获取特定层级的元素 [英] Get elements of specific hierarchy level

问题描述

有没有办法使用 XPath 检索特定层次结构级别的所有元素?

Is there any way to retrieve all elements of specific hierarchy level using XPath?

更新.

<A>
   <B>1</B>
   <B>2</B>
</A>
<C>
   <D>3</D>
   <D>4</D>
</C>

我需要检索所有 B 和 D 元素(层次级别 = 2)

I need to retrieve all B and D elements (hierarchy level = 2)

推荐答案

您的示例缺少根元素，因此我假设如下:

Your example lacks a root element, so I assume something like this:

<ROOT>
  <A>
    <B>1</B>
    <B>2</B>
  </A>
  <C>
    <D>3</D>
    <D>4</D>
  </C>
</ROOT>

有了这个，一个简单的版本就是使用适当数量的任何元素"通配符来获得结果:

With this, a simple version would be to just use the appropriate number of 'any element' wildcards to get your result:

xpath = '/*/*/*'

(意思是'选择任意根元素的任意子元素的任意子元素')

(Meaning 'select any child element of any child element of any root element')

或者，如果您想用数字表示级别，可以使用:

Alternatively, if you want to express the level numerically, you could use:

xpath = '//*[count(ancestor::*) = 2]'

(意思是选择任何具有 2 个祖先的元素")

(Meaning 'select any element with 2 ancestors')

编辑/注意:正如 Dimitre Novatchev 正确指出的，区分节点和元素很重要，我修正了我的答案因此.(虽然元素本身就是节点，还有六种其他类型的节点！)

Edit/Note: As Dimitre Novatchev correctly pointed out, it is important to differentiate between nodes and elements, and I fixed my answer accordingly. (While elements are nodes themselves, there are also six other types of nodes!)

通过将基于祖先的 xpath 稍微更改为:

The difference can be illustrated with the given example by altering the ancestor based xpath slightly to:

xpath = '//*[count(ancestor::node())=2]'

这将选择 A 和 B，因为根元素将算作一个祖先节点，而根节点 '/' 算作另一个！

This would select A and B, as the root element would count as one ancestor node, and the root node '/' as another!

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