我有一个如下的xml文件:
<?xml version="1.0" encoding="utf-8"?>
<EDoc CID="1000101" Cname="somename" IName="iname" CSource="e1" Version="1.0">
<RIGLIST>
<RIG RIGID="100001" RIGName="RgName1">
<ListID>
<nodeA nodeAID="1000011" nodeAName="node1A" nodeAExtID="9000011" />
<nodeA nodeAID="1000012" nodeAName="node2A" nodeAExtID="9000012" />
<nodeA nodeAID="1000013" nodeAName="node3A" nodeAExtID="9000013" />
<nodeA nodeAID="1000014" nodeAName="node4A" nodeAExtID="9000014" />
<nodeA nodeAID="1000015" nodeAName="node5A" nodeAExtID="9000015" />
<nodeA nodeAID="1000016" nodeAName="node6A" nodeAExtID="9000016" />
<nodeA nodeAID="1000017" nodeAName="node7A" nodeAExtID="9000017" />
</ListID>
</RIG>
<RIG RIGID="100002" RIGName="RgName2">
<ListID>
<nodeA nodeAID="1000021" nodeAName="node1B" nodeAExtID="9000021" />
<nodeA nodeAID="1000022" nodeAName="node2B" nodeAExtID="9000022" />
<nodeA nodeAID="1000023" nodeAName="node3B" nodeAExtID="9000023" />
</ListID>
</RIG>
</RIGLIST>
</EDoc>
我需要搜索Node值RIGName,如果找到匹配项,则打印出nodeAName的所有值
示例: 搜索RIGName =“ RgName2”会将所有值打印为node1B,node2B,node3B
到目前为止,我只能获得如下第一部分:
import xml.etree.ElementTree as eT
import re
xmlfilePath = "Path of xml file"
tree = eT.parse(xmlfilePath)
root = tree.getroot()
for elem in root.iter("RIGName"):
# print(elem.tag, elem.attrib)
if re.findall(searchtxt, elem.attrib['RIGName'], re.IGNORECASE):
print(elem.attrib)
count += 1
如何仅获取直接子节点值?
答案 0 :(得分:1)
由于lxml
,从xml.etree
切换到much better XPath query language support可以让您一次完成操作:
In [1]: from lxml import etree as ET
In [2]: tree = ET.parse('input.xml')
In [3]: root = tree.getroot()
In [4]: root.xpath('//RIG[@RIGName = "RgName2"]/ListID/nodeA/@nodeAName')
Out[4]: ['node1B', 'node2B', 'node3B']