Add a prefix to all Flask routes

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.wsgi 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:

1. modify PATH_INFO to handle the prefixed url.
2. 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.