Question

我是XSLT的新手。我知道我需要使用xsl：for-each-group，但除了基本列表之外我无法弄清楚。某种递归会更好吗？任何XSLT 1.0或2.0解决方案都可以。

以下是示例XML。请注意，将数据组织到树结构中的最重要属性是@taxonomy。其他属性@taxonomyName和@level作为可选的辅助属性提供。

<?xml version="1.0" encoding="utf-8"?>
<documents>
    <document level="0" title="Root document test" taxonomy="" taxonomyName="" />
    <document level="1" title="Level one document test" taxonomy="\CategoryI" taxonomyName="CategoryI" />
    <document level="1" title="Level one document test #2" taxonomy="\CategoryII" taxonomyName="CategoryII" />
    <document level="2" title="Level two document test" taxonomy="\CategoryII\SubcategoryA" taxonomyName="SubcategoryA" />
    <document level="2" title="Level two document test #2" taxonomy="\CategoryII\SubcategoryA" taxonomyName="SubcategoryA" />
    <document level="3" title="Level three document test" taxonomy="\CategoryII\SubcategoryA\Microcategory1" taxonomyName="Microcategory1" />
    <document level="2" title="Level two, no level one test" taxonomy="\CategoryIII\SubcategoryZ" taxonomyName="SubcategoryZ" />
</documents>

这是预期的输出。（请注意，输出中不需要缩进。为了便于阅读，我在这里做了。）

<ul>
    <li>Root document test</li>
    <li>CategoryI
        <ul>
            <li>Level one document test</li>
        </ul>
    </li>
    <li>CategoryII
        <ul>
            <li>Level one document test #2</li>
            <li>SubcategoryA
                <ul>
                    <li>Level two document test</li>
                    <li>Level two document test #2</li>
                    <li>Microcategory1
                        <ul>
                            <li>Level three document test</li>
                        </ul>
                    </li>
                </ul>
            </li>
        </ul>
    </li>
    <li>CategoryIII
        <ul>
            <li>SubcategoryZ
                <ul>
                    <li>Level two, no subcategory test</li>
                </ul>
            </li>
        </ul>
    </li>
</ul>

这是我能做的最好的事情。

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
    <xsl:key name="contacts-by-taxonomy" match="document" use="@taxonomy" />
    <xsl:template match="documents">
        <ul>
            <xsl:for-each-group select="document" group-by="@taxonomy">
                <xsl:sort select="@taxonomy" />
                <li>
                    <h3><xsl:value-of select="current-grouping-key()"/></h3>
                    <ul>
                        <xsl:for-each select="current-group()">
                            <li><xsl:value-of select="@title"/></li>
                        </xsl:for-each>
                    </ul>
                </li>
            </xsl:for-each-group>
        </ul>
    </xsl:template>
</xsl:stylesheet>

我会继续嘲笑它，但如果有人能给我一件救生衣，我会永远感激不尽。谢谢！

Answer 1

好的，这是我的解决方案。 :-)基本上它通过树进行递归，并且在每个级别上，它执行for-each-group-by =“@taxonomy的下一级”。

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

   <xsl:template match="documents">
      <ul>
         <xsl:call-template name="tree-depth-n">
            <xsl:with-param name="population" select="document"/>
            <xsl:with-param name="depth" select="0"/>
            <xsl:with-param name="taxonomy-so-far" select="''"/>
         </xsl:call-template>
      </ul>
   </xsl:template>

   <!-- This template is called with a population that are all descendants
      of the same ancestors up to level n. -->      
   <xsl:template name="tree-depth-n">
      <xsl:param name="depth" required="yes"/>
      <xsl:param name="population" required="yes"/>
      <xsl:param name="taxonomy-so-far" required="yes"/>
      <!-- output a <li> for each document that is a leaf at this level,
      and a <li> for each sub-taxon of this level. --> 
      <xsl:for-each-group select="$population"
                    group-by="string(tokenize(@taxonomy, '\\')[$depth + 2])">
         <xsl:sort select="@taxonomy" />
         <xsl:choose>
            <!-- process documents at this level. -->
            <xsl:when test="current-grouping-key() = ''">
              <xsl:for-each select="current-group()">
                 <li><xsl:value-of select="@title"/></li>
              </xsl:for-each>
            </xsl:when>
            <!-- process subcategories -->
            <xsl:otherwise>
               <li>
                  <h3><xsl:value-of select="current-grouping-key()"/></h3>
                  <ul>
                     <!-- recurse -->
                     <xsl:call-template name="tree-depth-n">
                        <xsl:with-param name="population" select="current-group()"/>
                        <xsl:with-param name="depth" select="$depth + 1"/>
                        <xsl:with-param name="taxonomy-so-far" 
                          select="concat($taxonomy-so-far, '\\', current-grouping-key())"/>
                     </xsl:call-template>                    
                  </ul>
              </li>
            </xsl:otherwise>
         </xsl:choose>
      </xsl:for-each-group>      
   </xsl:template>
</xsl:stylesheet>

使用给定输入，输出为：

<ul>
   <li>Root document test</li>
   <li>
      <h3>CategoryI</h3>
      <ul>
         <li>Level one document test</li>
      </ul>
   </li>
   <li>
      <h3>CategoryII</h3>
      <ul>
         <li>Level one document test #2</li>
         <li>
            <h3>SubcategoryA</h3>
            <ul>
               <li>Level two document test</li>
               <li>Level two document test #2</li>
               <li>
                  <h3>Microcategory1</h3>
                  <ul>
                     <li>Level three document test</li>
                  </ul>
               </li>
            </ul>
         </li>
      </ul>
   </li>
   <li>
      <h3>CategoryIII</h3>
      <ul>
         <li>
            <h3>SubcategoryZ</h3>
            <ul>
               <li>Level two, no level one test</li>
            </ul>
         </li>
      </ul>
   </li>
</ul>

我认为这就是你想要的。（我在您的XSL尝试中将<h3>放在那里，用于类别名称而不是文档标题。）

XSLT：扁平XML到嵌套HTML列表

1 个答案: