Python Flask, TypeError: 'dict' object is not callable
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.
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 jsonheaders = { "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 BytesIOfrom flask import jsondata = BytesIO(json.dumps(data))return send_file(data, mimetype="application/json", as_attachment=True, attachment_filename="data.json")