使用 xpath 查找倒数第二个节点 [英] find next-to-last node with xpath

问题描述

我有一个带有 chapters 和嵌套的 sections 的 XML 文档.我试图找到任何部分的第一个二级部分祖先.这是 ancestor-or-self 轴中的倒数第二部分.伪代码:

I have a XML document with chapters and nested sections. I am trying to find, for any section, the first second-level section ancestor. That is the next-to-last section in the ancestor-or-self axis. pseudo-code:

<chapter><title>mychapter</title>
  <section><title>first</title>
     <section><title>second</title>
       <more/><stuff/>
     </section>
  </section>
</chapter>

我的选择器:

<xsl:apply-templates 
    select="ancestor-or-self::section[last()-1]" mode="title.markup" />

当然在 last()-1 没有定义之前一直有效(当前节点是 first 部分).

Of course that works until last()-1 isn't defined (the current node is the first section).

如果当前节点低于 second 部分，我想要标题 second.否则我想要标题first.

If the current node is below the second section, i want the title second. Otherwise I want the title first.

推荐答案

用这个替换你的 xpath:

Replace your xpath with this:

ancestor-or-self::section[position()=last()-1 or count(ancestor::section)=0][1]

由于您已经可以在所有情况下找到正确的节点，除了一个，我将您的 xpath 更新为 也 找到 first 部分( 或 count(ancestor::section)=0)，然后选择 ([1]) 第一个匹配项(以相反的文档顺序，因为我们使用的是 ancestor-or-self代码>轴).

Since you can already find the right node in all cases except one, I updated your xpath to also find the first section (or count(ancestor::section)=0), and then select ([1]) the first match (in reverse document order, since we are using the ancestor-or-self axis).

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