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，我计划使用该工具来模拟在给定时间从一个点到另一个点的大量行程。到目前为止，我已经设法找到 URL，其中包含有关行程的所有信息的 XML 文件所在的位置。 XML 是根据要求构建的，因此 URL 不是静态的。 URL 看起来像这样：

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OpenTripPlanner服务器运行打开）

现在，我想读取这些 XML 文件并使用 python 3 进行一些数据分析，但我找不到读取这些文件的方法。我尝试使用 urllib.request 在本地下载文件，但我从中获得的文件格式很奇怪。它看起来像这样

{"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":[]}

当我尝试在浏览器中打开文件时，我收到一条错误消息

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

我使用的脚本很简单，看起来像这样

import urllib.request

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.xml")

如何使输出的 XML 文件格式正确？除了 urllib.request 之外，还有其他我可能想尝试的方法吗？

非常感谢

Answer 1

要将此文件导入为 JSON 数据（而不是 XML），您需要 JSON 库

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 "Pretty printing" JSON 数据 (https://docs.python.org/2/library/pprint.html)

从 Python 中的 URL 读取 XML 文件

Read XML file from URL in Python

python

xml

urllib

opentripplanner