flask restful: passing parameters to GET request

Question

I want to create a resource that supports GET request in following way:

/bar?key1=val1&key2=val2

I tried this code, but it is not worki

Answer 1

Edit: reqparse is no longer the recommended way to do this with flask-restful!, but there is another example using marshmallow below.

The reqparse object is deprecated. See the docs or the second example in this post for alternatives.

---

Use reqparse. You can see another example in the flask-restful docs.

It performs validation on the parameters and does not require jsonify.

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

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

class BarAPI(Resource):
    def get(self):

        parser = reqparse.RequestParser()
        parser.add_argument('key1', type=str)
        parser.add_argument('key2', type=str)

        return parser.parse_args()

api.add_resource(BarAPI, '/bar', endpoint='bar')

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

---

Another way is to use marshmallow.

You can use a Schema class,to validate request.args (for a PUT/POST request you might validate request.form)

from flask import Flask, request, abort
from flask_restful import Resource, Api
from marshmallow import Schema, fields


class BarQuerySchema(Schema):
    key1 = fields.Str(required=True)
    key2 = fields.Str(required=True)

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


class BarAPI(Resource):
    def get(self):
        errors = schema.validate(request.args)
        if errors:
            abort(400, str(errors))

        return 'ok'

api.add_resource(BarAPI, '/bar', endpoint='bar')

# omit of you intend to use `flask run` command
if __name__ == '__main__':
    app.run(debug=True)

This example requires that both parameters be present.