Python Flask, TypeError: 'dict' object is not callable
Question:
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)
auth.set_access_token(access_token,access_secret)
api = tweepy.API(auth)
statuses = api.user_timeline(screen_name=handle,
count= tweetnum,
include_rts=False)
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
saveFile.write(trat.encode('utf-8')+'n'+'n')
try:
contentFile = open("static/public/text/en.txt", "r")
fr = contentFile.read()
except Exception as e:
print "ERROR: couldn't read text file: %s" % e
finally:
contentFile.close()
return lookup.get_template("newin.html").render(content=fr)
def post(self):
try:
contentFile = open("static/public/text/en.txt", "r")
fd = contentFile.read()
except Exception as e:
print "ERROR: couldn't read text file: %s" % e
finally:
contentFile.close()
rety = requests.post('https://gateway.watsonplatform.net/personality-insights/api/v2/profile',
auth=('---', ''),
headers = {"content-type": "text/plain"},
data=fd
)
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/app.py", line 1836, in __call__
return self.wsgi_app(environ, start_response)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1820, in wsgi_app
response = self.make_response(self.handle_exception(e))
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1403, in handle_exception
reraise(exc_type, exc_value, tb)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1817, in wsgi_app
response = self.full_dispatch_request()
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1478, in full_dispatch_request
response = self.make_response(rv)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1577, in make_response
rv = self.response_class.force_type(rv, request.environ)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/wrappers.py", line 841, in force_type
response = BaseResponse(*_run_wsgi_app(response, environ))
File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/test.py", line 867, in run_wsgi_app
app_rv = app(environ, start_response)
Answers:
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.
Use the Flask.jsonify function to return the data.
from flask import jsonify
# ...
return jsonify(data)
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")
Instead of trying to jsonify the response, this worked.
return response.content
as for flask version 1.1.0 now you could return dict
flask will convert it automatically to json response.
https://flask.palletsprojects.com/en/1.1.x/quickstart/#apis-with-json
https://flask.palletsprojects.com/en/1.1.x/changelog/#version-1-1-0
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)
auth.set_access_token(access_token,access_secret)
api = tweepy.API(auth)
statuses = api.user_timeline(screen_name=handle,
count= tweetnum,
include_rts=False)
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
saveFile.write(trat.encode('utf-8')+'n'+'n')
try:
contentFile = open("static/public/text/en.txt", "r")
fr = contentFile.read()
except Exception as e:
print "ERROR: couldn't read text file: %s" % e
finally:
contentFile.close()
return lookup.get_template("newin.html").render(content=fr)
def post(self):
try:
contentFile = open("static/public/text/en.txt", "r")
fd = contentFile.read()
except Exception as e:
print "ERROR: couldn't read text file: %s" % e
finally:
contentFile.close()
rety = requests.post('https://gateway.watsonplatform.net/personality-insights/api/v2/profile',
auth=('---', ''),
headers = {"content-type": "text/plain"},
data=fd
)
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/app.py", line 1836, in __call__
return self.wsgi_app(environ, start_response)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1820, in wsgi_app
response = self.make_response(self.handle_exception(e))
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1403, in handle_exception
reraise(exc_type, exc_value, tb)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1817, in wsgi_app
response = self.full_dispatch_request()
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1478, in full_dispatch_request
response = self.make_response(rv)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/flask/app.py", line 1577, in make_response
rv = self.response_class.force_type(rv, request.environ)
File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/wrappers.py", line 841, in force_type
response = BaseResponse(*_run_wsgi_app(response, environ))
File "/Users/RZB/anaconda/lib/python2.7/site-packages/werkzeug/test.py", line 867, in run_wsgi_app
app_rv = app(environ, start_response)
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.
Use the Flask.jsonify function to return the data.
from flask import jsonify
# ...
return jsonify(data)
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")
Instead of trying to jsonify the response, this worked.
return response.content
as for flask version 1.1.0 now you could return dict
flask will convert it automatically to json response.
https://flask.palletsprojects.com/en/1.1.x/quickstart/#apis-with-json
https://flask.palletsprojects.com/en/1.1.x/changelog/#version-1-1-0