我正在尝试从这种输入：
<col title="one">
<cell>a</cell> <cell>b</cell> <cell>c</cell> <cell>d</cell>
</col>
<col title="two">
<cell>e</cell> <cell>f</cell> <cell>g</cell>
</col>
...使用 XSLT 输出 HTML：
<table>
<tr> <th>one</th> <th>two</th> </tr>
<tr> <td>a</td> <td>e</td> </tr>
<tr> <td>b</td> <td>f</td> </tr>
<tr> <td>c</td> <td>g</td> </tr>
<tr> <td>d</td> </tr>
</table>
换句话说，我想执行矩阵转置。我找不到一个简单的方法来做到这一点，我猜可能没有；复杂的怎么样？在 Google 上搜索时，我发现了一些提示，解决这个问题的方法是通过递归。任何想法表示赞赏。
一种可能性是找到<col>包含最多的单元格，然后在嵌套循环中迭代它们。这保证了生成结构上有效的 HTML 表。
<!-- this variable stores the unique ID of the longest <col> -->
<xsl:variable name="vMaxColId">
<xsl:for-each select="/root/col">
<xsl:sort select="count(cell)" data-type="number" order="descending" />
<xsl:if test="position() = 1">
<xsl:value-of select="generate-id()" />
</xsl:if>
</xsl:for-each>
</xsl:variable>
<!-- and this selects the children of that <col> for later iteration -->
<xsl:variable name="vIter" select="
/root/col[generate-id() = $vMaxColId]/cell
" />
<xsl:template match="root">
<xsl:variable name="columns" select="col" />
<table>
<!-- output the <th>s -->
<tr>
<xsl:apply-templates select="$columns/@title" />
</tr>
<!-- make as many <tr>s as there are <cell>s in the longest <col> -->
<xsl:for-each select="$vIter">
<xsl:variable name="pos" select="position()" />
<tr>
<!-- make as many <td>s as there are <col>s -->
<xsl:for-each select="$columns">
<td>
<xsl:value-of select="cell[position() = $pos]" />
</td>
</xsl:for-each>
</tr>
</xsl:for-each>
</table>
</xsl:template>
<xsl:template match="col/@title">
<th>
<xsl:value-of select="." />
</th>
</xsl:template>
应用于
<root>
<col title="one">
<cell>a</cell> <cell>b</cell> <cell>c</cell> <cell>d</cell>
</col>
<col title="two">
<cell>e</cell> <cell>f</cell> <cell>g</cell>
</col>
</root>
这会产生：
<table>
<tr>
<th>one</th> <th>two</th>
</tr>
<tr>
<td>a</td> <td>e</td>
</tr>
<tr>
<td>b</td> <td>f</td>
</tr>
<tr>
<td>c</td> <td>g</td>
</tr>
<tr>
<td>d</td> <td></td>
</tr>
</table>
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