从Python中的URL读取XML文件

Question

我正在使用一个名为OpenTripPlanner的开源项目，这是一个我计划用来模拟在给定时间从一个地点到另一个地点的许多行程的工具。到目前为止，我已经找到了包含行程所有信息的XML文件所在的URL。XML是根据请求构建的，因此URL不是静态的。URL如下所示：

http://localhost:8080/otp/routers/default/plan?fromPlace=48.40915,%20-71.04996&toPlace=48.41428,%20-71.06996&date=2017/12/04&time=8:00:00&mode=TRANSIT,WALK

（需要运行OpenTripPlanner服务器才能打开它）

现在，我想读取这些XML文件并使用python3进行一些数据分析，但是我找不到读取这些文件的方法。我试着用urllib.请求在本地下载文件，但是我从这里得到的文件格式很奇怪。看起来像这样

{"requestParameters":{"date":"2017/12/04","mode":"TRANSIT,WALK","fromPlace":"48.40915, -71.04996","toPlace":"48.41428, -71.06996","time":"8:00:00"},"plan":{"date":1512392400000,"from":{"name":"Origin","lon":-71.04996,"lat":48.40915,"orig":"","vertexType":"NORMAL"},"to":{"name":"Destination","lon":-71.06996,"lat":48.41428,"orig":"","vertexType":"NORMAL"},"itineraries":[{"duration":1538,"startTime":1512392809000,"endTime":1512394347000,"walkTime":934,"transitTime":602,"waitingTime":2,"walkDistance":1189.6595112715966,"walkLimitExceeded":false,"elevationLost":0.0,"elevationGained":0.0,"transfers":0,"legs":[{"startTime":1512392809000,"endTime":1512393537000,"departureDelay":0,"arrivalDelay":0,"realTime":false,"distance":926.553,"pathway":false,"mode":"WALK","route":"","agencyTimeZoneOffset":-18000000,"interlineWithPreviousLeg":false,"from":{"name":"Origin","lon":-71.04996,"lat":48.40915,"departure":1512392809000,"orig":"","vertexType":"NORMAL"},"to":{"name":"Roitelets / Martinets","stopId":"1:370","stopCode":"370","lon":-71.047688,"lat":48.401531,"arrival":1512393537000,"departure":1512393538000,"stopIndex":15,"stopSequence":16,"vertexType":"TRANSIT"},"legGeometry":{"points":"s{mfHb{spL|ExBp@sDl@V@@lB|@j@FL?j@GbCk@|A]vEsA^KBA|C{@pCeACS~CuA`@Q","length":19},"rentedBike":false,"transitLeg":false,"duration":728.0,"steps":[{"distance":131.991,"relativeDirection":"DEPART","streetName":"Rue D.-V.-Morrier","absoluteDirection":"SOUTH","stayOn":false,"area":false,"bogusName":false,"lon":-71.04961760502248,"lat":48.4090671692228,"elevation":[]},{"distance":72.319,"relativeDirection":"LEFT","streetName":"Rue Lorenzo-Genest","absoluteDirection":"EAST","stayOn":false,"area":false,"bogusName":false,"lon":-71.0502299,"lat":48.4079519,"elevation":[]}

当我试图在浏览器中打开文件时，我得到一个错误

XML Parsing Error: not well-formed
Location: http://localhost:63342/XML_reader/file.xml?_ijt=e1d6h53s4mh1ak94sqortejf9v
Line Number 1, Column 1: ...

我使用的脚本非常简单，看起来像这样

^{pr2}$

如何使输出的XML文件格式良好？还有别的办法吗urllib.请求我想试试吗？在

非常感谢

Answer 1

不需要将JSON（JSON）文件作为XML数据导入

import urllib.request
import json
from pprint import pprint

testfile = urllib.request.URLopener()
file_name = 'http://localhost:8080/otp/routers/default/plan?fromPlace=48.40915,%20-71.04996&toPlace=48.41428,%20-71.06996&date=2017/12/04&time=8:00:00&mode=TRANSIT,WALK'
testfile.retrieve(file_name, "file.json")

data = json.load(open('file.json'))
pprint(data)

• json.load读取JSON数据并转换为Python对象（https://docs.python.org/2/library/json.html?highlight=json%20load#json.load）
• pprint用于“漂亮地打印”JSON数据（https://docs.python.org/2/library/pprint.html）