When you return a dict in flask, is jsonify automatically applied?

Question

It appears that this works:

@app.route('/my/route/')
def my_route():
    return {
        'Something': True,
        'Else': None,
        'Thingy': 12345,
        'Blah': 'Blah'
    }

When I visit the route in a browser, I get valid JSON like this:

{
    "Something": true, 
    "Else": null, 
    "Thingy": 12345
    "Blah": "Blah", 
}

Everything is converted to valid JSON, but I haven't seen any documentation supporting this. I'm not even importing the jsonify module. Is this ok?

In current, vanilla versions of Flask this does not seem to be valid. It results in a TypeError and "The view function did not return a valid response. The return type must be a string, tuple, Response instance, or WSGI callable, but it was a dict." — jarmod, Aug 13 '19 at 01:33
@jarmod I think I'm using vanilla Flask. I'm not sure how it's returning the JSON in this case. I did it by accident really. — Glenn, Aug 13 '19 at 01:36
Typically you would import jsonify, and then return jsonify({...}). — jarmod, Aug 13 '19 at 01:40

score 2 · Answer 1 · answered Aug 13 '19 at 01:57

No, returning a dict in Flask will not apply jsonify automatically. In fact, Flask route cannot return dictionary.

Code:

from flask import Flask, render_template, json

app = Flask(__name__)

@app.route("/")
def index():
    return {
        'Something': True,
        'Else': None,
        'Thingy': 12345,
        'Blah': 'Blah'
    }

Output:

TypeError
TypeError: 'dict' object is not callable
The view function did not return a valid response. The return type must be a string, tuple, Response instance, or WSGI callable, but it was a dict.

Screenshot:

As the traceback indicates route's return type must be a string, tuple, Response instance, or WSGI callable.

As of Flask `1.1.0` you can return a dictionary, or still use `jsonify()` for other data types. See [this answer](https://stackoverflow.com/a/58917715/2052575) and [these docs](https://flask.palletsprojects.com/en/1.1.x/quickstart/#apis-with-json). — v25, Nov 23 '19 at 23:06

score 0 · Answer 2 · answered Aug 13 '19 at 01:46

define the custom json encoder! it will do the trick!

import flask
import datetime

app = flask.Flask(__name__)

class CustomJSONEncoder(flask.json.JSONEncoder):
    def default(self, obj):
        if   isinstance(obj, datetime.date): return obj.isoformat()
        try:
            iterable = iter(obj)
        except TypeError:
            pass
        else:
            return tuple(iterable)

        return flask.json.JSONEncoder.default(self, obj)

app.json_encoder = CustomJSONEncoder

When you return a dict in flask, is jsonify automatically applied?

2 Answers2