Hi,
In the toy problem offered, an XSLT 1.0 solution would simply perform
the filtering for the first item in the sorted list inside the
variable declaration:
That is, instead of
<xsl:variable name="sorted">
<xsl:apply-templates select="/response/data/result">
<xsl:sort select="COMPANY_SORT" data-type="text" order="descending"/>
</xsl:apply-templates>
</xsl:variable>
we would have
<xsl:variable name="first-sorted">
<xsl:for-each select="/response/data/result">
<xsl:sort select="COMPANY_SORT" data-type="text" order="descending"/>
<xsl:if test="position() = 1">
<xsl:apply-templates select="."/>
</xsl:if>
</xsl:apply-templates>
</xsl:variable>
But I bet there's something in the real-world case to prevent this (so
we may need to see more).
I agree with Mike that solving such problems in XSLT 1.0 seems
increasingly like a parlor game. Not that I'm against parlor games.
Especially in problems like this one ... processing a sorted list is
usually possible in XSLT 1.0 (harder or easier depending on the sort),
but then you discover you actually have to sort a processed list --
even harder.
Cheers,
Wendell
On Thu, Nov 1, 2012 at 5:14 PM, Darren Oh <darren(_at_)oh(_dot_)name> wrote:
Thanks for the suggestion. I got stuck when trying to produce a sorted
node-set. Instead of a node-set, I got a result tree fragment, for which no
node-set operations are possible. Here is a simplified example to illustrate
the problem. Whereas it is possible to select a node from $result, it is not
possible to select a node from $sorted. Any ideas?
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:output method="xml"/>
<xsl:template match="*">
<xsl:copy>
<xsl:apply-templates/>
</xsl:copy>
</xsl:template>
<xsl:variable name="result" select="/response/data/result"/>
<xsl:variable name="sorted">
<xsl:apply-templates select="/response/data/result">
<xsl:sort select="COMPANY_SORT" data-type="text" order="descending"/>
</xsl:apply-templates>
</xsl:variable>
<xsl:template match="/">
<xsl:copy-of select="$result[1]"/>
<xsl:copy-of select="$sorted[1]"/>
</xsl:template>
</xsl:stylesheet>
On Oct 19, 2012, at 1:27 PM, Michael Kay wrote:
Try:
1. define a global variable $v1 that selects the result of the path
expression in document order.
2. define another global variable $v2 that selects the sorted result of the
path expression
3. Use a base template rule that's the identity copy
4. Add a template rule that matches nodes in $v1 (match="node()[. intersect
$v1]). In this rule, determine the index position of this node in $v1 (count
($v1[. << $this]) + 1), and output the corresponding node from $v2 (copy-of
select="$v2[$n]").
Michael Kay
Saxonica
On 19/10/2012 18:03, Darren Oh wrote:
I am trying to generate a stylesheet that copies an XML source document.
The only change should be that nodes selected by an XPath expression are
sorted. I want this to work for any XML source document. The only
information available to generate the stylesheet is the XPath expression
and the sort criteria. I think this requires creating a template for the
parents of the nodes selected by the XPath expression. How can I do this?
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail:
<mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--
--
Wendell Piez | http://www.wendellpiez.com
XML | XSLT | electronic publishing
Eat Your Vegetables
_____oo_________o_o___ooooo____ooooooo_^
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--