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Re: muliple level sorting using xsl:sort

2005-09-05 02:43:03
Hi,

Tempore 11:19:01, die 09/05/2005 AD, hinc in 
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com scripsit Aravind J 
<aravindjp(_at_)yahoo(_dot_)com>:

when i tried to sort the data using
 <xsl:for-each select="Part">
<xsl:sort data-type="number"
select="descendant-or-self::code/name"/> it is doing e
first level sorting correctly, ie data is sorted in
1,2,5 order (code/name) . but sub parts of <part> 5 ie
5-3 and 5-4 are not getting sorted , similarly for
children of 5-4 also.

Any idea how we can sort sub parts ie (child <Part> )
also in this case .

Somehow, you'd need to extract the last number of the 'n(-n)*' sequence,
One possible solution is to rely on the fact that the sequence is the same as 
the previous level + '-n' :

"substring(code/name,string-length(ancestor::Part[1]/code/name) + 1 + 
boolean(ancestor::Part))"

e.g. this stylesheet:

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

<xsl:template match="AAAA|child">
<level>
<xsl:for-each select="Part">
        <xsl:sort data-type="number"
          select="substring(code/name,string-length(ancestor::Part[1]/code/name)
          +1+boolean(ancestor::Part))"/>
        <xsl:copy-of select="code/name"/>
        <xsl:apply-templates select="child"/>
</xsl:for-each>
</level>
</xsl:template>

</xsl:stylesheet>

Will return this result:
<level>
        <name>1</name>
        <name>2</name>
        <name>5</name>
        <level>
                <name>5-3</name>
                <name>5-4</name>
                <level>
                        <name>abc</name>
                        <name>def</name>
                </level>
        </level>
</level>


regards,
--
Joris Gillis (http://users.telenet.be/root-jg/me.html)
Ceterum censeo XML omnibus esse utendum

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