在foreach中对父节点结构进行XSLT复制

Question

我有一个输入XML，需要转换为下面给出的输出格式。我面临的问题是将确切的标签"IssuedList"复制到输出。因为我有一个for每个循环它不允许应用相同的变换。还有其他方法可以达到同样的效果吗?下面是我尝试过的XSLT

输入XML:

     <Books>
    <Book>
    <Id>1</Id>
    <Name>ABC</Name>
    <Categories RefId="1">
      <Priority>High</Priority>
    </Category>
    </Categories>
    <Categories RefId="2">
     <Category>
      <Priority>Low</Priority>
     <Category>
    </Categories>
    <IssuedList>
    <IssueList>
     <Number>1</Number>
     <Name>ABC</Name>
     </IssueList>
     <IssueList>
     <Number>1</Number>
     <Name>ABC</Name>
     </IssueList>
    </IssuedList>
    </Book>
    <Book>
    <Id>2</Id>
    <Name>DEF</Name>
    <Categories RefId="1">
     <Category>
      <Priority>High</Priority>
    </Category>
   </Categories>
    <Categories RefId="2">
    <Category>
      <Priority>Low</Priority>
    </Category>
    </Categories>
    <IssuedList>
    <IssueList>
     <Number>1</Number>
     <Name>DEF</Name>
     </IssueList>
    </IssuedList>
    </Book>   

并希望看到转换后的输出为

 <Books>
    <Book>
    <Id>1</Id>
    <Name>ABC</Name>
    <RefId>1</RefId>
    <Priority>High</Priority>
    <IssuedList>
     <IssueList>
     <Number>1</Number>
     <Name>ABC</Name>
     </IssueList>
     <IssueList>
     <Number>1</Number>
     <Name>ABC</Name>
     </IssueList>
    </IssuedList>
    </Book>
    <Book>
    <Id>1</Id>
    <Name>ABC</Name>
    <RefId>2</RefId>
    <Priority>Low</Priority>
    <IssuedList>
     <IssueList>
     <Number>1</Number>
     <Name>ABC</Name>
     </IssueList>
     <IssueList>
     <Number>1</Number>
     <Name>ABC</Name>
     </IssueList>
    </IssuedList>
    </Book>
    <Book>
    <Id>2</Id>
    <Name>DEF</Name>
    <RefId>1</RefId>
    <Priority>High</Priority>
    <IssuedList>
    <IssueList>
     <Number>1</Number>
     <Name>DEF</Name>
     </IssueList>
    </IssuedList>
    </Book>
    <Book>
    <Id>2</Id>
    <Name>DEF</Name>
    <RefId>2</RefId>
    <Priority>Low</Priority>
    <IssuedList>
    <IssueList>
     <Number>1</Number>
     <Name>DEF</Name>
     </IssueList>
    </IssuedList>
    </Book>
    </Books>

我将XSLT写成如下

<xsl:for-each select ="/Books/Book">
   <xsl:variable name="Id" select ="Id"></xsl:variable>
   <xsl:variable name="Name" select ="Name"></xsl:variable>
   <xsl:for-each select ="Categories/Category">
            <Book>                          
                <RefId>
                  <xsl:value-of select ="@RefId"></xsl:value-of>
                </RefId>
                 <Id>
                  <xsl:value-of select ="$Id"></xsl:value-of>
                </Id>
                <Name>
                  <xsl:value-of select ="$Name"></xsl:value-of>
                </Name>
 </Book>
</xsl:for-each>
</xsl:for-each>

Answer 1

我觉得你把事情弄得太复杂了。试一试:

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/">
    <Books>
        <xsl:for-each select ="Books/Book/Category">
            <Book>
                <xsl:copy-of select ="../Id"/>
                <xsl:copy-of select ="../Name"/>
                <RefId><xsl:value-of select ="@RefId"/></RefId>
                <xsl:copy-of select ="Priority"/>
                <xsl:copy-of select ="../IssuedList"/>
            </Book>
        </xsl:for-each>
    </Books>
</xsl:template>
</xsl:stylesheet>