Question

我刚开始尝试创建API，我很想制作一个仅在API发布请求后才返回XML格式数据的应用。

下面的烧瓶相关代码只是如何制作XML response from the git repo示例以进行查看的美化版本。

from simplexml import dumps
from flask import make_response, Flask
from flask_restful import Api, Resource

def output_xml(data, code, headers=None):
    """Makes a Flask response with a XML encoded body"""
    resp = make_response(dumps({'response' :data}), code)
    resp.headers.extend(headers or {})
    return resp


tou = [{
    "startTime": "2:00PM",
    "duration": "6 Hours",
    "numberOfIntervals" : 3,
    "intervalDuration" : "1 hour,4 hour,1 hour",
    "typicalIntervalValues" : "$0.50,$0.75,$0.50"
}]



app = Flask(__name__)
api = Api(app, default_mediatype='application/xml')
api.representations['application/xml'] = output_xml


class call(Resource):
    def post(self):
        callInfo = {"notification" : "True",
                "startTime" : "2:00PM",
                "duration" : "6 Hours",
                "randomization": "None",
                "rampUp" : "None",
                "recovery" : "None",
                "numberOfSignals" : "2",
                 "signalNameSimple" : [{
                    "signalType" : "level",
                    "units" : "None",
                    "numberOfIntervals" : "None",
                    "intervalDuration" : "None",
                    "typicalIntervalValues" : "1,2,1",
                    "signalTarget" : "None"}],
                 "signalNameElectricityPrice" : [{
                    "signalType" : "price",
                    "units" : "USD per kWh",
                    "numberOfIntervals" : "3",
                    "intervalDuration" : "1 hour,4 hour,1 hour",
                    "typicalIntervalValues" : "$0.50,$0.75,$0.50",
                    "signalTarget" : "None"}]
                        }
        return callInfo



api.add_resource(call, '/api/v1/cpp/scenario/three')


if __name__ == '__main__':
    app.run(debug=False)

我遇到的问题（也许这甚至不是问题）是Flask-restful应用程序仅在请求标头中定义了XML响应，否则返回JSON响应。例如，在下面的代码中，如果我没有headers={'Accept': 'application/xml'}，则此脚本将出错，因为Flask-restful应用程序响应为JSON格式。

import requests
import xml
from xml.etree import ElementTree


response = requests.post("http://127.0.0.1:5000/api/v1/cpp/scenario/three", headers={'Accept': 'application/xml'})
#response = requests.post("http://127.0.0.1:5000/api/v1/cpp/scenario/three")


root = ElementTree.fromstring(response.content)
for child in root.iter('*'):
    print(child.tag)

问题，如何修改仅返回XML的Flask App？如果这是一个愚蠢的问题，有人可以详细说明吗API为什么需要JSON和XML？我希望学习如何仅使用XML来实现的现实场景，并且我担心拥有多余的headers={'Accept': 'application/xml'}会造成问题，因为与API通信的设备只是“机器”，而不是人具有自定义的python编码功能。如果有人可以帮助我了解API行业的典型特征，那也可能会有所帮助：）

烧瓶仅宁静输出XML

0 个答案: