'How to return a dict as a JSON response from a Flask view?

I have a function that analyzes a CSV file with Pandas and produces a dict with summary information. I want to return the results as a response from a Flask view. How do I return a JSON response?

@app.route("/summary")
def summary():
    d = make_summary()
    # send it back as json

python json flask

Solution 1:^[1]

As of Flask 1.1.0 a view can directly return a Python dict and Flask will call jsonify automatically.

@app.route("/summary")
def summary():
    d = make_summary()
    return d

If your Flask version is less than 1.1.0 or to return a different JSON-serializable object, import and use jsonify.

from flask import jsonify

@app.route("/summary")
def summary():
    d = make_summary()
    return jsonify(d)

Solution 2:^[2]

jsonify serializes the data you pass it to JSON. If you want to serialize the data yourself, do what jsonify does by building a response with status=200 and mimetype='application/json'.

from flask import json

@app.route('/summary')
def summary():
    data = make_summary()
    response = app.response_class(
        response=json.dumps(data),
        status=200,
        mimetype='application/json'
    )
    return response

Solution 3:^[3]

Pass keyword arguments to flask.jsonify and they will be output as a JSON object.

@app.route('/_get_current_user')
def get_current_user():
    return jsonify(
        username=g.user.username,
        email=g.user.email,
        id=g.user.id
    )

{
    "username": "admin",
    "email": "admin@localhost",
    "id": 42
}

If you already have a dict, you can pass it directly as jsonify(d).

Solution 4:^[4]

If you don't want to use jsonify for some reason, you can do what it does manually. Call flask.json.dumps to create JSON data, then return a response with the application/json content type.

from flask import json

@app.route('/summary')
def summary():
    data = make_summary()
    response = app.response_class(
        response=json.dumps(data),
        mimetype='application/json'
    )
    return response

flask.json is distinct from the built-in json module. It will use the faster simplejson module if available, and enables various integrations with your Flask app.

Solution 5:^[5]

To return a JSON response and set a status code you can use make_response:

from flask import jsonify, make_response

@app.route('/summary')
def summary():
    d = make_summary()
    return make_response(jsonify(d), 200)

Inspiration taken from this comment in the Flask issue tracker.

Solution 6:^[6]

As of version 1.1.0 Flask, if a view returns a dict it will be turned into a JSON response.

@app.route("/users", methods=['GET'])
def get_user():
    return {
        "user": "John Doe",
    }

Solution 7:^[7]

If you want to analyze a file uploaded by the user, the Flask quickstart shows how to get files from users and access them. Get the file from request.files and pass it to the summary function.

from flask import request, jsonify
from werkzeug import secure_filename

@app.route('/summary', methods=['GET', 'POST'])
def summary():
    if request.method == 'POST':
        csv = request.files['data']
        return jsonify(
            summary=make_summary(csv),
            csv_name=secure_filename(csv.filename)
        )

    return render_template('submit_data.html')

Replace the 'data' key for request.files with the name of the file input in your HTML form.

Solution 8:^[8]

Flask 1.1.x supports returning a JSON dict without calling jsonify. If you want to return something besides a dict, you still need to call jsonify.

@app.route("/")
def index():
    return {
        "api_stuff": "values",
    }

is equivalent to

@app.route("/")
def index():
    return jsonify({
        "api_stuff": "values",
    })

See the pull request that added this: https://github.com/pallets/flask/pull/3111

Solution 9:^[9]

I use a decorator to return the result of jsonfiy. I think it is more readable when a view has multiple returns. This does not support returning a tuple like content, status, but I handle returning error statuses with app.errorhandler instead.

import functools
from flask import jsonify

def return_json(f):
    @functools.wraps(f)
    def inner(**kwargs):
        return jsonify(f(**kwargs))

    return inner

@app.route('/test/<arg>')
@return_json
def test(arg):
    if arg == 'list':
        return [1, 2, 3]
    elif arg == 'dict':
        return {'a': 1, 'b': 2}
    elif arg == 'bool':
        return True
    return 'none of them'

Solution 10:^[10]

Prior to Flask 0.11, jsonfiy would not allow returning an array directly. Instead, pass the list as a keyword argument.

@app.route('/get_records')
def get_records():
    results = [
        {
          "rec_create_date": "12 Jun 2016",
          "rec_dietary_info": "nothing",
          "rec_dob": "01 Apr 1988",
          "rec_first_name": "New",
          "rec_last_name": "Guy",
        },
        {
          "rec_create_date": "1 Apr 2016",
          "rec_dietary_info": "Nut allergy",
          "rec_dob": "01 Feb 1988",
          "rec_first_name": "Old",
          "rec_last_name": "Guy",
        },
    ]
    return jsonify(results=list)

Solution 11:^[11]

In Flask 1.1, if you return a dictionary and it will automatically be converted into JSON. So if make_summary() returns a dictionary, you can

from flask import Flask

app = Flask(__name__)

@app.route('/summary')
def summary():
    d = make_summary()
    return d

The SO that asks about including the status code was closed as a duplicate to this one. So to also answer that question, you can include the status code by returning a tuple of the form (dict, int). The dict is converted to JSON and the int will be the HTTP Status Code. Without any input, the Status is the default 200. So in the above example the code would be 200. In the example below it is changed to 201.

from flask import Flask

app = Flask(__name__)

@app.route('/summary')
def summary():
    d = make_summary()
    return d, 201  # 200 is the default

You can check the status code using

curl --request GET "http://127.0.0.1:5000/summary" -w "\ncode: %{http_code}\n\n"

Solution 12:^[12]

if its a dict, flask can return it directly (Version 1.0.2)

def summary():
    d = make_summary()
    return d, 200

Solution 13:^[13]

The answer is the same when using Flask's class-based views.

from flask import Flask, request, jsonify
from flask.views import MethodView

app = Flask(__name__)

class Summary(MethodView):
    def get(self):
        d = make_summary()
        return jsonify(d)

app.add_url_rule('/summary/', view_func=Summary.as_view('summary'))

Solution 14:^[14]

To serialize an object, use jsonify from flask module to jsonify the object, a dictionary gets serialized by default. Also, if you're dealing with files you can always use make_response.

Solution 15:^[15]

I like this way:

    @app.route("/summary")
    def summary():
        responseBody = { "message": "bla bla bla", "summary": make_summary() }
        return make_response(jsonify(responseBody), 200)

Sources

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Source: Stack Overflow

Solution	Source
Solution 1
Solution 2	davidism
Solution 3	davidism
Solution 4	davidism
Solution 5	davidism
Solution 6	davidism
Solution 7	davidism
Solution 8	davidism
Solution 9	davidism
Solution 10	davidism
Solution 11
Solution 12	c8999c 3f964f64
Solution 13	davidism
Solution 14	bkoiki950
Solution 15	Amin Shojaei

'How to return a dict as a JSON response from a Flask view?

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