Question

两个问题：

如何基于完美匹配的ID将第一个与对应的相互链接？ XML源示例：


＆LT; DIV＆GT;



＆lt; div id =“fr”＆gt;



＆lt; seg id =“fr_1”＆gt; abc＆lt; / seg＆gt;
      ＆lt; seg id =“fr_2”＆gt; def＆lt; / seg＆gt;
      ＆lt; seg id =“fr_3”＆gt; ghi＆lt; / seg＆gt;
      ＆lt; seg id =“fr_4”＆gt; jkl＆lt; / seg＆gt;
      ＆lt; seg id =“fr_5”＆gt; mno＆lt; / seg＆gt;
      ＆LT; / DIV＆GT;
      ＆lt; div id =“en”＆gt;
      ＆lt; seg id =“en_1”＆gt; AAA＆lt; / seg＆gt;
      ＆lt; seg id =“en_2”＆gt; BBB＆lt; / seg＆gt;
      ＆lt; seg id =“en_3”＆gt; CCC＆lt; / seg＆gt;
      ＆lt; seg id =“en_4”＆gt; DDD＆lt; / seg＆gt;
      ＆lt; seg id =“en_5”＆gt; EEE＆lt; / seg＆gt;
      ＆LT; / DIV＆GT;
      ＆LT; / DIV＆GT;
如何基于与@corresp建立的链接链接来自两个或多个div的segs？ XML源示例：


＆lt;？xml version =“1.0”encoding =“ISO-8859-1”？＆gt;


＆LT; DIV＆GT;


＆lt; div id =“fr”＆gt;


＆lt; seg id =“fr_1”corresp =“＃en_1”＆gt; abc＆lt; / seg＆gt;

        ＆lt; seg id =“fr_2”corresp =“＃en_2＃en3”＆gt; def＆lt; / seg＆gt;

        ＆lt; seg id =“fr_3”corresp =“＃en_3＃en_4”＆gt; ghi＆lt; / seg＆gt;

        ＆lt; seg id =“fr_4”corresp =“＃en_4＃en_5”＆gt; jkl＆lt; / seg＆gt;

        ＆lt; seg id =“fr_5”corresp =“＃en_6”＆gt; mno＆lt; / seg＆gt;

        ＆LT; / DIV＆GT;
        ＆lt; div id =“en”＆gt;
        ＆lt; seg id =“en_1”corresp =“＃fr_1”＆gt; ab＆lt; / seg＆gt;

        ＆lt; seg id =“en_2”corresp =“＃fr_1＃fr_2”＆gt; cde＆lt; / seg＆gt;

        ＆lt; seg id =“en_3”corresp =“＃fr_2＃fr_3”＆gt; fg＆lt; / seg＆gt;

        ＆lt; seg id =“en_4”corresp =“＃fr_3 fr_4”＆gt; hij＆lt; / seg＆gt;

        ＆lt; seg id =“en_5”corresp =“＃fr_4”＆gt; kl＆lt; / seg＆gt;

        ＆lt; seg id =“en_6”corresp =“＃fr_5”＆gt; mno＆lt; / seg＆gt;

        ＆LT; / DIV＆GT;
        ＆LT; / DIV＆GT;

感谢您的帮助。

Answer 1

这是一个分组问题，使用Muenchian分组http://www.jenitennison.com/xslt/grouping/muenchian.xml解决它的XSLT 1.0;那样的XSLT

<xsl:stylesheet
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:xs="http://www.w3.org/2001/XMLSchema"
  exclude-result-prefixes="xs">

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

<xsl:key name="id" match="seg" use="substring-after(@id, '_')"/>

<xsl:template match="DIV">
  <xsl:copy>
    <ol>
      <xsl:apply-templates select="div/seg[generate-id() = generate-id(key('id', substring-after(@id, '_'))[1])]"/>
    </ol>
  </xsl:copy>
</xsl:template>

<xsl:template match="seg">
  <li>
    <xsl:apply-templates select="key('id', substring-after(@id, '_'))" mode="list"/>
  </li>
</xsl:template>

<xsl:template match="seg" mode="list">
  <xsl:if test="position() > 1">, </xsl:if>
  <xsl:value-of select="."/>
</xsl:template>

</xsl:stylesheet>

转换输入

<DIV>

    <div id="fr">

        <seg id="fr_1">abc</seg>
        <seg id="fr_2">def</seg>
        <seg id="fr_3">ghi</seg>
        <seg id="fr_4">jkl</seg>
        <seg id="fr_5">mno</seg>
        </div>
        <div id="en">
        <seg id="en_1">AAA</seg>
        <seg id="en_2">BBB</seg>
        <seg id="en_3">CCC</seg>
        <seg id="en_4">DDD</seg>
        <seg id="en_5">EEE</seg>
        </div>
        </DIV>

进入输出

<DIV>
   <ol>
      <li>abc, AAA</li>
      <li>def, BBB</li>
      <li>ghi, CCC</li>
      <li>jkl, DDD</li>
      <li>mno, EEE</li>
   </ol>
</DIV>

使用XSLT迭代和链接XML源代码中的元素

1 个答案: