Is it possible to directly declare a flask URL optional parameter?

Currently I’m proceeding the following way:

@user.route('/<userId>')
@user.route('/<userId>/<username>')
def show(userId, username=None):
    pass

How can I directly say that username is optional?

Another way is to write

@user.route('/<user_id>', defaults={'username': None})
@user.route('/<user_id>/<username>')
def show(user_id, username):
    pass

But I guess that you want to write a single route and mark username as optional? If that’s the case, I don’t think it’s possible.

Almost the same as Audrius cooked up some months ago, but you might find it a bit more readable with the defaults in the function head – the way you are used to with python:

@app.route('/<user_id>')
@app.route('/<user_id>/<username>')
def show(user_id, username='Anonymous'):
    return user_id + ':' + username

If you are using Flask-Restful like me, it is also possible this way:

api.add_resource(UserAPI, '/<userId>', '/<userId>/<username>', endpoint = 'user')

a then in your Resource class:

class UserAPI(Resource):

  def get(self, userId, username=None):
    pass

@user.route('/<userId>/')  # NEED '/' AFTER LINK
@user.route('/<userId>/<username>')
def show(userId, username=None):
    pass

https://flask.palletsprojects.com/en/1.1.x/quickstart/#unique-urls-redirection-behavior

@app.route('/', defaults={'path': ''})
@app.route('/< path:path >')
def catch_all(path):
    return 'You want path: %s' % path

http://flask.pocoo.org/snippets/57/

@user.route('/<user_id>', defaults={'username': default_value})
@user.route('/<user_id>/<username>')
def show(user_id, username):
   #
   pass

Almost the same as skornos, but with variable declarations for a more explicit answer. It can work with Flask-RESTful extension:

from flask import Flask
from flask_restful import Resource, Api

app = Flask(__name__)
api = Api(app)

class UserAPI(Resource):
    def show(userId, username=None):
    pass

api.add_resource(UserAPI, '/<userId>', '/<userId>/<username>', endpoint='user')

if __name__ == '__main__':
    app.run()

The add_resource method allows pass multiples URLs. Each one will be routed to your Resource.

I know this post is really old but I worked on a package that does this called flask_optional_routes. The code is located at: https://github.com/sudouser2010/flask_optional_routes.

from flask import Flask

from flask_optional_routes import OptionalRoutes


app = Flask(__name__)
optional = OptionalRoutes(app)

@optional.routes('/<user_id>/<user_name>?/')
def foobar(user_id, user_name=None):
    return 'it worked!'

if __name__ == "__main__":
    app.run(host='0.0.0.0', port=5000)

You can write as you show in example, but than you get build-error.

For fix this:

1. print app.url_map () in you root .py
2. you see something like:

<Rule '/<userId>/<username>' (HEAD, POST, OPTIONS, GET) -> user.show_0>

and

<Rule '/<userId>' (HEAD, POST, OPTIONS, GET) -> .show_1>

3. than in template you can {{ url_for('.show_0', args) }} and {{ url_for('.show_1', args) }}

Since Flask 0.10 you can`t add multiple routes to one endpoint. But you can add fake endpoint

@user.route('/<userId>')
def show(userId):
   return show_with_username(userId)

@user.route('/<userId>/<username>')
def show_with_username(userId,username=None):
   pass

I think you can use Blueprint and that’s will make ur code look better and neatly.

example:

from flask import Blueprint

bp = Blueprint(__name__, "example")

@bp.route("/example", methods=["POST"])
def example(self):
   print("example")