How do I set response headers in Flask?

Question:

This is my code:

@app.route('/hello', methods=["POST"])
def hello():
    resp = make_response(render_template('hello.html'))
    resp.headers['Access-Control-Allow-Origin'] = '*'
    return resp

However, when I make a request from the browser to my server I get this error:

XMLHttpRequest cannot load http://localhost:5000/hello. 
No 'Access-Control-Allow-Origin' header is present on the requested resource.

I have also tried this approach, setting the response headers "after" the request:

@app.after_request
def add_header(response):
    response.headers['Access-Control-Allow-Origin'] = '*'
    return response

No dice. I get the same error. Is there a way to just set the response headers in the route function? Something like this would be ideal:

@app.route('/hello', methods=["POST"])
    def hello(response): # is this a thing??
        response.headers['Access-Control-Allow-Origin'] = '*'
        return response

but I cant find anyway to do this. Please help.

EDIT

if I curl the url with a POST request like so:

curl -iX POST http://localhost:5000/hello

I get this response:

HTTP/1.0 500 INTERNAL SERVER ERROR
Content-Type: text/html
Content-Length: 291
Server: Werkzeug/0.9.6 Python/2.7.6
Date: Tue, 16 Sep 2014 03:58:42 GMT

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<title>500 Internal Server Error</title>
<h1>Internal Server Error</h1>
<p>The server encountered an internal error and was unable to complete your request.  Either the server is overloaded or there is an error in the application.</p>

Any ideas?

Asked By: dopatraman

||

Answers:

You can do this pretty easily:

@app.route("/")
def home():
    resp = flask.Response("Foo bar baz")
    resp.headers['Access-Control-Allow-Origin'] = '*'
    return resp

Look at flask.Response and flask.make_response()

But something tells me you have another problem, because the after_request should have handled it correctly too.

EDIT
I just noticed you are already using make_response which is one of the ways to do it. Like I said before, after_request should have worked as well. Try hitting the endpoint via curl and see what the headers are:

curl -i http://127.0.0.1:5000/your/endpoint

You should see

> curl -i 'http://127.0.0.1:5000/'
HTTP/1.0 200 OK
Content-Type: text/html; charset=utf-8
Content-Length: 11
Access-Control-Allow-Origin: *
Server: Werkzeug/0.8.3 Python/2.7.5
Date: Tue, 16 Sep 2014 03:47:13 GMT

Noting the Access-Control-Allow-Origin header.

EDIT 2
As I suspected, you are getting a 500 so you are not setting the header like you thought. Try adding app.debug = True before you start the app and try again. You should get some output showing you the root cause of the problem.

For example:

@app.route("/")
def home():
    resp = flask.Response("Foo bar baz")
    user.weapon = boomerang
    resp.headers['Access-Control-Allow-Origin'] = '*'
    return resp

Gives a nicely formatted html error page, with this at the bottom (helpful for curl command)

Traceback (most recent call last):
...
  File "/private/tmp/min.py", line 8, in home
    user.weapon = boomerang
NameError: global name 'boomerang' is not defined
Answered By: sberry

This work for me

from flask import Flask
from flask import Response

app = Flask(__name__)

@app.route("/")
def home():
    return Response(headers={'Access-Control-Allow-Origin':'*'})

if __name__ == "__main__":
    app.run()
Answered By: German Lopez

Use make_response of Flask something like

@app.route("/")
def home():
    resp = make_response("hello") #here you could use make_response(render_template(...)) too
    resp.headers['Access-Control-Allow-Origin'] = '*'
    return resp

From flask docs,

flask.make_response(*args)

Sometimes it is necessary to set additional headers in a view. Because views do not have to return response objects but can return a value that is converted into a response object by Flask itself, it becomes tricky to add headers to it. This function can be called instead of using a return and you will get a response object which you can use to attach headers.

Answered By: Devi

This was how added my headers in my flask application and it worked perfectly

@app.after_request
def add_header(response):
    response.headers['X-Content-Type-Options'] = 'nosniff'
    return response
Answered By: Cole.E

We can set the response headers in Python Flask application using Flask application context using flask.g

This way of setting response headers in Flask application context using flask.g is thread safe and can be used to set custom & dynamic attributes from any file of application, this is especially helpful if we are setting custom/dynamic response headers from any helper class, that can also be accessed from any other file ( say like middleware, etc), this flask.g is global & valid for that request thread only.

Say if i want to read the response header from another api/http call that is being called from this app, and then extract any & set it as response headers for this app.

Sample Code: file: helper.py

import flask
from flask import request, g
from multidict import CIMultiDict
from asyncio import TimeoutError as HttpTimeout
from aiohttp import ClientSession

    def _extract_response_header(response)
      """
      extracts response headers from response object 
      and stores that required response header in flask.g app context
      """
      headers = CIMultiDict(response.headers)
      if 'my_response_header' not in g:
        g.my_response_header= {}
        g.my_response_header['x-custom-header'] = headers['x-custom-header']


    async def call_post_api(post_body):
      """
      sample method to make post api call using aiohttp clientsession
      """
      try:
        async with ClientSession() as session:
          async with session.post(uri, headers=_headers, json=post_body) as response:
            responseResult = await response.read()
            _extract_headers(response, responseResult)
            response_text = await response.text()
      except (HttpTimeout, ConnectionError) as ex:
        raise HttpTimeout(exception_message)

file: middleware.py

import flask
from flask import request, g

class SimpleMiddleWare(object):
    """
    Simple WSGI middleware
    """

    def __init__(self, app):
        self.app = app
        self._header_name = "any_request_header"

    def __call__(self, environ, start_response):
        """
        middleware to capture request header from incoming http request
        """
        request_id_header = environ.get(self._header_name)
        environ[self._header_name] = request_id_header

        def new_start_response(status, response_headers, exc_info=None):
            """
            set custom response headers
            """
            # set the request header as response header
            response_headers.append((self._header_name, request_id_header))
            # this is trying to access flask.g values set in helper class & set that as response header
            values = g.get(my_response_header, {})
            if values.get('x-custom-header'):
                response_headers.append(('x-custom-header', values.get('x-custom-header')))
            return start_response(status, response_headers, exc_info)

        return self.app(environ, new_start_response)

Calling the middleware from main class

file : main.py

from flask import Flask
import asyncio
from gevent.pywsgi import WSGIServer
from middleware import SimpleMiddleWare

app = Flask(__name__)
app.wsgi_app = SimpleMiddleWare(app.wsgi_app)
Answered By: src3369

According to the documentation, you can return headers from your view function together with the response.

If a tuple is returned the items in the tuple can provide extra
information. Such tuples have to be in the form (response, status),
(response, headers), or (response, status, headers). The status value
will override the status code and headers can be a list or dictionary
of additional header values.

For example:

@app.route('/hello', methods=["POST"])
def hello():
   return flask.make_response(), {"Access-Control-Allow-Origin": "*"}

Or:

@app.route('/hello', methods=["POST"])
def hello():
   return {"foo": "bar"}, 200, {"Access-Control-Allow-Origin": "*"}
Answered By: J. Klima
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