Muenchian 分组 - 在节点内分组，而不是在整个文档内分组

Question

我尝试在 XSLT 中使用 Muenchian 分组来对匹配节点进行分组，但我只想在父节点内进行分组，而不是在整个源 XML 文档中进行分组。

给定 XSLT 和 XML 如下（对示例代码的长度表示歉意）：

XSLT

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"> 
 <xsl:output method="html" indent="yes"/>

 <xsl:key name="contacts-by-surname" match="contact" use="surname" />
 <xsl:template match="records">
  <xsl:for-each select="contact[count(. | key('contacts-by-surname', surname)[1]) = 1]">
   <xsl:sort select="surname" />
   <xsl:value-of select="surname" />,<br />
   <xsl:for-each select="key('contacts-by-surname', surname)">
    <xsl:sort select="forename" />
    <xsl:value-of select="forename" /> (<xsl:value-of select="title" />)<br />
   </xsl:for-each>
  </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

XML

<root>
 <records>
  <contact id="0001">
   <title>Mr</title>
   <forename>John</forename>
   <surname>Smith</surname>
  </contact>
  <contact id="0002">
   <title>Dr</title>
   <forename>Amy</forename>
   <surname>Jones</surname>
  </contact>
  <contact id="0003">
   <title>Mrs</title>
   <forename>Mary</forename>
   <surname>Smith</surname>
  </contact>
  <contact id="0004">
   <title>Ms</title>
   <forename>Anne</forename>
   <surname>Jones</surname>
  </contact>
  <contact id="0005">
   <title>Mr</title>
   <forename>Peter</forename>
   <surname>Smith</surname>
  </contact>
  <contact id="0006">
   <title>Dr</title>
   <forename>Indy</forename>
   <surname>Jones</surname>
  </contact>
 </records>
 <records>
  <contact id="0001">
   <title>Mr</title>
   <forename>James</forename>
   <surname>Smith</surname>
  </contact>
  <contact id="0002">
   <title>Dr</title>
   <forename>Mandy</forename>
   <surname>Jones</surname>
  </contact>
  <contact id="0003">
   <title>Mrs</title>
   <forename>Elizabeth</forename>
   <surname>Smith</surname>
  </contact>
  <contact id="0004">
   <title>Ms</title>
   <forename>Sally</forename>
   <surname>Jones</surname>
  </contact>
  <contact id="0005">
   <title>Mr</title>
   <forename>George</forename>
   <surname>Smith</surname>
  </contact>
  <contact id="0006">
   <title>Dr</title>
   <forename>Harry</forename>
   <surname>Jones</surname>
  </contact>
 </records>
</root>

RESULT

Jones,
Amy (Dr)
Anne (Ms)
Harry (Dr)
Indy (Dr)
Mandy (Dr)
Sally (Ms)

Smith,
Elizabeth (Mrs)
George (Mr)
James (Mr)
John (Mr)
Mary (Mrs)
Peter (Mr)

如何在每个 内进行分组 并实现此结果：

Jones,
Amy (Dr)
Anne (Ms)
Indy (Dr)

Smith,
John (Mr)
Mary (Mrs)
Peter (Mr)

Jones,
Harry (Dr)
Mandy (Dr)
Sally (Ms)

Smith,
Elizabeth (Mrs)
George (Mr)
James (Mr)

Answer 1

花了我一些时间......我正要放弃，但还是继续了:)

key 函数的缺点是生成的密钥将始终针对整个 xml。因此，您应该在密钥中连接其他信息以使其更加具体。在下面的示例中，我连接了记录节点的位置，以便获得每个记录的不同姓氏的键。

这是 xslt：

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
  <xsl:output method="html" indent="yes"/>
  <xsl:key name="distinct-surname" match="contact" use="concat(generate-id(..), '|', surname)"/>
  <xsl:template match="records">
    <xsl:for-each select="contact[generate-id() = generate-id(key('distinct-surname', concat(generate-id(..), '|', surname))[1])]">
      <xsl:sort select="surname" />
      <xsl:value-of select="surname" />,<br />
      <xsl:for-each select="key('distinct-surname', concat(generate-id(..), '|', surname))">
        <xsl:sort select="forename" />
        <xsl:value-of select="forename" /> (<xsl:value-of select="title" />)<br />
      </xsl:for-each>
    </xsl:for-each>
  </xsl:template>  
</xsl:stylesheet>

这是结果：

Jones,
Amy (Dr)
Anne (Ms)
Indy (Dr)
Smith,
John (Mr)
Mary (Mrs)
Peter (Mr)
Jones,
Harry (Dr)
Mandy (Dr)
Sally (Ms)
Smith,
Elizabeth (Mrs)
George (Mr)
James (Mr)

请注意，结果也按名字排序。如果您不想按名字排序，则需要删除行

Answer 2

有一种更简单的方法，通过添加一个谓词来确保 muench 测试中涉及的联系人是当前记录的子项。

<xsl:key name="contacts-by-surname" match="contact" use="surname" />
<xsl:template match="records">
  <xsl:for-each select="contact[count(. | key('contacts-by-surname', surname)[generate-id(parent::records) = generate-id(current())][1]) = 1]">
   <xsl:sort select="surname" />
   <xsl:value-of select="surname" />,<br />
   <xsl:for-each select="key('contacts-by-surname', surname)[generate-id(parent::records) = generate-id(current()/parent::records)]">
    <xsl:sort select="forename" />
    <xsl:value-of select="forename" /> (<xsl:value-of select="title" />)<br />
   </xsl:for-each>
  </xsl:for-each>
</xsl:template>