xsl reordering of elements

new2xsl · March 31st, 2010, 10:37 PM

I have a large merged dataset from c# dataset merge class as:

<message>
<block1>
<name1>bob</name1>
<field2><2</field2>
<field3>3</field3>
</block1>
<block2>
<name2>bob</name2>
<field5>2</field5>
<field6>3</field6>
</block2>
message

desired output from xsl is:
<group1>
<name1>bob</name1>
<name2>bob</name2>
</group1>
<group2>
<field2><2</field2>
<field5>2</field5>
</group2>
<group3>
<field3>3</field3>
<field6>3</field6>
</group3>

i am needing to reoder the xml and group tags based on equal attribue values
having different element paths in the same xml dataset.

thoughts?

samjudson · April 1st, 2010, 03:42 AM

In XSLT 2.0 you can use the xsl:for-each-group instruction to group on the text of the element:

Code:

   <xsl:template match="message">
   	  <xsl:for-each-group select="*/*" group-by="text()">
			      <xsl:element name="group{position()}">
					       <xsl:copy-of select="current-group()"/>
				     </xsl:element>
   		</xsl:for-each-group>
   </xsl:template>

In XSLT 1.0 (which if you are using C# then you will probably have, although both Saxon and Altova will work from within .Net) then you need to use muenchian grouping. http://www.jenitennison.com/xslt/gro...muenchian.html.

Code:

   <xsl:key name="block" match="/message/*/*" use="text()"/>
   <xsl:template match="message">
   	  <xsl:for-each select="*/*[generate-id() = generate-id(key('block', text())[1])]">
			      <xsl:element name="group{position()}">
							     <xsl:copy-of select="key('block', text())"/>
				     </xsl:element>
   		</xsl:for-each>
   	</xsl:template>

new2xsl · April 2nd, 2010, 08:44 AM

Thanks for the reply, lets say i have this issue:

<tag>
<misc1>
<matchfield>A</matchfield>
<another>1</another>
</misc1>
<misc1>
<matchfield>B</matchfield>
<another>5</another>
</misc1>
<misc2>
<matchfield>A</matchfield>
<anotherone>2</anotherone>
</misc2>
<misc2>
<matchfield>B</matchfield>
<anotherone>9</anotherone>
</misc2>
</tag>

Output needs to group nodes where value of matchfield equals: So where matchfield equals i need to group the nodes under a group tag

<tag>
<group>
<misc1>
<matchfield>A</matchfield>
<another>1</another>
</misc1>
<misc2>
<matchfield>A</matchfield>
<anotherone>2</anotherone>
</misc2>
</group>
<group>
<misc1>
<matchfield>B</matchfield>
<another>5</another>
</misc1>
<misc2>
<matchfield>B</matchfield>
<anotherone>9</anotherone>
</misc2>
</group>
</tag>

Advice would be greatly appreciated.

Martin Honnen · April 2nd, 2010, 08:50 AM

Do you use XSLT 1.0 or 2.0?

new2xsl · April 2nd, 2010, 10:02 AM

XSLT version 1.0

I should have specified.

Thanks.

Martin Honnen · April 2nd, 2010, 11:39 AM

Read the link posted on Muechian grouping. Here is it that technique applied to your sample:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">
  
  <xsl:strip-space elements="*"/>
  <xsl:output indent="yes"/>
  
  <xsl:key name="k1" match="tag/*" use="matchfield"/>
  
  <xsl:template match="tag">
    <xsl:copy>
      <xsl:for-each select="*[generate-id() = generate-id(key('k1', matchfield)[1])]">
        <group>
          <xsl:copy-of select="key('k1', matchfield)"/>
        </group>
      </xsl:for-each>
    </xsl:copy>
  </xsl:template>
  
</xsl:stylesheet>

new2xsl · April 2nd, 2010, 12:45 PM

that was some great info, I think i am getting real close.

How would i handle the added level to the mix where they are not named the same. When i apply the last thread result to the added tag below, i get empty results.

Please let me know what you think.

<tag>
<misc1>
<addedlevel1>
<matchfield>A</matchfield>
<another>1</another>
</addedlevel1>
</misc1>
<misc1>
<addedlevel1>
<matchfield>B</matchfield>
<another>5</another>
</addedlevel1>
</misc1>
<misc2>
<addedlevel2>
<matchfield>A</matchfield>
<anotherone>2</anotherone>
</addedlevel2>
</misc2>
<misc2>
<addedlevel2>
<matchfield>B</matchfield>
<anotherone>9</anotherone>
</addedlevel2>
</misc2>
</tag>

output expected is:

<tag>
<group>
<misc1>
<addedlevel1>
<matchfield>A</matchfield>
<another>1</another>
</addedlevel1>
</misc1>
<misc2>
<addedlevel2>
<matchfield>A</matchfield>
<anotherone>2</anotherone>
</addedlevel2>
</misc2>
</group>
<group>
<misc1>
<addedlevel1>
<matchfield>B</matchfield>
<another>5</another>
</addedlevel1>
</misc1>
<misc2>
<addedlevel2>
<matchfield>B</matchfield>
<anotherone>9</anotherone>
</addedlevel2>
</misc2>
</group>
</tag>

If you get this figured out. I so hope you have an answer to this. Your input has been great.

Martin Honnen · April 2nd, 2010, 12:51 PM

You can use * as a wildcard for the level with different element names:

Code:

<xsl:stylesheet
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="1.0">
  
  <xsl:strip-space elements="*"/>
  <xsl:output indent="yes"/>
  
  <xsl:key name="k1" match="tag/*" use="*/matchfield"/>
  
  <xsl:template match="tag">
    <xsl:copy>
      <xsl:for-each select="*[generate-id() = generate-id(key('k1', */matchfield)[1])]">
        <group>
          <xsl:copy-of select="key('k1', */matchfield)"/>
        </group>
      </xsl:for-each>
    </xsl:copy>
  </xsl:template>
  
</xsl:stylesheet>

new2xsl · April 2nd, 2010, 01:08 PM

If you had to go 2 levels would it be */*/

Martin Honnen · April 2nd, 2010, 01:23 PM

Yes, just try it. And read an XPath tutorial, that is fundamental to doing XSLT.