用etree python解析XML
-
21-12-2019 - |
题
为此XML
<locations>
<location>
<locationid>1</locationid>
<homeID>281</homeID>
<buildingType>Added</buildingType>
<address>A</address>
<address2>This is address2</address2>
<city>This is city/city>
<state>State here</state>
<zip>1234</zip>
</location>
<location>
<locationid>2</locationid>
<homeID>81</homeID>
<buildingType>Added</buildingType>
<address>B</address>
<address2>This is address2</address2>
<city>This is city/city>
<state>State here</state>
<zip>1234</zip>
</location>
.
.
.
.
<location>
<locationid>10</locationid>
<homeID>21</homeID>
<buildingType>Added</buildingType>
<address>Z</address>
<address2>This is address2</address2>
<city>This is city/city>
<state>State here</state>
<zip>1234</zip>
</location>
</locations>
.
如何使用locationID
获取地址生成的A
。
这是我的代码,
import urllib2
import lxml.etree as ET
url="url for the xml"
xmldata = urllib2.urlopen(url).read()
# print xmldata
root = ET.fromstring(xmldata)
for target in root.xpath('.//location/address[text()="A"]'):
print target.find('LocationID')
.
作为etree
获取输出,我在这里做什么?
解决方案
首先,您的
xml
不是很好的。发布它并尝试避免其他用户来修复数据时,您应该更加小心。
您可以搜索前面的兄弟,如:
import urllib2
import lxml.etree as ET
url="..."
xmldata = urllib2.urlopen(url).read()
root = ET.fromstring(xmldata)
for target in root.xpath('.//location/address[text()="A"]'):
for location in [e for e in target.itersiblings(preceding=True) if e.tag == "locationid"]:
print location.text
.
或直接从xpath
表达式表达式,如:
import urllib2
import lxml.etree as ET
url="..."
xmldata = urllib2.urlopen(url).read()
root = ET.fromstring(xmldata)
print root.xpath('.//location/address[text()="A"]/preceding-sibling::locationid/text()')[0]
.
运行它们中的任何一个:
python2 script.py
.
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