Python Flask: Error "The method is not allowed for the requested URL"

Question

I am very new to working with Python Flask and i wanted to try a simple API-example:

from flask import Flask, jsonify, request
from flask_restful import Resource, Api

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

class HelloWorld(Resource):
    def get(self):
        some_json = request.get_json()
        return {'you sent': some_json}, 201

class Multi(Resource):
    def get(self,num):
        return {'result': num*10}

api.add_resource(HelloWorld, '/')
api.add_resource(Multi,'/multi/<int:num>')

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

and if I type in the terminal

-H "Content-Type: application/json" -X POST -d '{"name":"xyz","address":"myaddress"}' http://127.0.0.1:5000/

I get the following message:

{
    "message": "The method is not allowed for the requested URL."
}

I hope someone can help me with this...

filani · Accepted Answer · 2020-03-18 12:39:20Z

1

Since your are calling the POST HTTP method. You should rename 'get' function in class HelloWorld to 'post'. 'HelloWorld' class can also have both 'get' and a 'post' functions if '/' endpoint should serves both.

from flask import Flask, jsonify, request
from flask_restful import Resource, Api

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

class HelloWorld(Resource):
    def post(self):
        some_json = request.get_json()
        return {'you sent': some_json}, 201

class Multi(Resource):
    def get(self,num):
        return {'result': num*10}

api.add_resource(HelloWorld, '/')
api.add_resource(Multi,'/multi/<int:num>')

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

answered Mar 18, 2020 at 12:39

filani

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1 Comment

Phobos Over a year ago

but then I get the message: "Failed to decode JSON object: Expecting value: line 1 column 1 (char 0)"

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