Re: [xsl] Sorting network addresses

[Craig W <codecraig@xxxxxxxxx>]

Hmm...I must have gotten some stuff mixed up when pasting into the
email... Here is the XML, and XSLT exactly as I have it.

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE networks [
<!ELEMENT networks (network+)>
<!ELEMENT network (address)>
<!ELEMENT address (#PCDATA)>
]>
<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>


<?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="$newline" />	
							<address>
								<xsl:value-of select="address" />
							</address>
							<xsl:value-of select="$newline" />
						</network>
						<xsl:value-of select="$newline" />
				</xsl:for-each>
		</networks>
	</xsl:template>
</xsl:stylesheet>


and here is the output

<?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 am using XMLSpy to generate the output.  To repeat how I did it,
open the XML document, go to the "XSL/XQuery" menu and choose "XSL
Transformation", point it to the XSLT and thats it.


....now do u get what i get?

thanks

On Wed, 2 Mar 2005 13:33:22 GMT, David Carlisle <davidc@xxxxxxxxx> wrote:
> 
> 
> > the result i am getting now is like this
> Not from upur posted code, you can't: it generates srcaddr elements
> there can't be any address elements in the output.
> 
> Running your posted stylesheet on your posted input I get:
> 
> $ saxon ip.xml ip.xsl
> <?xml version="1.0" encoding="UTF-8"?>
>                 <networks>
>                 <network>
>                 <srcaddr>1.1.1.1</srcaddr>
>                 </network>
>                 <network>
>                 <srcaddr>2.3.1.2</srcaddr>
>                 </network>
>                 <network>
>                 <srcaddr>170.5.2.4</srcaddr>
>                 </network>
>                 </networks>
> 
> Note by the way your $newline is actually a newline and two tabs, hence
> the strange indentaion in the result.
> 
> David
> 
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