Find maximum value of all child elements and get i

2020-03-04 02:21发布

Using the below XML, I need to figure out which person worked more hours in each site. For instance in the below XML, person 1 worked 8 hours in site 1 but person 2 worked only 6 hours. So result should contain person 1 and site 1 in transformed XML. If the hours are equal, select first person.

EDIT: I want this to be implemented using XSLT 1.0.

<root>
    <WorkSite Person="P1" Site="S1">
        <Hours>8</Hours>
    </WorkSite>
    <WorkSite Person="P1" Site="S2">
        <Hours>2</Hours>
    </WorkSite>
    <WorkSite Person="P1" Site="S3">
        <Hours>9</Hours>
    </WorkSite>
    <WorkSite Person="P2" Site="S1">
        <Hours>6</Hours>
    </WorkSite>
    <WorkSite Person="P2" Site="S2">
        <Hours>10</Hours>
    </WorkSite>
    <WorkSite Person="P2" Site="S3">
        <Hours>2</Hours>
    </WorkSite>
</root>

XSLT transform result should be like this:

<root> 
    <WorkSite Person="P1" Site="S1"/>  
    <WorkSite Person="P2" Site="S2"/> 
    <WorkSite Person="P1" Site="S3"/> 
</root>

标签: xml xslt
3条回答
我欲成王,谁敢阻挡
2楼-- · 2020-03-04 02:43

An XSLT 1.0 solution. This stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:key name="BySite" match="WorkSite" use="@Site"/>
 <xsl:template match="root">
  <root>
   <xsl:for-each select="/*/WorkSite[count(.|key('BySite',@Site)[1])=1]">
    <WorkSite Person="{key('BySite',@Site)
                       [not(key('BySite',@Site)/Hours 
                            > Hours)]/@Person}"
              Site="{@Site}" />
   </xsl:for-each>
  </root>
 </xsl:template>
</xsl:stylesheet>

Output:

<root>
 <WorkSite Person="P1" Site="S1" />
 <WorkSite Person="P2" Site="S2" />
 <WorkSite Person="P1" Site="S3" />
</root>
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Ridiculous、
3楼-- · 2020-03-04 02:52
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

   <xsl:template match="/">
      <root>
         <xsl:for-each-group select="*/WorkSite" group-by="@Site">
            <WorkSite Person="{(current-group()[Hours = max(current-group()/Hours)])[1]/@Person}" Site="{current-grouping-key()}" />
         </xsl:for-each-group>
      </root>
   </xsl:template>

</xsl:stylesheet>
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\"骚年 ilove
4楼-- · 2020-03-04 03:10

This XSLT 1.0 transformation:

<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="kSiteByName" match="@Site" use="."/>

 <xsl:key name="kWorksiteBySite"
   match="WorkSite" use="@Site"/>

 <xsl:variable name="vSites" select=
  "/*/*/@Site[generate-id()
             =
              generate-id(key('kSiteByName',.)[1])
              ]"
  />

 <xsl:template match="/">
  <root>
    <xsl:for-each select="$vSites">
      <xsl:for-each select="key('kWorksiteBySite', .)">
        <xsl:sort select="Hours" data-type="number"
         order="descending"/>
        <xsl:if test="position()=1">
         <xsl:copy>
           <xsl:copy-of select="@*"/>
         </xsl:copy>
        </xsl:if>
      </xsl:for-each>
    </xsl:for-each>
  </root>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document:

<root>
    <WorkSite Person="P1" Site="S1">
        <Hours>8</Hours>
    </WorkSite>
    <WorkSite Person="P1" Site="S2">
        <Hours>2</Hours>
    </WorkSite>
    <WorkSite Person="P1" Site="S3">
        <Hours>9</Hours>
    </WorkSite>
    <WorkSite Person="P2" Site="S1">
        <Hours>6</Hours>
    </WorkSite>
    <WorkSite Person="P2" Site="S2">
        <Hours>10</Hours>
    </WorkSite>
    <WorkSite Person="P2" Site="S3">
        <Hours>2</Hours>
    </WorkSite>
</root>

produces the wanted, correct result:

<root>
    <WorkSite Person="P1" Site="S1"/>
    <WorkSite Person="P2" Site="S2"/>
    <WorkSite Person="P1" Site="S3"/>
</root>

Do note:

  1. The use of the Muenchian method for grouping to find all different Site values.

  2. The way maximum is found by sorting in descending order and getting the first result from the sorted node-list.

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