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RE: [xsl] complex sorting

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> This should parse the numeric value and sort, which it does 
> fine, except that all the attributes are stripped off the output.
> 
> <xsl:template match ="I1/@* |I1">
> 	<xsl:if test="I1/@i44='N'">
> 		<xsl:copy>
> 			<xsl:apply-templates select="@* | I1">
> 				<xsl:sort select="@i29" 
> data-type="number" order="descending"/>
> 			</xsl:apply-templates>
> 		</xsl:copy>
> </xsl:if>
> </xsl:template>

Atrtibute nodes will match the above template rule but fail the <xsl:if>
test, so they will produce no output.
> 
> I need to perform a different sort if test ='Y' for the same 
> attribute node, moving the entire record to another section 
> of the document.  that would look something like this.
> 

You may need to use the substring() trick: 

concat(substring(@i44, 1 div (test='N')), 
       substring(@i25, 1 div (test='Y'))

If test='Y' this is
concat(subtring(@i44, infinity), substring(@i25, 0))
which is string(@i25)

while if test='N' it is string(@i44). 

Michael Kay
Software AG
home: Michael.H.Kay@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx 


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