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Python Flask REST API Resourceful Blueprints Parse the Request

Andre Hammons
Andre Hammons
9,265 Points

Flask REST API - Parse the Request code challenge

I've been looking at this for a while and I'm stumped. I keep getting "Bummer: Try again." every time I recheck. I've tried excluding the super()__init__() and tried explicitly declaring type=str despite it being the default.

resources/ingredients.py
from flask import Blueprint

from flask.ext.restful import Resource, Api, reqparse, inputs

import models


class IngredientList(Resource):
    def __init__(self):
        self.reqparse = reqparse.ReqParser()
        self.reqparse.add_argument(
            "name",
            required=True,
            location=["form", "json"]
        )
        self.reqparse.add_argument(
            "description,
            required=True,
            location=["form", "json"]
        )
        self.reqparse.add_argument(
            "measurement_type",
            required=True,
            location=["form", "json"]
        )
        self.reqparse.add_argument(
            "quantity",
            required=True,
            location=["form", "json"],
            type=float
        )
        self.reqparse.add_argument(
            "recipe",
            required=True,
            location=["form", "json"],
            type=inputs.positive
        )
        super().__init__()

    def get(self):
        return 'IngredientList'


class Ingredient(Resource):
    def get(self, id):
        return 'Ingredient'

ingredients_api = Blueprint('resources.ingredients', __name__)
api = Api(ingredients_api)
api.add_resource(IngredientList, '/api/v1/ingredients')
api.add_resource(Ingredient, '/api/v1/ingredients/<int:id>')

1 Answer

Andre Hammons
Andre Hammons
9,265 Points

I figured it out. I misspelled reqparse.RequestParser()