XSLT/XPath :从所选内容中排除节点

Question

我希望选择满足特定条件的节点列表，在处理这些节点之后，我希望选择其余的节点。我如何在XSLT和XPath中做到这一点。

下面是一个场景，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<type>为“<book>”的所有教程节点，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>

然后，

1. 选择其他<book>节点，这些节点的<type>不是"Tutorial“，并且与在#1中选择的节点的@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']"/>

然后，您将需要适当的模板来处理它们：

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

最后一步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

这里的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>

在所提供的XML文档上应用此转换时为

<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>

生成所需的正确结果

<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>