Question: optional/wildcard url parameter

[bastianb]

Hi,
I would like to have a route with a param in it but not having to have to specify it in the method level, like:

Class ComponentList(Resource):
    def get(self):  # note that there is no some_random_string param here
        ...
api.add_resource(ComponentList, '/api/<some_random_string>/components')

For the moment I couldn't find any keyword in the code, * doesn't work either. Is it possible?

[HomiGrotas]

Hi @bastianb , fortunately there isn't an option to do so.
I would suggest to create a middleware that catches this endpoint and then returns your resource response:

from flask_restful import Api, Resource 
from flask import Flask, request


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


class ComponentList(Resource):
    def get(self):  # note that there is no some_random_string param here
        ...
        return "my_response"


@app.route('/api/<string:ignored>/components')
def middleware(ignored):
    method = request.method.lower()
    if hasattr(ComponentList, method):
        return getattr(ComponentList(), method)()


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

Notice that if you need a scaleable solution you can use a class decorator