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Change response based on content type of request in Flask

I have a Flask application that returns both HTML pages and JSON responses to API requests. I want to change what an error handler returns based on the content type of the request. If the client requests application/json, I want to return a jsonify response, otherwise I want to return a render_template response. How can I detect what was requested and change the response appropriately?
The current error handlers I have only return an HTML response.
def register_errorhandlers(app):
"""Register error handlers."""
def render_error(error):
"""Render error template."""
# If a HTTPException, pull the `code` attribute; default to 500
error_code = getattr(error, 'code', 500)
return render_template('{0}.html'.format(error_code)), error_code
for errcode in [401, 404, 500]:
app.errorhandler(errcode)(render_error)

Use request.content_type to get the content type the client sent with the request. Use request.accept_mimetypes the get the mimetypes the client indicated it can accept in a response. Use these to determine what to return.
from flask import request, jsonify, render_template
if request.accept_mimetypes.accept_json:
return jsonify(...)
else:
return render_template(...)

I used the after_request decorator to do this and checked the content type:
#app.after_request
def after_request_helper(resp):
if resp.content_type == "text/html":
# If a HTTPException, pull the `code` attribute; default to 500
error_code = getattr(error, 'code', 500)
return render_template('{0}.html'.format(error_code)), error_code
else:
return app.errorhandler(errcode)(render_error)

A more detailed answer:
def wants_json_response():
return request.accept_mimetypes['application/json'] >= \
request.accept_mimetypes['text/html']
The wants_json_response() helper function compares the preference for JSON or HTML selected by the client in their list of preferred formats. If JSON rates higher than HTML, then it is necessary to return a JSON response.
Otherwise, return the original HTML responses based on templates.
For the JSON responses would slightly supplement the function with one condition:
if wants_json_response(): which is what you need. So the answer is in that.
If the condition is true we could write a function that would generate a response:
def api_error_response(status_code, message=None):
payload = {'error': HTTP_STATUS_CODES.get(status_code, 'Unknown error')}
if message:
payload['message'] = message
response = jsonify(payload)
response.status_code = status_code
return response
This function uses the handy HTTP_STATUS_CODES dictionary from Werkzeug (a core dependency of Flask) that provides a short descriptive name for each HTTP status code.
For easier and faster understanding, 'error' is used to represent errors, so you only need to worry about the numeric status code and the optional long description.
The jsonify() function returns a Flask Response object with a default status code of 200, so after the response is created, it is necessary to set the status code to the correct one for the error.
So if we put it all together now it would look like this:
# app/__init__.py
import requests
def register_errorhandlers(app):
from .errors import render_error
for e in [
requests.codes.INTERNAL_SERVER_ERROR,
requests.codes.NOT_FOUND,
requests.codes.UNAUTHORIZED,
]:
app.errorhandler(e)(render_error)
and
# app/errors.py
import requests
from flask import render_template, request, jsonify
from werkzeug.http import HTTP_STATUS_CODES
from .extensions import db
def api_error_response(status_code, message=None):
payload = {'error': HTTP_STATUS_CODES.get(status_code, 'Unknown error')}
if message:
payload['message'] = message
response = jsonify(payload)
response.status_code = status_code
return response
def wants_json_response():
return request.accept_mimetypes['application/json'] >= \
request.accept_mimetypes['text/html']
def render_error(e):
if requests.codes.INTERNAL_SERVER_ERROR == e.code:
db.session.rollback()
if wants_json_response():
return api_error_response(e.code)
else:
return render_template(f'{e.code}.html'), e.code
Additionally
Then they could use the response generation for other cases as well.
The most common error that the API is going to return is going to be
the code 400, which is the error for “bad request”. This is the error
that is used when the client sends a request that has invalid data in it.
In order to generate messages to the function below even easier in these cases, we forward only the required description - message.
def bad_request(message):
return api_error_response(400, message)
I hope this will help in approaching with errors :)

Ensure the POST data is valid JSON

I am developping a JSON API with Python Flask.
What I want is to always return JSON, with a error message indicating any error that occured.
That API also only accept JSON data in the POST body, but Flask by default return a HTML error 400 if it can't read the data as JSON.
Preferably, I d also like to not force the user to send the Content-Type header, and if raw or text content-type, try to parse the body as JSON nonetheless.
In short, I need a way to validate that the POST body's is JSON, and handle the error myself.
I've read about adding decorator to request to do that, but no comprehensive example.

You have three options:
Register a custom error handler for 400 errors on the API views. Have this error return JSON instead of HTML.
Set the Request.on_json_loading_failed method to something that raises a BadRequest exception subclass with a JSON payload. See Custom Errors in the Werkzeug exceptions documentation to see how you can create one.
Put a try: except around the request.get_json() call, catch the BadRequest exception and raise a new exception with a JSON payload.
Personally, I'd probably go with the second option:
from werkzeug.exceptions import BadRequest
from flask import json, Request, _request_ctx_stack
class JSONBadRequest(BadRequest):
def get_body(self, environ=None):
"""Get the JSON body."""
return json.dumps({
'code': self.code,
'name': self.name,
'description': self.description,
})
def get_headers(self, environ=None):
"""Get a list of headers."""
return [('Content-Type', 'application/json')]
def on_json_loading_failed(self):
ctx = _request_ctx_stack.top
if ctx is not None and ctx.app.config.get('DEBUG', False):
raise JSONBadRequest('Failed to decode JSON object: {0}'.format(e))
raise JSONBadRequest()
Request.on_json_loading_failed = on_json_loading_failed
Now, every time request.get_json() fails, it'll call your custom on_json_loading_failed method and raise an exception with a JSON payload rather than a HTML payload.

Combining the options force=True and silent=True make the result of request.get_json be None if the data is not parsable, then a simple if allow you to check the parsing.
from flask import Flask
from flask import request
#app.route('/foo', methods=['POST'])
def function(function = None):
print "Data: ", request.get_json(force = True, silent = True);
if request.get_json() is not None:
return "Is JSON";
else:
return "Nope";
if __name__ == "__main__":
app.run()
Credits to lapinkoira and Martijn Pieters.

You can try to decode JSON object using python json library.
The main idea is to take plain request body and try to convert to JSON.E.g:
import json
...
# somewhere in view
def view():
try:
json.loads(request.get_data())
except ValueError:
# not a JSON! return error
return {'error': '...'}
# do plain stuff

python function .json(): No JSON object could be decoded

the function looks like this:
import requests
import json
def parse(s):
r = requests.post('http://166.111.139.15:9000/?properties%3d%7b%22annotators%22%3a%22tokenize%2cssplit%2cpos%2clemma%2cparse%22%2c%22outputFormat%22%3a%22json%22%7d%0a', data=s)
return r.json()
print parse("I am a student")
And when I use it in Django, the web page shows:
"No JSON object could be decoded".Why?

No JSON object could be decoded is an exception message raised at r.json(). If your response is not a valid json object you can still retrieve it with r.text. Even if you are sure your response is always a valid json object, you should still check whether server returned a success code. If there's an internal server error (code 500), you won't get a valid json response!
import requests
def parse(s)
r = requests.post('http://someserver.com', data=s)
if r.status_code !== 200:
return "There was a problem: {} !".format(r.text)
return r.json()

How do I parse a JSON response from Python Requests?

I am trying to parse a response.text that I get when I make a request using the Python Requests library. For example:
def check_user(self):
method = 'POST'
url = 'http://localhost:5000/login'
ck = cookielib.CookieJar()
self.response = requests.request(method,url,data='username=test1&passwd=pass1', cookies=ck)
print self.response.text
When I execute this method, the output is:
{"request":"POST /login","result":"success"}
I would like to check whether "result" equals "success", ignoring whatever comes before.

The manual suggests: if self.response.status_code == requests.codes.ok:
If that doesn't work:
if json.loads(self.response.text)['result'] == 'success':
whatever()

Since the output, response, appears to be a dictionary, you should be able to do
result = self.response.json().get('result')
print(result)
and have it print
'success'

If the response is in json you could do something like (python3):
import json
import requests as reqs
# Make the HTTP request.
response = reqs.get('http://demo.ckan.org/api/3/action/group_list')
# Use the json module to load CKAN's response into a dictionary.
response_dict = json.loads(response.text)
for i in response_dict:
print("key: ", i, "val: ", response_dict[i])
To see everything in the response you can use .__dict__:
print(response.__dict__)

import json
def check_user(self):
method = 'POST'
url = 'http://localhost:5000/login'
ck = cookielib.CookieJar()
response = requests.request(method,url,data='username=test1&passwd=pass1', cookies=ck)
#this line converts the response to a python dict which can then be parsed easily
response_native = json.loads(response.text)
return self.response_native.get('result') == 'success'

I found another solution. It is not necessary to use json module. You can create a dict using dict = eval(whatever) and return, in example, dict["result"]. I think it is more elegant. However, the other solutions also work and are correct

Put in the return of your method like this:
return self.response.json()
If you wanna looking for more details, click this following link:
https://www.w3schools.com/python/ref_requests_response.asp
and search for json() method.
Here is an code example:
import requests
url = 'https://www.w3schools.com/python/demopage.js'
x = requests.get(url)
print(x.json())

In some cases, maybe the response would be as expected. So It'd be great if we can built a mechanism to catch and log the exception.
import requests
import sys
url = "https://stackoverflow.com/questions/26106702/how-do-i-parse-a-json-response-from-python-requests"
response = requests.get(url)
try:
json_data = response.json()
except ValueError as exc:
print(f"Exception: {exc}")
# to find out why you have got this exception, you can see the response content and header
print(str(response.content))
print(str(response.headers))
print(sys.exc_info())
else:
if json_data.get('result') == "success":
# do whatever you want
pass

Return a requests.Response object from Flask

I'm trying to build a simple proxy using Flask and requests. The code is as follows:
#app.route('/es/<string:index>/<string:type>/<string:id>',
methods=['GET', 'POST', 'PUT']):
def es(index, type, id):
elasticsearch = find_out_where_elasticsearch_lives()
# also handle some authentication
url = '%s%s%s%s' % (elasticsearch, index, type, id)
esreq = requests.Request(method=request.method, url=url,
headers=request.headers, data=request.data)
resp = requests.Session().send(esreq.prepare())
return resp.text
This works, except that it loses the status code from Elasticsearch. I tried returning resp (a requests.models.Response) directly, but this fails with
TypeError: 'Response' object is not callable
Is there another, simple, way to return a requests.models.Response from Flask?

Ok, found it:
If a tuple is returned the items in the tuple can provide extra information. Such tuples have to be in the form (response, status, headers). The status value will override the status code and headers can be a list or dictionary of additional header values.
(Flask docs.)
So
return (resp.text, resp.status_code, resp.headers.items())
seems to do the trick.

Using text or content property of the Response object will not work if the server returns encoded data (such as content-encoding: gzip) and you return the headers unchanged. This happens because text and content have been decoded, so there will be a mismatch between the header-reported encoding and the actual encoding.
According to the documentation:
In the rare case that you’d like to get the raw socket response from the server, you can access r.raw. If you want to do this, make sure you set stream=True in your initial request.
and
Response.raw is a raw stream of bytes – it does not transform the response content.
So, the following works for gzipped data too:
esreq = requests.Request(method=request.method, url=url,
headers=request.headers, data=request.data)
resp = requests.Session().send(esreq.prepare(), stream=True)
return resp.raw.read(), resp.status_code, resp.headers.items()
If you use a shortcut method such as get, it's just:
resp = requests.get(url, stream=True)
return resp.raw.read(), resp.status_code, resp.headers.items()

Flask can return an object of type flask.wrappers.Response.
You can create one of these from your requests.models.Response object r like this:
from flask import Response
return Response(
response=r.reason,
status=r.status_code,
headers=dict(r.headers)
)

I ran into the same scenario, except that in my case my requests.models.Response contained an attachment. This is how I got it to work:
return send_file(BytesIO(result.content), mimetype=result.headers['Content-Type'], as_attachment=True)

My use case is to call another API in my own Flask API. I'm just propagating unsuccessful requests.get calls through my Flask response. Here's my successful approach:
headers = {
'Authorization': 'Bearer Muh Token'
}
try:
response = requests.get(
'{domain}/users/{id}'\
.format(domain=USERS_API_URL, id=hit['id']),
headers=headers)
response.raise_for_status()
except HTTPError as err:
logging.error(err)
flask.abort(flask.Response(response=response.content, status=response.status_code, headers=response.headers.items()))