XSLT/XPath : 从所选内容中排除节点

Question

我想选择满足特定条件的节点列表，在处理完这些节点后，我想选择剩余的节点。如何在 XSLT 和 XPath 中做到这一点。

下面是场景，我有这个xml

<books>
<book name="Basic XML">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
<book name="Basic XML">
    <type>Tutorial</type>
    <grouping>A</grouping>
</book>
<book name="Basic XSLT">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
<book name="Basic XSLT">
    <type>Tutorial</type>
    <grouping>A</grouping>
</book>
<book name="Basic Java">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
<book name="Basic Java">
    <type>Tutorial</type>
    <grouping>A</grouping>
</book>
<book name="Web Service">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
<book name="C Programming">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
</books>

1。选择“教程”的<type>的所有<book>节点，下面是输出

<books>
<book name="Basic XML">
    <type>Tutorial</type>
    <grouping>A</grouping>
</book>   
<book name="Basic XSLT">
    <type>Tutorial</type>
    <grouping>A</grouping>
</book>
<book name="Basic Java">
    <type>Tutorial</type>
    <grouping>A</grouping>
</book>
</books>

2。然后选择其他<book> 节点没有“教程”的<type> 并且与#1 中选择的@name 不同，输出仅为：

<books>
<book name="Web Service">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
<book name="C Programming">
    <type>Educational</type>
    <grouping>A</grouping>
</book>
</books>

Answer 1

对于第一个查询：

<xsl:apply-templates select="/books/book[type='Tutorial']"/>

对于第二个查询：

<xsl:apply-templates select="/books/book[type!='Tutorial']"/>

那么你需要合适的模板来处理它们：

<xsl:template match="/books/book[type='Tutorial']">
   Do Something...
</xsl:template>

最后是检查当前节点是否也有教程节点

<xsl:template match="/books/book[type!='Tutorial']">
   <xsl:variable name="bookname">
      <xsl:value-of select="@name"/>
   </xsl:variable>
   <xsl:if test="count('/books/book[@name=$bookname and type='Tutorial']')=0">
      Do Something...
   </xsl:if>
</xsl:template>

Answer 2

这是一个完整的解决方案：

<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="node()|@*">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="/*">
  <xsl:variable name="vTutBooks" select="book[type = 'Tutorial']"/>
  <books>
   <xsl:apply-templates select="$vTutBooks"/>
  </books>

  <books>
   <xsl:apply-templates select=
   "book[not(type = 'Tutorial')
       and
         not(@name = $vTutBooks/@name)
        ]"/>
  </books>
 </xsl:template>
</xsl:stylesheet>

当此转换应用于提供的 XML 文档时：

<books>
    <book name="Basic XML">
        <type>Educational</type>
        <grouping>A</grouping>
    </book>
    <book name="Basic XML">
        <type>Tutorial</type>
        <grouping>A</grouping>
    </book>
    <book name="Basic XSLT">
        <type>Educational</type>
        <grouping>A</grouping>
    </book>
    <book name="Basic XSLT">
        <type>Tutorial</type>
        <grouping>A</grouping>
    </book>
    <book name="Basic Java">
        <type>Educational</type>
        <grouping>A</grouping>
    </book>
    <book name="Basic Java">
        <type>Tutorial</type>
        <grouping>A</grouping>
    </book>
    <book name="Web Service">
        <type>Educational</type>
        <grouping>A</grouping>
    </book>
    <book name="C Programming">
        <type>Educational</type>
        <grouping>A</grouping>
    </book>
</books>

产生想要的正确结果：

<books>
   <book name="Basic XML">
      <type>Tutorial</type>
      <grouping>A</grouping>
   </book>
   <book name="Basic XSLT">
      <type>Tutorial</type>
      <grouping>A</grouping>
   </book>
   <book name="Basic Java">
      <type>Tutorial</type>
      <grouping>A</grouping>
   </book>
</books>
<books>
   <book name="Web Service">
      <type>Educational</type>
      <grouping>A</grouping>
   </book>
   <book name="C Programming">
      <type>Educational</type>
      <grouping>A</grouping>
   </book>
</books>