如何在Flask中设置响应头?


99

这是我的代码:

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

但是,当我从浏览器向服务器发出请求时,出现此错误:

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

我也尝试过这种方法,在请求之后设置响应头:

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

没有骰子。我犯了同样的错误。有没有一种方法可以只在route函数中设置响应头?这样的事情将是理想的:

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

但我还是找不到这样做。请帮忙。

编辑

如果我使用POST请求卷曲网址,如下所示:

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

我得到这个回应:

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>

有任何想法吗?

Answers:


96

您可以很容易地做到这一点:

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

查看flask.Responseflask.make_response()

但是有些事情告诉我您还有另一个问题,因为after_request应当也正确地处理了它。

编辑
我刚刚注意到您已经在使用make_response它,这是执行此操作的方法之一。就像我之前说过的,after_request应该也可以。尝试通过curl到达端点,看看标题是什么:

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

你应该看到

> 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

注意Access-Control-Allow-Origin标头。

编辑2
正如我所怀疑的,您得到500,所以您没有按照您的想法设置标题。app.debug = True在启动应用程序之前尝试添加,然后重试。您应该获得一些输出,以显示问题的根本原因。

例如:

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

给出格式正确的html错误页面,该页面位于底部(有助于curl命令)

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

24

使用make_responseFlask之类的东西

@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

flask文档

flask.make_response(* args)

有时有必要在视图中设置其他标题。由于视图不必返回响应对象,而是可以返回Flask本身将其转换为响应对象的值,因此向其添加标头变得很棘手。可以调用此函数而不是使用return,您将获得一个响应对象,可用于附加标题。



5

这对我有用

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()

2
还有一种在return Response(headers={'Access-Control-Allow-Origin':'*'})我看来更干净的符号。
赫尔曼

4

这就是在我的烧瓶应用程序中添加我的标头的方式,并且效果很好

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

0

我们可以在Flask应用程序上下文中使用以下命令在Python Flask应用程序中设置响应标头: flask.g

这种在Flask应用程序上下文中使用设置响应标头的方法flask.g是线程安全的,可用于从任何应用程序文件设置自定义和动态属性,如果我们从任何帮助程序类设置自定义/动态响应标头,则该方法特别有用。也可以从任何其他文件(例如中间件等)进行访问,这flask.g是全局的,仅对该请求线程有效。

说我是否要从正在从此应用程序调用的另一个api / http调用中读取响应标头,然后提取任何内容并将其设置为该应用程序的响应标头。

示例代码:文件: 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)

文件: 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)

从主类调用中间件

文件: 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)
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