Python Flask, TypeError: 'dict' object is not callable


Having an issue that seems to be common yet I have done my research and don’t see it being exactly recreated anywhere. When I print json.loads(rety.text), I am seeing the output I need. Yet when I call return, it shows me this error. Any ideas? Help is greatly appreciated and thank you. I am using the Flask MethodHandler.

class MHandler(MethodView):
    def get(self):
        handle = ''
        tweetnum = 100

        consumer_token = '' 
        consumer_secret = ''
        access_token = '-'
        access_secret = ''

        auth = tweepy.OAuthHandler(consumer_token,consumer_secret)

        api  = tweepy.API(auth)

        statuses = api.user_timeline(screen_name=handle,
                          count= tweetnum,

        pi_content_items_array = map(convert_status_to_pi_content_item, statuses)
        pi_content_items = { 'contentItems' : pi_content_items_array }

        saveFile = open("static/public/text/en.txt",'a') 
        for s in pi_content_items_array: 
            stat = s['content'].encode('utf-8')
            print stat

            trat = ''.join(i for i in stat if ord(i)<128)
            print trat

            contentFile = open("static/public/text/en.txt", "r")
            fr =
        except Exception as e:
            print "ERROR: couldn't read text file: %s" % e
        return lookup.get_template("newin.html").render(content=fr) 

    def post(self):
            contentFile = open("static/public/text/en.txt", "r")
            fd =
        except Exception as e:
            print "ERROR: couldn't read text file: %s" % e
        rety ='', 
                auth=('---', ''),
                headers = {"content-type": "text/plain"},

        print json.loads(rety.text)
        return json.loads(rety.text)

    user_view = MHandler.as_view('user_api')
    app.add_url_rule('/results2', view_func=user_view, methods=['GET',])
    app.add_url_rule('/results2', view_func=user_view, methods=['POST',])

Here is the Traceback(Keep in mind results are printing above):

Traceback (most recent call last):
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/", line 1478, in full_dispatch_request
    response = self.make_response(rv)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/", line 1577, in make_response
    rv = self.response_class.force_type(rv, request.environ)
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/", line 841, in force_type
    response = BaseResponse(*_run_wsgi_app(response, environ))
  File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/", line 867, in run_wsgi_app
    app_rv = app(environ, start_response)
Asked By: puhtiprince



Flask only expects views to return a response-like object. This means a Response, a string, or a tuple describing the body, code, and headers. You are returning a dict, which is not one of those things. Since you’re returning JSON, return a response with the JSON string in the body and a content type of application/json.

return app.response_class(rety.content, content_type='application/json')

In your example, you already have a JSON string, the content returned by the request you made. However, if you want to convert a Python structure to a JSON response, use jsonify:

data = {'name': 'davidism'}
return jsonify(data)

Behind the scenes, Flask is a WSGI application, which expects to pass around callable objects, which is why you get that specific error: a dict isn’t callable and Flask doesn’t know how to turn it into something that is.

Answered By: davidism

Use the Flask.jsonify function to return the data.

from flask import jsonify 
# ...
return jsonify(data)
Answered By: Sahil Shah

If you return a data, status, headers tuple from a Flask view, Flask currently ignores the status code and content_type header when the data is already a response object, such as what jsonify returns.

This doesn’t set the content-type header:

headers = {
    "Content-Type": "application/octet-stream",
    "Content-Disposition": "attachment; filename=foobar.json"
return jsonify({"foo": "bar"}), 200, headers

Instead, use flask.json.dumps to generate the data (which is what jsonfiy uses internally).

from flask import json

headers = {
    "Content-Type": "application/octet-stream",
    "Content-Disposition": "attachment; filename=foobar.json"
return json.dumps({"foo": "bar"}), 200, headers

Or work with the response object:

response = jsonify({"foo": "bar"})
response.headers.set("Content-Type", "application/octet-stream")
return response

However, if you want to literally do what these examples show and serve JSON data as a download, use send_file instead.

from io import BytesIO
from flask import json
data = BytesIO(json.dumps(data))
return send_file(data, mimetype="application/json", as_attachment=True, attachment_filename="data.json")
Answered By: firelynx

Instead of trying to jsonify the response, this worked.

return response.content
Answered By: l3o

as for flask version 1.1.0 now you could return dict

flask will convert it automatically to json response.

Answered By: buncis
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