与其他答案不同，

>

 < xsl：stylesheet version =1.0xmlns：xsl =http://www.w3.org/1999/XSL/变换&GT; 
< xsl：output omit-xml-declaration =yesindent =yes/> 
< xsl：strip-space elements =*/> 

< xsl：key name =kRowByDocIdmatch =Rowuse =doc_id/> 

< xsl：template match =/ *> 
Row [generate-id（）= generate-id（key（'kRowByDocId'，doc_id）[1]）]/> 
< / xsl：template> 

< xsl：template match =Row> 
< xsl：for-each select =key（'kRowByDocId'，doc_id）> 
< xsl：sort select =doc_versiondata-type =numberorder =descending/> 

< xsl：if test =position（）= 1>< xsl：copy-of select =。/>< / xsl：if> 
< / xsl：for-each> 
< / xsl：template> 
< / xsl：stylesheet> 
 

当应用于提供的XML文档时：

 < RowSet> 
< Row> 
< msg_id> 1< / msg_id> 
< doc_id> 1< / doc_id> 
< doc_version> 1< / doc_version> 
< / Row> 
< Row> 
< msg_id> 2< / msg_id> 
< doc_id> 1< / doc_id> 
< doc_version> 2< / doc_version> 
< / Row> 
< Row> 
< msg_id> 3< / msg_id> 
< doc_id> 1< / doc_id> 
< doc_version> 3< / doc_version> 
< / Row> 
< Row> 
< msg_id> 4< / msg_id> 
< doc_id> 2< / doc_id> 
< doc_version> 1< / doc_version> 
< / Row> 
< / RowSet> 
 

想要的，正确的结果会产生：
$ b

 < Row> 
< msg_id> 3< / msg_id> 
< doc_id> 1< / doc_id> 
< doc_version> 3< / doc_version> 
< / Row> 
< Row> 
< msg_id> 4< / msg_id> 
< doc_id> 2< / doc_id> 
< doc_version> 1< / doc_version> 
< / Row> 
 

解释：

1. 正确使用 Muenchian分组方法来找到属于每个不同组的一个项目。

正确使用 排序 以查找组中的最大项目。


2. 正确使用 key（） 函数 - 用于选择给定组中的所有项目。

My xml is:

<RowSet>
  <Row>
     <msg_id>1</msg_id>
     <doc_id>1</doc_id>
     <doc_version>1</doc_version>
  </Row>
  <Row>
     <msg_id>2</msg_id>
     <doc_id>1</doc_id>
     <doc_version>2</doc_version>
  </Row>
    <Row>
     <msg_id>3</msg_id>
     <doc_id>1</doc_id>
     <doc_version>3</doc_version>
  </Row>
      <Row>
     <msg_id>4</msg_id>
     <doc_id>2</doc_id>
     <doc_version>1</doc_version>
  </Row>
  <RowSet>

What I need to do:

If there are Rows with the same doc_id, I need to select only node with the bigger doc_version number.

Expected output:

 <RowSet>
    <Row>
     <msg_id>3</msg_id>
     <doc_id>1</doc_id>
     <doc_version>3</doc_version>
   </Row>
      <Row>
     <msg_id>4</msg_id>
     <doc_id>2</doc_id>
     <doc_version>1</doc_version>
  </Row>
  <RowSet>

May be it might be helpful: msg_id is unique, so Row with bigger msg_id for the same doc_id hold the last doc_version.

This transformation works, unlike some other answers:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:key name="kRowByDocId" match="Row" use="doc_id"/>

 <xsl:template match="/*">
    <xsl:apply-templates select=
      "Row[generate-id()=generate-id(key('kRowByDocId', doc_id)[1])]"/>
 </xsl:template>

 <xsl:template match="Row">
     <xsl:for-each select="key('kRowByDocId',doc_id)">
      <xsl:sort select="doc_version" data-type="number" order="descending"/>

      <xsl:if test="position() = 1"><xsl:copy-of select="."/></xsl:if>
     </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

When applied on the provided XML document:

<RowSet>
    <Row>
        <msg_id>1</msg_id>
        <doc_id>1</doc_id>
        <doc_version>1</doc_version>
    </Row>
    <Row>
        <msg_id>2</msg_id>
        <doc_id>1</doc_id>
        <doc_version>2</doc_version>
    </Row>
    <Row>
        <msg_id>3</msg_id>
        <doc_id>1</doc_id>
        <doc_version>3</doc_version>
    </Row>
    <Row>
        <msg_id>4</msg_id>
        <doc_id>2</doc_id>
        <doc_version>1</doc_version>
    </Row>
</RowSet>

the wanted, correct result is produced:

<Row>
   <msg_id>3</msg_id>
   <doc_id>1</doc_id>
   <doc_version>3</doc_version>
</Row>
<Row>
   <msg_id>4</msg_id>
   <doc_id>2</doc_id>
   <doc_version>1</doc_version>
</Row>

Explanation:

1. Proper use of the Muenchian Grouping method for finding one item belonging to each different group.

2. Proper use of sorting for finding a maximum item in a group.

3. Proper use of the key() function -- for selecting all items in a given group.