Return HTTP status code 201 in flask

Question

We're using Flask for one of our API's and I was just wondering if anyone knew how to return a HTTP response 201?

For errors such as 404 we can call:

from flask import abort
abort(404)

But for 201 I get

LookupError: no exception for 201

Do I need to create my own exception like this in the docs?

Answer 1

You can read about it here.

return render_template('page.html'), 201

Answer 2

You can use Response to return any http status code.

> from flask import Response
> return Response("{'a':'b'}", status=201, mimetype='application/json')

Answer 3

As lacks suggested send status code in return statement and if you are storing it in some variable like

notfound = 404
invalid = 403
ok = 200

and using

return xyz, notfound

than time make sure its type is int not str. as I faced this small issue also here is list of status code followed globally http://www.w3.org/Protocols/HTTP/HTRESP.html

Hope it helps.

Answer 4

You can do

result = {'a': 'b'}
return jsonify(result), 201

if you want to return a JSON data in the response along with the error code You can read about responses here and here for make_response API details

Answer 5

In your flask code, you should ideally specify the MIME type as often as possible, as well:

return html_page_str, 200, {'ContentType':'text/html'}

return json.dumps({'success':True}), 200, {'ContentType':'application/json'}

...etc

Answer 6

In my case I had to combine the above in order to make it work

return Response(json.dumps({'Error': 'Error in payload'}), 
status=422, 
mimetype="application/json")

Answer 7

you can also use flask_api for sending response

from flask_api import status

@app.route('/your-api/')
def empty_view(self):
    content = {'your content here'}
    return content, status.HTTP_201_CREATED

you can find reference here http://www.flaskapi.org/api-guide/status-codes/