XSLT转换成表

Question

我已经下面的XML文件

<root> 
 
    <entries> 
 
    <entry> 
 
     <id>1</id> 
 
     <title>entry A</title> 
 
     <group>1</group> 
 
     <category>a</category> 
 
    </entry> 
 
    <entry> 
 
     <id>2</id> 
 
     <title>entry B</title> 
 
     <group>1</group> 
 
     <category>c</category> 
 
    </entry> 
 
    <entry> 
 
     <id>3</id> 
 
     <title>entry C</title> 
 
     <group>2</group> 
 
     <category>b</category> 
 
    </entry> 
 
    <entry> 
 
     <id>4</id> 
 
     <title>entry D</title> 
 
     <group>2</group> 
 
     <category>c</category> 
 
    </entry> 
 
    <entry> 
 
     <id>5</id> 
 
     <title>entry E</title> 
 
     <group>3</group> 
 
     <category>a</category> 
 
    </entry> 
 
    <entry> 
 
     <id>6</id> 
 
     <title>entry F</title> 
 
     <group>4</group> 
 
     <category>c</category> 
 
    </entry> 
 
    </entries> 
 
    <groups> 
 
    <group id="1"> 
 
     <title>Group 1</title> 
 
    </group> 
 
    <group id="2"> 
 
     <title>Group 2</title> 
 
    </group> 
 
    <group id="3"> 
 
     <title>Group 3</title> 
 
    </group> 
 
    <group id="4"> 
 
     <title>Group 4</title> 
 
    </group> 
 
    </groups> 
 
    <categories> 
 
    <category id="a"> 
 
     <title>A</title> 
 
    </category> 
 
    <category id="b"> 
 
     <title>B</title> 
 
    </category> 
 
    <category id="c"> 
 
     <title>C</title> 
 
    </category> 
 
    </categories> 
 
</root>

包含表定义（基团节点对应于行和类别节点对应于列）和表条目。每个条目由组和列ID标识，并且不为所有单元定义条目。

我需要为输出表，像这样的：

<table> 
 
    <thead> 
 
    <tr> 
 
     <th>Groups</th> 
 
     <th>A</th> 
 
     <!-- category name --> 
 
     <th>B</th> 
 
     <!-- category name --> 
 
     <th>C</th> 
 
     <!-- category name --> 
 
    </tr> 
 
    </thead> 
 
    <tbody> 
 
    <tr> 
 
     <th group-id="1">Group 1</th> 
 
     <!-- group name --> 
 
     <th group-id="1" category-id="a"><span>entry A</span></th> 
 
     <th group-id="1" category-id="b"></th> 
 
     <th group-id="1" category-id="c"><span>entry B</span></th> 
 
    </tr> 
 
    <tr> 
 
     <th group-id="3">Group 2</th> 
 
     <!-- group name --> 
 
     <th group-id="2" category-id="a"></th> 
 
     <th group-id="2" category-id="b"><span>entry C</span></th> 
 
     <th group-id="2" category-id="c"><span>entry D</span></th> 
 
    </tr> 
 
    <tr> 
 
     <th group-id="3">Group 3</th> 
 
     <!-- group name --> 
 
     <th group-id="3" category-id="a"><span>entry E</span></th> 
 
     <th group-id="3" category-id="b"></th> 
 
     <th group-id="3" category-id="c"></th> 
 
    </tr> 
 
    <tr> 
 
     <th group-id="4">Group 4</th> 
 
     <!-- group name --> 
 
     <th group-id="4" category-id="a"></th> 
 
     <th group-id="4" category-id="b"></th> 
 
     <th group-id="4" category-id="c"><span>entry F</span></th> 
 
    </tr> 
 
    </tbody> 
 
</table>

这将填补该表的所有单元格空跨度，如果在最初的XML没有数据。

任何帮助表示赞赏！

Answer 1

根据您的意见，你可能需要使用钥匙交叉引用从您给出组：

<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> 

<xsl:output method="html" indent="yes"/> 

<xsl:key name="group-title" match="groups/group/title" use="../@id"/> 

<xsl:key name="entry-group" match="entries/entry" use="group"/> 

<xsl:template match="root"> 
    <table> 
     <thead> 
      <tr> 
       <th>Groups</th> 
       <xsl:for-each select="categories/category"> 
        <th> 
         <xsl:value-of select="title"/> 
        </th> 
       </xsl:for-each> 
      </tr> 
     </thead> 
     <tbody> 
      <xsl:variable name="categories" select="categories/category"/> 
      <xsl:for-each select="groups/group"> 
       <tr> 
        <td group-id="{@id}"> 
         <xsl:value-of select="title"/> 
        </td> 
        <xsl:variable name="group-id" select="@id"/> 
        <xsl:for-each select="$categories"> 
         <xsl:variable name="group" select="key('entry-group', $group-id)"/> 
         <td group-id="{$group-id}" category-id="{@id}"> 
          <xsl:value-of select="$group[category = current()/@id]/title"/> 
         </td> 
        </xsl:for-each>     
       </tr> 
      </xsl:for-each> 
     </tbody> 
    </table> 
</xsl:template> 

</xsl:transform> 

http://xsltransform.net/bEzjRKP/1