Python tornado中烧瓶API的替代方案（Dialogflow webhook）

我需要在Tornado服务器中为Dialogflow创建一个webhook。我以前是用烧瓶做的。这两方面我都是初学者。烧瓶的代码 Flask webhook代码：

from flask import render_template
import os
from flask import Flask
from flask import request
from flask import make_response
import json
import time

app = Flask(__name__)

@app.route("/")
def index():
    return render_template("index.html")

@app.route('/webhook', methods=['POST', 'GET'])
def webhook():
    req = request.get_json(silent=True, force=True)
    print(json.dumps(req, indent=4))
    res = makeWebhookResult(req)
    res = json.dumps(res, indent=4)
    print("+++++++++++++++RESPONSE+++++++++++++++++", res)
    r = make_response(res)
    r.headers['Content-Type'] = 'application/json'
    return r

# Right now I'm just printing a response to see if it works properly

def makeWebhookResult(req):
      queryResult = req.get('queryResult').get("queryText")
      speech =  queryResult       
      return {
            "fulfillmentText": 'YOLO',
            "source": 'App'
        }

#./ngrok http 8090
if __name__ == '__main__':
    port = int(os.getenv('PORT', 8090))
    print("Starting app on port %d" % (port))
    app.run(debug=True, port=port, host='localhost')

现在我在《龙卷风》中试过这个：

import tornado.ioloop
import tornado.web as web
import tornado
import json
import os

static_root = os.path.join(os.path.dirname(__file__), 'static')


class MainHandler(tornado.web.RequestHandler):
    def get(self):
        template = "./templates/index.html"
        self.render(template)


class Webhook(tornado.web.RequestHandler):
    def prepare(self):
        if self.request.headers.get("Content-Type", "").startswith("application/json"):
            self.req = json.loads(self.request.body)
            print(self.req)
            print(json.dumps(self.req, indent=4))
            response = self.webhook_result(self.req)
            response = json.dumps(response, indent=4)
            response.headers['Content-Type'] = 'application/json'
            return response

    def webhook_result(self, *args):
            queryResult = self.req.get('queryResult').get("queryText")
            speech = queryResult
            print(speech)
            return {
                "fulfillmentText": 'YOLO',
                "source": 'App'
            }

handlers = [
    (r'/', MainHandler),
    (r'(.*)', web.StaticFileHandler, {'path': static_root}),
    (r'/webhook', Webhook,)

]
settings = dict(
    debug=True,
    static_path=static_root
)

application = web.Application(handlers, **settings)
if __name__ == "__main__":
    port = 8090
    application.listen(port)
    print(f"Server running on port : {port}")
    tornado.ioloop.IOLoop.instance().start()

它与烧瓶配合使用效果很好。 当我尝试在tornado中运行此程序（使用ngrok进行隧道挖掘）时，我得到一个警告：tornado.access:405 POST/webhook（127.0.0.1）0.83ms

---

我读过《龙卷风》的文档，但我似乎仍然不知道该怎么做。我假设问题在于Webhook类。我做错了什么？

警告显示了

POST

请求的问题-要处理

POST

您需要在类

Webhook

中使用

def POST（）

方法

它应该是

post（）

，而不是

prepare（）

（用于不同的东西）。
您可以使用

self.write（dictionary）

将其作为

'application/json'

class Webhook(tornado.web.RequestHandler):

    def post(self):
        if self.request.headers.get("Content-Type", "").startswith("application/json"):
            data_input = json.loads(self.request.body)

            print('data_input:', data_input)
            print('data_input json.dumps:', json.dumps(data_input, indent=4))

            data_output = self.webhook_result(data_input) # get as normal dict, not string

            print('data_output:', data_output)
            print('data_output json.dumps:', json.dumps(data_output, indent=4))

            self.write(data_output) # it will send as JSON
        else:
            self.write({'error': 'Wrong Content-Type'}) # it will send as JSON

顺便说一句：如果您将值发送到

webhook\u result（）

，那么您可以获取该值-即作为

data

-并使用它代替

self.req

    def webhook_result(self, data):
        speech = data.get('queryResult').get("queryText")

        print('speech:', speech)

        return {
            "fulfillmentText": 'YOLO',
            "source": 'App'
        }

---

我测试的代码

import tornado
import tornado.web
import json
import os


static_root = os.path.join(os.path.dirname('.'), 'static')


class MainHandler(tornado.web.RequestHandler):

    def get(self):
        #self.render("./templates/index.html")

        # to test POST request but with wrong Content-Type
        self.write('''<form action="/webhook" method="POST"><button>SUBMIT</button></form>''')


class Webhook(tornado.web.RequestHandler):

    def post(self):
        if self.request.headers.get("Content-Type", "").startswith("application/json"):
            data_input = json.loads(self.request.body)

            print('data_input:', data_input)
            print('data_input json.dumps:', json.dumps(data_input, indent=4))

            data_output = self.webhook_result(data_input) # get as normal dict, not string

            print('data_output:', data_output)
            print('data_output json.dumps:', json.dumps(data_output, indent=4))

            self.write(data_output) # it will send as JSON
        else:
            self.write({'error': 'Wrong Content-Type'}) # it will send as JSON

    def webhook_result(self, data):
        speech = data.get('queryResult').get("queryText")

        print('speech:', speech)

        return {
            "fulfillmentText": 'YOLO',
            "source": 'App'
        }


handlers = [
    (r'/', MainHandler),
    (r'/webhook', Webhook),
    # probably it should be as last
    #(r'(.*)', web.StaticFileHandler, {'path': static_root}),
]

settings = dict(
    debug=True,
    static_path=static_root
)

application = tornado.web.Application(handlers, **settings)
if __name__ == "__main__":
    port = 8090
    application.listen(port)
    print(f"Running: http://127.0.0.1:{port}")
    tornado.ioloop.IOLoop.instance().start()

---

顺便说一句：在烧瓶里你可以做

@app.route('/webhook', methods=['POST', 'GET'])
def webhook():
    data_input = request.get_json(silent=True, force=True)
    data_output = makeWebhookResult(data_input)
    return jsonify(data_output)

---

也许你应该在类

Webhook

中创建方法

def post（）

-类似于

MainHandler

中的

def get（）

在测试一些代码后，我确信你必须使用name

def post（）

而不是

def prepare（）

来处理

post

请求。谢谢。这帮了大忙。在最小化和改进代码方面：D

@app.route('/webhook', methods=['POST', 'GET'])
def webhook():
    data_input = request.get_json(silent=True, force=True)
    data_output = makeWebhookResult(data_input)
    return jsonify(data_output)