获取没有特定祖先xml xpath的节点 [英] Obtain nodes that don't have specific ancestor xml xpath

问题描述

我想要 xpath，它获取没有祖先的节点，它是特定节点的第一个后代.

I would like to have xpath, that obtains nodes, that don't have ancestor, which is first descendant of specific node.

假设我们有这样的 xml 文档:

Let's assume we have xml document like this:

<a>
  <b>This node</b>
  <c>
    <a>
      <b>not this</b>
      <g>
        <b>not this</b>
      </g>
    </a>
    <a>
      <b>This node</b>
      <c/>
    </a>
  </c>
</a>


<a>
  <c>
    <a>
      <b>not this</b>
    </a>
    <a>
      <b>This node</b> 
    </a>
    <a>
      <b>This node</b> 
    </a>
    <a>
      <b>This node</b> 
    </a>
  </c>
</a>


<d>
  <b>This node</b>
</d>

我想选择文档中所有没有作为祖先节点//a/c/a[1] 的 b 节点.

I would like to select all b nodes in document that don't have as their ancestor node //a/c/a[1].

推荐答案

我想选择所有 b 节点没有作为他们的文件祖先节点//a/c/a[1]

使用这个 XPath 表达式:

//b[not(ancestor::a
             [parent::c[parent::a]
            and
              not(preceding-sibling::a)
             ]
       )
   ]

这将选择文档中所有没有祖先 a 有父 c 父 的文档中的元素a and 具有父 c 的 a 祖先不是其父的第一个 a 子.

This selects all b elements in the document that don't have ancestor a that has a parent c that has parent a and the a ancestor that has parent c is not the first a child of its parent.

给定以下 XML 文档(基于提供的，但格式正确，并且还在应选择的节点中放置了标识文本):

Given the following XML document (based on the provided, but made well-formed and also put identifying text in the nodes that should be selected):

<t>
    <a>
        <b>This node 1</b>
        <c>
            <a>
                <b>not this</b>
                <g>
                    <b>not this</b>
                </g>
            </a>
            <a>
                <b>This node 2</b>
                <c/>
            </a>
        </c>
    </a>
    <a>
        <c>
            <a>
                <b>not this</b>
            </a>
            <a>
                <b>This node 3</b>
            </a>
            <a>
                <b>This node 4</b>
            </a>
            <a>
                <b>This node 5</b>
            </a>
        </c>
    </a>
    <d>
        <b>This node 6</b>
    </d>
</t>

正好选择了想要的 6 个 b 元素.

exactly the wanted 6 b elements are selected.

使用 XSLT 进行验证:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/">
 <xsl:copy-of select=
 "//b[not(ancestor::a
             [parent::c[parent::a]
            and
              not(preceding-sibling::a)
             ]
         )
     ]

 "/>
 </xsl:template>
</xsl:stylesheet>

当此转换应用于上述 XML 文档时，会选择所需的 b 元素并将其复制到输出中.产生了想要的、正确的结果:

when this transformation is applied on the above XML document, exactly the wanted b elements are selected and copied to the output. The wanted, correct result is produced:

<b>This node 1</b>
<b>This node 2</b>
<b>This node 3</b>
<b>This node 4</b>
<b>This node 5</b>
<b>This node 6</b>

这篇关于获取没有特定祖先xml xpath的节点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！