Cannot return 404 error as json instead of html from a Flask-Restful app

Question

I am working on a simple Flask REST API test and when I call the {{url}}/items for example I get the items list. However if a call is passed to an endpoint that does not exist for example {{url}}/itemsss then I get the error 404 in html.

I would like to make the error handling more friendly and return json instead of html for certain errors such as 400, 404,405...

For the 404 for example i tried this:

@app.errorhandler(404)
def not_found(e):
    response = jsonify({'status': 404,'error': 'not found',
                        'message': 'invalid resource URI'})
    response.status_code = 404
    return response

However it does not work.

My issue is similar to this one: Python Flask - Both json and html 404 error

I wanted to know, if using the blueprints the only way to accomplish this?

If there a simpler way to output the 404 error as json?

For example instead of this:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">

<title>404 Not Found</title>

<h1>Not Found</h1>

<p>The requested URL was not found on the server.  If you entered the URL manually please check your spelling and try again.</p>

Something like this:

{

error: true,

status: 404,

code: "error.notFound",

message: "API endpoint not found",

data: { }

}

I appreciate your help with this.

Miguel Machado · Accepted Answer · 2019-03-10 18:20:05Z

14

Usually when I need to return a custom error message with Flask-RESTful I would do something like:

from flask import make_response, jsonify

def custom_error(message, status_code): 
    return make_response(jsonify(message), status_code)

answered Mar 10, 2019 at 18:20

Miguel Machado

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Community · Accepted Answer · 2022-08-08 06:57:23Z

7

I think I found the solution in the official documentation:

from flask import json
from werkzeug.exceptions import HTTPException

@app.errorhandler(HTTPException)
def handle_exception(e):
    """Return JSON instead of HTML for HTTP errors."""
    # start with the correct headers and status code from the error
    response = e.get_response()
    # replace the body with JSON
    response.data = json.dumps({
        "code": e.code,
        "name": e.name,
        "description": e.description,
    })
    response.content_type = "application/json"
    return response

edited Aug 8, 2022 at 6:57

CommunityBot

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answered Jan 23, 2021 at 14:38

Andrea

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Praveen · Accepted Answer · 2022-03-06 07:08:47Z

3

Code and description are there in HTTP anyways, need only the message. And using jsonify from flask.

from flask import jsonify
from werkzeug.exceptions import HTTPException

@app.errorhandler(HTTPException)
def handle_exception(e):
    return jsonify({"message": e.description}), e.code

answered Mar 6, 2022 at 7:08

Praveen

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Comments

minou · Accepted Answer · 2023-09-20 14:04:48Z

1

This works and is simpler than all of the other answers:

return {}, 404

answered Sep 20, 2023 at 14:04

minou

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