1

对于这个 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>

我怎样才能得到locationID地址A，使用etree。

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')

得到输出None，我在这里做错了什么？

4

1 回答 1

2

首先，你xml的格式不正确。您在发布时应更加小心，并尽量避免让其他用户修复您的数据。

您可以搜索前面的兄弟，例如：

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

或者直接从xpath表达式中执行，例如：

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]

像这样运行它们中的任何一个：

python2 script.py

那个产量：

1
于 2014-02-25T07:29:51.553 回答