Re: [xsl] Sorting network addresses

[Craig W <codecraig@xxxxxxxxx>]

ok my xml looks like this

<networks>

    <network>
        <address>1.1.1.1</address>
    </network>
    <network>
        <address>170.5.2.4</address>
    </network>
    <network>
        <address>2.3.1.2</address>
    </network>

</networks>


my XSLT looks like

<?xml version="1.0" encoding="UTF-8"?>
<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="no"/>
	<xsl:strip-space elements="*" />
	<xsl:variable name="newline">
		<xsl:text>
		</xsl:text>
	</xsl:variable>

	<xsl:template match="/">
		<xsl:value-of select="$newline" />
		<networks>
		<xsl:value-of select="$newline" />
				<xsl:for-each select="networks/network">
					<xsl:sort data-type="number" select="substring-before(address,'.')"/>
					<xsl:sort data-type="number"
select="substring-before(substring-after(address,'.'),'.')"/>
					<xsl:sort data-type="number"
select="substring-before(substring-after(substring-after(address,'.'),'.'),'.')"/>
					<xsl:sort data-type="number"
select="substring-after(substring-after(substring-after(address,'.'),'.'),'.')"/>
						<network>
							<xsl:value-of select="$newline" />	
							<srcaddr>
								<xsl:value-of select="address" />
							</srcaddr>
							<xsl:value-of select="$newline" />
						</network>
						<xsl:value-of select="$newline" />
				</xsl:for-each>
		</networks>
	</xsl:template>
</xsl:stylesheet>


...essentially i want to take the source XML sort it by the address
element, and write out the result (as a sorted version of the
source)..

the result i am getting now is like this

<?xml version="1.0" encoding="UTF-8"?>
		<networks>
			<network>
				<address>1.1.1.1</address>
			</network>
			<network>
				<address>1.1.1.1</address>
			</network>
			<network>
				<address>1.1.1.1</address>
			</network>
			<network>
				<address>1.1.1.1</address>
			</network>
			<network>
				<address>2.3.1.2</address>
			</network>
			<network>
				<address>170.5.2.4</address>
			</network>
			<network>
				<address>2.3.1.2</address>
			</network>
			<network>
				<address>2.3.1.2</address>
			</network>
			<network>
				<address>2.3.1.2</address>
			</network>
			<network>
				<address>170.5.2.4</address>
			</network>
			<network>
				<address>170.5.2.4</address>
			</network>
			<network>
				<address>170.5.2.4</address>
			</network>
		</networks>

I should only have 3 networks not twelve....i am guessing something is
wrong with my sort/for-each.

thanks

On Wed, 2 Mar 2005 13:05:35 GMT, David Carlisle <davidc@xxxxxxxxx> wrote:
> 
> > Is there a way to extend the sort function provided by XSLT?
> 
> the xslt spec allows you to use any name (in a namespace of your choice)
> in addition to the two predefined names of number and text as sorting
> datatypes.
> 
> How you associate the name with the java that specifies the sorting, and
> what java would be needed would be processor specific but I'd be
> surprised if xalan doesn't offer a hook to this (saxon does)
> 
> That said, you could do this just in xslt as
> 
> <xsl:sort data-type="number" select="substring-before(address,'.')"/>
> <xsl:sort data-type="number" select="substring-before(substring-after(address,'.'),'.')"/>
> <xsl:sort data-type="number" select="substring-before(substring-after(substring-after(address,'.'),'.'),'.')"/>
> <xsl:sort data-type="number" select="substring-after(substring-after(substring-after(address,'.'),'.'),'.')"/>
> 
> (modulo typing errors:-)
> 
> david
> 
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