Flask Restful add resource parameters

Question

I am looking to pass an object instance as a parameter into a Flask-RESTfull Resource.

Here is my setup:

# in main.py
from flask import Flask
from flask.ext.restful import Api
from bar import Bar
from foo import views

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

my_bar = Bar()

api.add_resource(views.ApiPage, "/api/my/end/point/")

Then in views.py I have the resource set up as follows:

# In views.py
from flask.ext.restful import Resource

class ApiPage(Resource):
    def get(self):
        serialized = str(my_bar)
        return serialized

So the issue that I am having is that I need to pass my instance of Bar() into the api resource. Is there any way to pass it in through the add_resource method like api.add_resource(views.ApiPage, "/api/my/end/point/", instance=bar)?

Answer 1

Since version 0.3.3 (released May 22, 2015), add_resource() is able to pass parameters to your Resource constructor.

Following the original example, here is the views.py:

from flask.ext.restful import Resource

class ApiPage(Resource):
    def __init__(self, bar):
        self.bar = bar

    def get(self):
        serialized = str(my_bar)
        return serialized

And the relevant code for main.py:

# ...
my_bar = Bar()
api.add_resource(views.ApiPage, '/api/my/end/point/',
                 resource_class_kwargs={'bar': my_bar})

Answer 2

In version 0.3.5 documentation you can do it this way

# in main.py
...
my_bar = Bar()
api.add_resource(views.ApiPage, '/api/my/end/point/',
             resource_class_kwargs={'my_bar': my_bar})

pass your object in the add_resource method and use it in the resource init method

and in your class init

class ApiPage(Resource):
    def __init__(self, **kwargs):
       self.my_bar= kwargs['my_bar']