I'm working on the redesign of an API with Flask using Flask-restx, but I've a problem: We need a legacy version of the API that accepts the old URLs, for compatibility reasons, but I'm not understanding how to do this since Flask-restx requires a namespace to be declared. Urls should be something like this: {{host}}/api/v1/art/savegallery <- new one {{host}}/savegallery <- legacy In Flask I've something like this: app/__init__.py db = SQLAlchemy() migrate = Migrate() cors = CORS() def create_app(config_class=DevelopmentConfig): app = Flask(__name__) app.config.from_object(config_class) db.init_app(app=app) migrate.init_app(app=app, db=db) cors.init_app(app=app) from app.api import api_bp, legacy_bp app.register_blueprint(api_bp, url_prefix='/api/v1') app.register_blueprint(legacy_bp) return app /app/api/__init__.py api_bp = Blueprint('v1', __name__) legacy_bp = Blueprint('legacy', __name__) api_v1 = Api( app=api_bp, version='1.00', title='Art', description=( "API" ), ) api_lgc = Api( app=legacy_bp, version='1.00', title='Art Legacy', description=( "API Legacy" ), ) from app.art.routes import art_ns api_v1.add_namespace(art_ns) api_lgc.add_namespace(art_ns) app/art/routes.py art_ns = Namespace(name='art', description='Art Storage') #artlegacy_ns = Namespace(name='legacy', description='Art Storage') @art_ns.route('/savegallery') class GalleryAPI(Resource): def get(self): try: #data = request.json data = {} return {"foo":"bar"}, 200 except Exception as e: print(e) return {"error": "Something happened"}, 500 With this, I can access {{host}}/api/v1/art/savegallery correctly, but I'm not finding a way to declare the legacy one, since creating a URL this way would require at least the namespace part of the URL. Does Flask-restx have a way to declare those URLs and/or redirect the flow to the new ones? Continue reading...