xpath - XPath：如何检查相似节点的多个属性

Question

如果我有一些 xml，例如：

    <root>
   <customers>
        <customer firstname="Joe" lastname="Bloggs" description="Member of the Bloggs family"/>
        <customer firstname="Joe" lastname="Soap" description="Member of the Soap family"/>
        <customer firstname="Fred" lastname="Bloggs" description="Member of the Bloggs family"/>
        <customer firstname="Jane" lastname="Bloggs" description="Is a member of the Bloggs family"/>
   </customers>
 </root>

我如何在纯 XPath（而不是 XSLT）中获得一个 xpath 表达式，该表达式检测 lastname 相同但描述不同的行？那么它会拉出上面的最后一个节点吗？

Answer 1

我如何在纯 XPath（而不是 XSLT）中获得一个 xpath 表达式，该表达式检测 lastname 相同但描述不同的行？

以下是使用单个 XPath 表达式执行此操作的方法：

   "/*/*/customer
        [@lastname='Bloggs'
       and
        not(@description
           = preceding-sibling::*[@lastname='Bloggs']/@description
            )
        ]"

此表达式选择属性等于“Bloggs”且属性值不同的所有<customer>元素。lastnamedescription

选择的节点是：

<customer firstname="Joe" lastname="Bloggs" description="Member of the Bloggs family"/>
<customer firstname="Jane" lastname="Bloggs" description="Is a member of the Bloggs family"/>

Answer 2

/root/customers/customer[@lastname='Bloggs'
  and not(@description = preceding-sibling::*[@lastname='Bloggs']/@description)
  and not(@description = following-sibling::*[@lastname='Bloggs']/@description)]

不过，分步执行会更好。