Xslt递归转换

Question

我试过了所有方法，但没有结果。我如何使用XSLT转换 - 这样的XML：

<root> 
    <row>topic</row> 
    <row>Topic1</row> 
    <row>words</row> 
    <row>word1</row> 
    <row>word2</row> 
    <row>word3</row> 
    <row>words</row> 
    <row>word4</row> 
    <row>word5</row> 
    <row>word6</row> 
    <row>topic</row> 
    <row>Topic2</row> 
    <row>words</row> 
    <row>word7</row> 
    <row>word8</row> 
    <row>word9</row> 
    <row>topic</row> 
    <row>Topic3</row> 
    <row>words</row> 
    <row>word10</row> 
    <row>word11</row> 
    <row>word12</row> 
</root> 

在这样的XML中？

<topic>Topic1 
    <words>word1word2word3word4word5word6</words> 
</topic> 
<topic>Topic2 
    <words>word7word8word9</words> 
</topic> 
<topic>Topic3 
    <words>word10word11word12</words> 
</topic> 

“主题”和“字”值经常发生在XML和我会第一组由主题，每个主题组中的字由“字” ...提前

非常感谢！

Answer 1

在XSLT 2.0，你可以如此简单：

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/> 

<xsl:template match="root"> 
    <xsl:copy> 
     <xsl:for-each-group select="row" group-starting-with="row[.='topic']"> 
      <topic> 
       <xsl:value-of select="current-group()[2]"/> 
       <words> 
        <xsl:value-of select="current-group()[position() gt 3]" separator=""/> 
       </words> 
      </topic> 
     </xsl:for-each-group> 
    </xsl:copy> 
</xsl:template> 

</xsl:stylesheet>