xml 将基于同级值的节点与 XPath 匹配
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Matching a node based on a sibling's value with XPath
提问by Flavius
Having a XML document like this:
有一个像这样的 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>
How to match all the /records/record/namehaving as sibling /records/record/record-type with the value "A"?
如何将所有/records/record/名称作为兄弟/records/record/record-type 与值“A”匹配？
回答by Flavius
Found it:
找到了：
/records/record/name[../record-type/text() = "A"]
回答by kjhughes
Surprisingly, none of the answers to date on this old question provide the simplestXPath solution.
令人惊讶的是，迄今为止关于这个老问题的答案都没有提供最简单的XPath 解决方案。
This simple XPath
这个简单的 XPath
/records/record[record-type = "A"]/name
selects
选择
<name>svn</name>
<name>blog</name>
as requested.
按照要求。
回答by thomasb
You can also filter a parent element by its children :
您还可以通过其子元素过滤父元素：
/records/record[record-type[text()='A']]/name
/records/record[record-type[text()='A']]/name
