XSL Sort specifc value last - xslt

Question

Is it possible to sort nodes as follows:
Example XML
<record>
<id>0</id>
<sku>0</sku>
<name>Title</name>
<prop>456</prop>
<number>99</number>
</record>
If I apply this template
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="record/*">
<xsl:param select="." name="value"/>
<div>
<xsl:value-of select="concat(local-name(), ' - ', $value)"/>
</div>
</xsl:template>
</xsl:stylesheet>
Ouput:
<div>id - 0</div>
<div>sku - 0</div>
<div>name - Title</div>
<div>prop - 456</div>
<div>number - 99</div>
However, I would like all 0 values to be outputted last, as so:
<div>name - Title</div>
<div>prop - 456</div>
<div>number - 99</div>
<div>id - 0</div>
<div>sku - 0</div>
Is this possible by applying a sort to the <xsl:apply-templates/>?

Answer 1

There is an easy way of achieving this with XSLT-1.0. Just use a predicate on xsl:apply-templates checking if the content is zero:
<xsl:template match="record/*">
<div>
<xsl:value-of select="concat(local-name(), ' - ', .)"/>
</div>
</xsl:template>
<xsl:template match="/record">
<xsl:apply-templates select="*[normalize-space(.) != '0']" />
<xsl:apply-templates select="*[normalize-space(.) = '0']" />
</xsl:template>
This does not sort the output, but groups it the way you want it. The xsl:param is unnecessary.

Answer 2

As far as I see your question, the issue is not any sort at all.
You even didn't write what is the specific value, which you mentioned
it the title. It is rather an arbitrary sequence of child elements of record.
Try the following script, performing just that:
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" encoding="UTF-8" indent="yes" />
<xsl:template match="record">
<xsl:copy>
<xsl:apply-templates select="name, prop, number, id, sku"/>
</xsl:copy>
</xsl:template>
<xsl:template match="record/*">
<div><xsl:value-of select="concat(local-name(), ' - ', .)"/></div>
</xsl:template>
<xsl:template match="#*|node()">
<xsl:copy><xsl:apply-templates select="#*|node()"/></xsl:copy>
</xsl:template>
</xsl:transform>
I used XSLT 2.0, because initially you did't specify the XSLT version.
Could you move on to version 2.0? As you can see, it allows to write quite
an elegent solution (impossible in version 1.0).
I also changed your template matching record/*. You actually don't
need any param. It is enough to use . - the value of the current
element.
Edit
Another possibility is that you want the following sort:
First, elements with non-numeric value (in your case, only name),
maybe without any sort.
Then elements with numeric value, sorted descending on this value.
If this is the case, then change the template matching record to the
following:
<xsl:template match="record">
<xsl:copy>
<xsl:apply-templates select="*[not(. castable as xs:integer)]"/>
<xsl:apply-templates select="*[. castable as xs:integer]">
<xsl:sort select="." order="descending" data-type="number"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
And add:
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="#all"
to transform tag.
But I still can't see anything, which can be called the specific value.