Re: [xsl] rearranging nodes using XSLT

["G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>]

At 2010-02-17 10:32 -0600, a kusa wrote:
> How can I rearrange nodes in xslt?

Remember you are creating a new arrangement of nodes, you are not 
rearranging the old nodes.

> My source XML looks something like this: I am trying to rearrance
> tbd1, tbd2, tbd3 in that order

Then simply process them in that order where you find them 
grouped.  I hope the answer below helps.  It creates the structure 
you cite as your desired output.

. . . . . . . . . . . Ken

T:\ftemp>type akusa.xml
<root>
  <list1>
    <item1>
      <para>sample</para>
    </item1>
    <tbd2><para>test1</para></tbd2>
    <tbd1><para>test1</para></tbd1>
    <list2>
      <item1>
        <para>sample</para>
      </item1>
      <tbd3><para>test1</para></tbd3>
      <tbd1><para>test1</para></tbd1>
      <item1>
        <para>sample</para>
      </item1>
    </list2>
  </list1>
</root>

T:\ftemp>call xslt2 akusa.xml akusa.xsl
<?xml version="1.0" encoding="UTF-8"?>
<root>
  <list1>
      <item1>
         <para>sample</para>
      </item1>
      <tbd1>
         <para>test1</para>
      </tbd1>
      <tbd2>
         <para>test1</para>
      </tbd2>
      <list2>
         <item1>
            <para>sample</para>
         </item1>
         <tbd1>
            <para>test1</para>
         </tbd1>
         <tbd3>
            <para>test1</para>
         </tbd3>
         <item1>
            <para>sample</para>
         </item1>
      </list2>
   </list1>
</root>
T:\ftemp>type akusa.xsl
<?xml version="1.0" encoding="US-ASCII"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
                version="2.0">

<xsl:output indent="yes"/>

<xsl:template match="*[tbd1|tbd2|tbd3]">
  <xsl:copy>
    <xsl:apply-templates select="@*"/>

    <!--find each group of elements in which a sequence of target elements
        are found-->
    <xsl:for-each-group select="*"
                        group-starting-with="*[not(self::tbd1 | self::tbd2 |
                                                   self::tbd3 )]">
     <!--first put out all of those that are not in the group, then put out
         the group of target elements in the desired order-->
     <xsl:apply-templates select="current-group()
                                  [not(self::tbd1 | self::tbd2 | 
self::tbd3 )],
                                  current-group()[self::tbd1],
                                  current-group()[self::tbd2],
                                  current-group()[self::tbd3]"/>
    </xsl:for-each-group>
  </xsl:copy>
</xsl:template>

<xsl:template match="@*|node()"><!--identity for all other nodes-->
  <xsl:copy>
    <xsl:apply-templates select="@*,node()"/>
  </xsl:copy>
</xsl:template>

</xsl:stylesheet>


> <root>
> <list1>
> <item1>
> <para>sample</para>
> </item1>
> <tbd2><para>test1</tbd2>
> <tbd1><para>test1</tbd1>
> <list2>
> <item1>
> <para>sample</para>
> </item1>
> <tbd3><para>test1</tbd3>
> <tbd1><para>test1</tbd1>
> <item1>
> <para>sample</para>
> </item1>
> </list2>
> </list1>
> </root>
>
> My desired output:
>
> <root>
> <list1>
> <item1>
> <para>sample</para>
> </item1>
> <tbd1><para>test1</tbd1>
> <tbd2><para>test1</tbd2>
> <list2>
> <item1>
> <para>sample</para>
> </item1>
>
> <tbd1><para>test1</tbd1>
> <tbd3><para>test1</tbd3>
> <item1>
> <para>sample</para>
> </item1>
> </list2>
> </list1>
> </root>
>
> Thanks in advance for your help.


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