在 XSLT 2.0 中创建具有固定列数的字母索引

问题描述

我有一个这样的 XML:

I have an XML like this:

<xml version="1.0" encoding="UTF-8"?>
<countries>
    <country>
       <name>Latvia</name>
    </country>
    <country>
       <name>USA</name>
    </country>
    <country>
       <name>Australia</name>
    </country>
    <country>
       <name>Indonesia</name>
    </country>
    <country>
       <name>UK</name>
    </country>
    <country>
       <name>India</name>
    </country>
    <country>
       <name>Argentina</name>
    </country>
    <country>
       <name>Chile</name>
    </country>
    <country>
       <name>Singapore</name>
    </country>
    <country>
       <name>New Zeland</name>
    </country>
    <country>
       <name>Kenya</name>
    </country>
    <country>
       <name>Zambia</name>
    </country>
    <country>
       <name>Tunisia</name>
    </country>
</countries>

现在我想在 3 列中创建按字母顺序排列的国家/地区索引.每列将包含作为起始字母的三分之一的字母数量及其对应的国家.

Now I want to create alphabetical index of countries in 3 columns. Each column will contain the number of alpha as one third of the starting alphabets present and its corresponding countries.

如果出现的起始字母数恰好不能被 3 整除，则最后一列可以包含其余部分.

Last column can have the rest of them if number of starting alphabets present happens to be not divisible by 3.

例如，这里我们有以 L、C、U、A、I、S、N、K、Z 和 T 开头的国家/地区名称.

For example, here we have country names starting with L, C, U, A, I, S, N, K, Z and T.

整理后:A C I K L N S T U Z

After arranging: A C I K L N S T U Z

现在，我的索引将有:

         Column1: A, C and I countries

         Column2: K, L and N countries

         Column3: S, T, U and Z countries

因此所需的输出是:

<countries>
  <column1>
     <A>
        <country>
           <name>Argentina</name>
        </country>
        <country>
          <name>Australia</name>
        </country>
     </A>
     <C>
        <country>
           <name>Chile</name>
        </country>
     </C>
     <I>
        <country>
           <name>India</name>
        </country>
        <country>
          <name>Indonesia</name>
        </country>
     </I>
  </column1>

  <column2>
     <K>
        <country>
           <name>Kenya</name>
        </country>
     </K>
     <L>
        <country>
           <name>Latvia</name>
        </country>

     </L>
     <N>
        <country>
           <name>New Zeland</name>
        </country>
     </N>
  </column2>

  <column3>
     <S>
        <country>
           <name>Singapore</name>
        </country>
     </S>
     <T>
        <country>
           <name>Tunisia</name>
        </country>

     </T>
     <U>
        <country>
           <name>UK</name>
        </country>
        <country>
           <name>USA</name>
        </country>
     </U>
     <Z>
        <country>
           <name>Zambia</name>
        </country>
     </Z>
  </column3>
</countries>

请帮忙.我使用的是 xslt 2.0.

Please help. I am using xslt 2.0.

推荐答案

我想出了

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">

<xsl:output indent="yes"/>

<xsl:param name="size" select="3"/>

<xsl:template match="countries">
    <xsl:copy>
      <xsl:variable name="groups">
        <xsl:for-each-group select="country" group-by="substring(name, 1, 1)">
            <xsl:sort select="current-grouping-key()"/>
            <xsl:element name="{current-grouping-key()}">
                <xsl:copy-of select="current-group()"/>
            </xsl:element>
        </xsl:for-each-group>
      </xsl:variable>
      <xsl:for-each select="$groups/*[position() mod $size eq 1]">
        <xsl:if test="position() le $size">
          <xsl:variable name="pos" select="position()"/>
          <xsl:element name="column{position()}">
            <xsl:copy-of select="., following-sibling::*[if ($pos eq $size) then true() else (position() lt $size)]"/>
          </xsl:element>
        </xsl:if>
      </xsl:for-each>
    </xsl:copy>
</xsl:template>


</xsl:transform>

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