找到一个只有另一种子元素的元素 [英] Find an element that only has one other kind of child

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问题描述

我想使用 XPath 查找每个

元素，该元素至少有一个

 子元素，没有其他类型的子元素，并且可选的文本节点作为子节点:
<blockquote><pre>YES</pre></blockquote><blockquote><p>NO</p></blockquote><blockquote><pre>NO</pre><p>NO</p></blockquote><blockquote><p>NO</p><pre>NO</pre></blockquote><blockquote><pre>YES</pre><pre>YES</pre></blockquote><blockquote>NO</blockquote></div></body>
这个 XPath 似乎有效，但我怀疑它过于复杂:
//blockquote[pre][not(*[not(name()="pre")])]
是否有更好(更少代码、更高效、更 DRY)的方式来选择我想要的东西?
 解决方案
使用:
//blockquote[* and not(*[not(self::pre)])]
这将选择 XML 文档中的所有 blockquote 元素，这些元素至少有一个子元素，并且没有任何不是 pre 元素的子元素.
这只是双重否定法则 :).
请注意，此表达式比计算所有元素子元素的表达式更有效(因为选择会在找到非 pre 子元素的那一刻停止).
基于 XSLT 的验证:
<xsl:output omit-xml-declaration="yes" indent="yes"/><xsl:strip-space elements="*"/><xsl:template match="/"><xsl:copy-of select="//blockquote[* and not(*[not(self::pre)])]"/></xsl:模板></xsl:stylesheet>
当此转换应用于提供的 XML 文档时:
<blockquote><pre>YES</pre></blockquote><blockquote><p>NO</p></blockquote><blockquote><pre>NO</pre><p>NO</p></blockquote><blockquote><p>NO</p><pre>NO</pre></blockquote><blockquote><pre>YES</pre><pre>YES</pre></blockquote><blockquote>NO</blockquote></div></body>
计算 XPath 表达式并将所选节点复制到输出:
<pre>是</pre></blockquote><blockquote><pre>是</pre><pre>是</pre></blockquote>
I want to use XPath to find every <blockquote> element that has at least one child <pre> element, no other kinds of child elements, and optionally text nodes as children:
<body><div><!-- arbitrary nesting -->
  <blockquote><pre>YES</pre></blockquote>
  <blockquote><p>NO</p></blockquote>
  <blockquote><pre>NO</pre><p>NO</p></blockquote>
  <blockquote><p>NO</p><pre>NO</pre></blockquote>
  <blockquote><pre>YES</pre> <pre>YES</pre></blockquote>
  <blockquote>NO</blockquote>
</div></body>
This XPath appears to work, but I suspect that it's overly complicated:
//blockquote[pre][not(*[not(name()="pre")])]
Is there a better (less code, more efficient, more DRY) way to select what I want?
 解决方案 
Use:
//blockquote[* and not(*[not(self::pre)])]
This selects all blockquote elements in the XML document that have at least one element child and don't have any element child that isn't a pre element.


This is just an application of the double negation law :).

Do note, that this expression is more efficient than one that counts all element children (because the selection stops right at the moment a non-pre child is found).

XSLT - based verification:
<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="/">
     <xsl:copy-of select="//blockquote[* and not(*[not(self::pre)])]"/>
 </xsl:template>
</xsl:stylesheet>
When this transformation is applied on the provided XML document:
<body><div><!-- arbitrary nesting -->
      <blockquote><pre>YES</pre></blockquote>
      <blockquote><p>NO</p></blockquote>
      <blockquote><pre>NO</pre><p>NO</p></blockquote>
      <blockquote><p>NO</p><pre>NO</pre></blockquote>
      <blockquote><pre>YES</pre> <pre>YES</pre></blockquote>
      <blockquote>NO</blockquote>
</div></body>
the XPath expression is evaluated and the selected nodes are copied to the output:
<blockquote>
   <pre>YES</pre>
</blockquote>
<blockquote>
   <pre>YES</pre>
   <pre>YES</pre>
</blockquote>


                        
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找到一个只有另一种子元素的元素 [英] Find an element that only has one other kind of child

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找到一个只有另一种子元素的元素 [英] Find an element that only has one other kind of child

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