XSLT - 相较于同级发表排序

Question

我有，我希望下面的XML结构进行排序

<ESTABLISHMENTS> 
    <ESTABLISHMENT> 
     <NAME>A</NAME> 
     <LOCATION> 
      <LOCALITY>A</LOCALITY> 
     </LOCATION> 
     <RATING> 
      <LEVEL>1</LEVEL> 
      <SCORE>50</SCORE> 
     </RATING> 
    </ESTABLISHMENT> 
    <ESTABLISHMENT> 
     <NAME>C</NAME> 
     <LOCATION> 
      <LOCALITY>B</LOCALITY> 
     </LOCATION> 
     <RATING> 
      <LEVEL>2</LEVEL> 
      <SCORE>50</SCORE> 
     </RATING> 
    </ESTABLISHMENT> 
    <ESTABLISHMENT> 
     <NAME>B</NAME> 
     <LOCATION> 
      <LOCALITY>B</LOCALITY> 
     </LOCATION> 
     <RATING> 
      <LEVEL>2</LEVEL> 
      <SCORE>50</SCORE> 
     </RATING> 
    </ESTABLISHMENT> 
</ESTABLISHMENTS> 

我的目标是通过地方先进行排序，然后按等级，然后通过功能评分，最后通过名字。 我设法得到排序使用下面的代码工作

<?xml version="1.0" encoding="UTF-8"?> 
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> 
    <xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes" omit-xml-declaration="no"/> 
    <xsl:key name="establishment-by-locality" match="ESTABLISHMENTS/ESTABLISHMENT/LOCATION" 
     use="LOCALITY"/> 
    <xsl:template match="EXPORT"> 
    <xsl:for-each select="ESTABLISHMENTS/ESTABLISHMENT/LOCATION[count(.| key('establishment-by-locality', LOCALITY)[1]) = 1]"> 
     <xsl:sort select="LOCALITY"/> 
     <xsl:for-each select="key('establishment-by-locality', LOCALITY)"> 
       <xsl:sort select="../RATING/LEVEL" data-type="number" order="descending"/> 
       <xsl:sort select="../RATING/MERIT_SCORE" data-type="number" order="descending"/> 
       <xsl:sort select="../NAME"/> 

       <xsl:apply-templates select="ancestor::ESTABLISHMENT" mode="test"/> 
      </xsl:for-each> 
    </xsl:for-each> 
    </xsl:template> 
    <xsl:template match="node()" mode="test">  
      <hotel>  
      <name> 
      <xsl:value-of select="NAME"/> 
      </name> 
      <town> 
      <xsl:value-of select="LOCATION/LOCALITY"/> 
      </town> 
      <level> 
      <xsl:value-of select="RATING/LEVEL"/> 
      </level> 
      <score> 
      <xsl:value-of select="RATING/MERIT_SCORE"/> 
     </score> 
    </hotel> 
</xsl:template></xsl:stylesheet> 

然而，一旦XML已经整理我想在“地方”价值观对新程序建立比较，但我的XPath仍然受到限制使用原始树。不幸的是，我仅限于XSLT 1.0，因为目标是Adobe InDesign。

任何帮助非常感谢。

EDIT

比较的目的是为了决定是否插入和附加的元件，即如果所述地点的不同添加一个节头。

的最终期望的输出是大致如下

<establishments> 
    <town> 
     <townName>A</townName> 
     <hotel> 
      <name>A</name> 
      <town>A</town> 
      <level>1</level> 
      <score>50</score> 
     </hotel> 
    </town> 
    <town> 
     <townName>B</townName> 
     <hotel> 
      <name>B</name> 
      <town>B</town> 
      <level>2</level> 
      <score>50</score> 
     </hotel> 
     <hotel> 
      <name>C</name> 
      <town>B</town> 
      <level>2</level> 
      <score>50</score> 
     </hotel> 
    </town> 
</establishments> 

Answer 1

我看到的是不需要这种比较。期望的输出可以通过应用以下的样式表到您的示例输入可以容易地获得：

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/> 
<xsl:strip-space elements="*"/> 

<xsl:key name="establishment-by-locality" match="ESTABLISHMENT" use="LOCATION/LOCALITY"/> 

<xsl:template match="/ESTABLISHMENTS"> 
    <establishments> 
     <xsl:for-each select="ESTABLISHMENT[count(.| key('establishment-by-locality', LOCATION/LOCALITY)[1]) = 1]"> 
      <xsl:sort select="LOCATION/LOCALITY"/> 
      <town> 
       <townName> 
        <xsl:value-of select="LOCATION/LOCALITY"/> 
       </townName> 
       <xsl:apply-templates select="key('establishment-by-locality', LOCATION/LOCALITY)"> 
        <xsl:sort select="RATING/LEVEL" data-type="number" order="descending"/> 
        <xsl:sort select="RATING/SCORE" data-type="number" order="descending"/> 
        <xsl:sort select="NAME"/> 
       </xsl:apply-templates> 
      </town> 
     </xsl:for-each> 
    </establishments> 
</xsl:template> 

<xsl:template match="ESTABLISHMENT">  
    <hotel>  
     <name> 
      <xsl:value-of select="NAME"/> 
     </name> 
     <town> 
      <xsl:value-of select="LOCATION/LOCALITY"/> 
     </town> 
     <level> 
      <xsl:value-of select="RATING/LEVEL"/> 
     </level> 
     <score> 
      <xsl:value-of select="RATING/SCORE"/> 
     </score> 
    </hotel> 
</xsl:template> 

</xsl:stylesheet> 

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注：显然你的实际输入具有略微不同的结构。你将不得不自己做出调整。