API REST 猎鹰 python POST 方法

Question

我尝试使用 falcon 编写 API REST。

on_get方法很好用，但是在使用on_post时，我无法获取POST请求的主体，我不知道为什么

class ProfileUpdate(object):
    def on_post(self, req, resp):
        data = json.load(req.stream)
        print(data)
        resp.body = {"test": "answer"}
        resp.status = falcon.HTTP_200
        return resp


def setup_profile():
    app = falcon.API()

    profile_update = ProfileUpdate()
    app.add_route('/profiles', profile_update)

    return app

我收到以下错误

Traceback (most recent call last):
  File "/usr/local/lib/python3.7/site-packages/gunicorn/workers/sync.py", line 135, in handle
    self.handle_request(listener, req, client, addr)
  File "/usr/local/lib/python3.7/site-packages/gunicorn/workers/sync.py", line 176, in handle_request
    respiter = self.wsgi(environ, resp.start_response)
  File "/usr/local/lib/python3.7/site-packages/falcon/api.py", line 318, in __call__
    body, length = self._get_body(resp, env.get('wsgi.file_wrapper'))
  File "/usr/local/lib/python3.7/site-packages/falcon/api.py", line 820, in _get_body
    body = body.encode('utf-8')
AttributeError: 'dict' object has no attribute 'encode'

我正在使用 Postman 测试 API。我尝试在 POSTMAN (raw -> JSON)

中使用以下正文

{
"email":"test"
}

我错过了什么？

Answer 1

首先，使用req.media检索数据（建议）。
其次，对于响应，使用 resp.body = json.dumps({"test": "answer"})

Answer 2

如documentation

所述

您可以使用 media 使用请求原始正文 JSON 并将 media 附加到响应。

Zero configuration is needed if you’re creating a JSON API. Just access or set the media attribute as appropriate and let Falcon do the heavy lifting for you.

class EchoResource:
    def on_post(self, req, resp):
        message = req.media.get("message")
        resp.media = {
            "message": message
        }
        resp.status = falcon.HTTP_200

Warning:

Once media is called on a request, it’ll consume the request’s stream.

API REST 猎鹰 python POST 方法

API REST falcon python POST METHOD

python

api

post

falcon