根据兄弟节点的值与 XPath 匹配节点

拥有这样的 XML 文档:

<?xml version="1.0" encoding="UTF-8"?>
<records type="array">
<record>
<name>svn</name>
<record-type>A</record-type>
<ttl type="integer">86400</ttl>
<zone-id type="integer">69075</zone-id>
<aux type="integer">0</aux>
<id type="integer">xxx</id>
<active>Y</active>
<data>xxx.xxx.xxx.xxx</data>
</record>
<record>
<name>domain.tld.</name>
<record-type>NS</record-type>
<ttl type="integer">86400</ttl>
<zone-id type="integer">xxx</zone-id>
<aux type="integer">0</aux>
<id type="integer">xxx</id>
<active>Y</active>
<data>domain.tld.</data>
</record>
<record>
<name>blog</name>
<record-type>A</record-type>
<ttl type="integer">86400</ttl>
<zone-id type="integer">xxx</zone-id>
<aux type="integer">0</aux>
<id type="integer">xxx</id>
<active>Y</active>
<data>xxx.xxx.xxx.xxx</data>
</record>
</records>

如何匹配所有的 /records/record/name作为兄弟 /records/record/record-type与值 A？

xml
xpath

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最佳答案

Found it:

/records/record/name[../record-type/text() = "A"]

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You can also filter a parent element by its children :

/records/record[record-type[text()='A']]/name

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Surprisingly, none of the answers to date on this old question provide the simplest XPath solution.

This simple XPath

/records/record[record-type = "A"]/name

selects

<name>svn</name>
<name>blog</name>

as requested.