
srvr = make_server('0.0.0.0', 6543, app, handler_class=MyReqHandler)
It works!
Straight-up substitution is all I needed for the problem at hand. Extending it to use regular expressions and exposing it via a nice API would be pretty straightforward.
I have another solution, that is straight-up pyramid, so it will work with some other wsgi server:
from pyramid.events import NewRequest, subscriber
#subscriber(NewRequest)
def mysubscriber(event):
req = event.request
if req.path_info.startswith('/~cfuller'):
req.path_info = req.path_info.replace('foo', 'bar', 1)
That's the declarative way, and it requires a config.scan(). Imperitively, you'd do something like
config.add_subscriber(mysubscriber, NewRequest)
See http://docs.pylonsproject.org/projects/pyramid/en/1.5-branch/narr/events.html for the skinny on events.