使用XSL在XML中获取不同的节点值

<NewDataSet>
  <SearchResult>
    <Name>HAREDIN </Name>
    <Surname>FEIMI</Surname>
    <FathersName>QAMIL</FathersName>
    <Birthdate>1949-06-13T00:00:00+02:00</Birthdate>
    <CustomerSegment>Individe Standart </CustomerSegment>
  </SearchResult>
  <SearchResult>
    <Name>HARMENAK</Name>
    <Surname>BADEJAN</Surname>
    <FathersName>VARAHAN  </FathersName>
    <Birthdate>1943-10-02T00:00:00+02:00</Birthdate>
    <CustomerSegment>Individe Standart </CustomerSegment>
  </SearchResult>
   <SearchResult>
    <Name>HARMENAK</Name>
    <Surname>BADEJAN</Surname>
    <FathersName>VARAHAN  </FathersName>
    <Birthdate>1943-10-02T00:00:00+02:00</Birthdate>
    <CustomerSegment>Individe Standart </CustomerSegment>
  </SearchResult>
</NewDataSet>

我不会所有的不同名称节点,输出像这样：

<root>

<Name>HAREDIN </Name>

<Name>HARMENAK</Name>

</root>

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

 <xsl:key name="kNamesByVal" match="Name" use="."/>

 <xsl:template match="/">
  <t>
    <xsl:copy-of select=
    "*/*/Name[generate-id()
             =
              generate-id(key('kNamesByVal', .)[1])
              ]
    "/>
  </t>
 </xsl:template>
</xsl:stylesheet>

当应用于提供的XML文档时,生成所需的正确结果：

<t>
   <Name>HAREDIN </Name>
   <Name>HARMENAK</Name>
</t>

不使用密钥的XSLT 2.0解决方案：

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>  

 <xsl:template match="/">
  <t>
   <xsl:for-each-group select="*/*/Name" group-by=".">
     <xsl:copy-of select="."/>
   </xsl:for-each-group>
  </t>
 </xsl:template>
</xsl:stylesheet>