通过Web响应进行XML解析 [英] Xml parsing from web response
本文介绍了通过Web响应进行XML解析的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试让nominatim响应,对数千个城市进行地理编码.
I'm trying to get response from nominatim to geo-code few thousands of cities.
import os
import requests
import xml.etree.ElementTree as ET
txt = open('input.txt', 'r').readlines()
for line in txt:
lp, region, district, municipality, city = line.split('\t')
baseUrl = 'http://nominatim.openstreetmap.org/search/gb/'+region+'/'+district+'/'+municipality+'/'+city+'/?format=xml'
# eg. http://nominatim.openstreetmap.org/search/pl/podkarpackie/stalowowolski/Bojan%C3%B3w/Zapu%C5%9Bcie/?format=xml
resp = requests.get(baseUrl)
resp.encoding = 'UTF-8' # special diacritics
msg = resp.text
# parse response to get lat & long
tree = ET.parse(msg)
root = tree.getroot()
print tree
但结果是:
Traceback (most recent call last):
File "geo_miasta.py", line 17, in <module>
tree = ET.parse(msg)
File "/usr/lib/python2.7/xml/etree/ElementTree.py", line 1182, in parse
tree.parse(source, parser)
File "/usr/lib/python2.7/xml/etree/ElementTree.py", line 647, in parse
source = open(source, "rb")
IOError: [Errno 2] No such file or directory: u'<?xml version="1.0" encoding="UTF-8" ?>\n<searchresults timestamp=\'Tue, 11 Feb 14 21:13:50 +0000\' attribution=\'Data \xa9 OpenStreetMap contributors, ODbL 1.0. http://www.openstreetmap.org/copyright\' querystring=\'\u015awierczyna, Drzewica, opoczy\u0144ski, \u0142\xf3dzkie, gb\' polygon=\'false\' more_url=\'http://nominatim.openstreetmap.org/search?format=xml&exclude_place_ids=&q=%C5%9Awierczyna%2C+Drzewica%2C+opoczy%C5%84ski%2C+%C5%82%C3%B3dzkie%2C+gb\'>\n</searchresults>'
这有什么问题?
想要@rob我的解决方案是:
Thant to @rob my solution is:
#! /usr/bin/env python2.7
# -*- coding: utf-8 -*-
import os
import requests
import xml.etree.ElementTree as ET
txt = open('input.txt', 'r').read().split('\n')
for line in txt:
lp, region, district, municipality, city = line.split('\t')
baseUrl = 'http://nominatim.openstreetmap.org/search/pl/'+region+'/'+district+'/'+municipality+'/'+city+'/?format=xml'
resp = requests.get(baseUrl)
msg = resp.content
tree = ET.fromstring(msg)
for place in tree.findall('place'):
location = '{:5f}\t{:5f}'.format(
float(place.get('lat')),
float(place.get('lon')))
f = open('result.txt', 'a')
f.write(location+'\t'+region+'\t'+district+'\t'+municipality+'\t'+city)
f.close()
推荐答案
You are using xml.etree.ElementTree.parse()
, which takes a filename or a file object as an argument. But, you are not passing a file or file object in, you are passing a unicode string.
尝试 xml.etree.ElementTree.fromstring(text)
.
赞:
tree = ET.fromstring(msg)
这是一个完整的示例程序:
Here is a complete sample program:
import os
import requests
import xml.etree.ElementTree as ET
baseUrl = 'http://nominatim.openstreetmap.org/search/pl/podkarpackie/stalowowolski/Bojan%C3%B3w/Zapu%C5%9Bcie\n/?format=xml'
resp = requests.get(baseUrl)
msg = resp.content
tree = ET.fromstring(msg)
for place in tree.findall('place'):
print u'{:s}: {:+.2f}, {:+.2f}'.format(
place.get('display_name'),
float(place.get('lon')),
float(place.get('lat'))).encode('utf-8')
这篇关于通过Web响应进行XML解析的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
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