Grouping Nodes by hardcoding node values in XSLT

问题

<root>
   <Entry>
      <ID>1</ID>
      <Details>
         <Code>A1</Code>
         <Value>1000</Value>
      </Details>
   </Entry>

   <Entry>
      <ID>2</ID>
      <Details>
         <Code>A2</Code>
         <Value>2000</Value>
      </Details>
   </Entry>

   <Entry>
      <ID>3</ID>
      <Details>
         <Code>B1</Code>
         <Value>3000</Value>
      </Details>
   </Entry>

   <Entry>
      <ID>4</ID>
      <Details>
         <Code>B2</Code>
         <Value>4000</Value>
      </Details>
   </Entry>
</root>

I have this input XML which I am looking to group via XSLT wherein the grouping happens by hardcoding node values. Let me explain that in detail:

The grouping needs to happen based on the Code parameter appearing in the node <Code> as follows:

Codes 'A1' and 'A2' needs to be grouped together
Codes 'B1' and 'B2' needs to be grouped together

I am eventually summing the values coming out of <Value> nodes in these groups. So the Output would be as follows:

<Output>
   <Code-group> A </Code-group>
   <Sum> 3000 </Sum>

   <Code-group> B </Code-group>
   <Sum> 7000 </Sum>
</Output>

For this requirement there needs to be hardcoding of the grouping values (to group A1,A2 as A and B1, B2 as B). I am using the word 'hardcoded' because the Codes (A1,A2,B1,B2) can come in any order so I want to rather hardcode the values to look for grouping than looking for Node indices.

I looked at for-each-group method as well as the Muenchian Grouping method but wasn't able to achieve the above group mapping. Any help appreciated!

Thanks in advance

EDIT: The mapping A1,A2 --> A & B1,B2 --> B is a generic example, the actual node values are entirely different than this, so a substring solution wouldn't work. That is why I was focusing on hardcoding to achieve that mapping.

回答1:

The hardcoding requirement is difficult to understand. Perhaps you want to do something like:

XSLT 2.0

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:key name="entry" match="Entry" use="Details/Code"/>

<xsl:template match="/root">
    <Output>
        <Code-group> A </Code-group>
        <Sum>
            <xsl:value-of select="sum(key('entry', ('A1', 'A2'))/Details/Value)" />
        </Sum>
        <Code-group> B </Code-group>
        <Sum>
            <xsl:value-of select="sum(key('entry', ('B1', 'B2'))/Details/Value)" />
        </Sum>
    </Output>
</xsl:template>

</xsl:stylesheet>

回答2:

If you have the option of using for-each-group, I'd definitely use that over Muenchian grouping...

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="/*">
    <Output>
      <xsl:for-each-group select="Entry" group-by="substring(Details/Code,1,1)">
        <Code-group>
          <xsl:value-of select="current-grouping-key()"/>
        </Code-group>
        <Sum>
          <xsl:value-of select="sum(current-group()/Details/Value)"/>
        </Sum>
      </xsl:for-each-group>
    </Output>
  </xsl:template>
  
</xsl:stylesheet>

Fiddle: http://xsltfiddle.liberty-development.net/jxNakzX

If you truly want to hard code the "A" -> A1, A2 and "B" -> B1, B2 mapping, you could use xsl:key and not group at all...

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:key name="A" match="Entry[Details/Code=('A1','A2')]" use="'A'"/>
  <xsl:key name="B" match="Entry[Details/Code=('B1','B2')]" use="'B'"/>

  <xsl:template match="/*">
    <xsl:variable name="ctx" select="."/>
    <Output>
      <xsl:for-each select="('A','B')">
        <xsl:variable name="key" select="."/>
        <Code-group>
          <xsl:value-of select="$key"/>
        </Code-group>
        <Sum>
          <xsl:value-of select="sum($ctx/key($key,$key)/Details/Value)"/>
        </Sum>
      </xsl:for-each>
    </Output>
  </xsl:template>
  
</xsl:stylesheet>

Fiddle: http://xsltfiddle.liberty-development.net/jxNakzX/1

来源：https://stackoverflow.com/questions/64321960/grouping-nodes-by-hardcoding-node-values-in-xslt

标签

xml

group-by

xslt-1.0

xslt-2.0

muenchian-grouping