Move Flask-Restplus Swagger API Docs

With Flask-Restplus <= 0.8.0 you should write:

from flask import Flaskfrom flask.ext.restplus import Api, apidocapp = Flask(__name__)api = Api(app, ui=False)@app.route('/doc/', endpoint='doc')def swagger_ui():    return apidoc.ui_for(api)

Note the use of a @app instead of @api

Starting from v0.8.1 (soon to be released), you will simply have to write:

from flask import Flaskfrom flask.ext.restplus import Api, apidocapp = Flask(__name__)api = Api(app, doc='/doc/')

See: http://flask-restplus.readthedocs.org/en/latest/swagger.html#swagger-ui

After recently struggling with this myself, I've had luck with this approach:

from flask import Flask, Blueprintfrom flask.ext.restplus import Api, apidocapp = Flask(__name__)blueprint = Blueprint('api', __name__)api = Api(blueprint, ui=False)@blueprint.route('/doc/', endpoint='doc')def swagger_ui():   return apidoc.ui_for(api)app.register_blueprint(blueprint)

It looks like @api will need Resource, so I modified the code a bit to get around the error. The following will work only at /doc/, not the default root level.

from flask import Flask, make_responsefrom flask.ext.restplus import Api, apidoc, Resourceapp = Flask(__name__)api = Api(app, ui=False)@api.route('/doc/', endpoint='doc', doc=False)class ApiDoc(Resource):    def get(self):        return make_response(apidoc.ui_for(api))app.register_blueprint(apidoc.apidoc)if __name__ == '__main__':    app.run(debug=True)