RE: [xsl] Complex sorting problem (looking for an XSLT outer join ?)

Subject: RE: [xsl] Complex sorting problem (looking for an XSLT outer join ?)
From: Taras Tielkes <taras@xxxxxxx>
Date: Thu, 16 Jan 2003 21:02:40 +0100
Hi Philippe,

If we assume that I want to restrict myself to XSLT 1.0, do you think it
would be possible to write something with more elegance than the following
monstrosity?

tt

<?xml version="1.0"?> 
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
	
	<xsl:output method="html"/>
	
	<xsl:key name="weight" match="/root/categories/category/@weight"
use="../@id"/>

	<xsl:variable name="match2"
select="/root/items/item[key('weight',@category)=2]"/>
	<xsl:variable name="match1"
select="/root/items/item[key('weight',@category)=1]"/>
	<xsl:variable name="match0"
select="/root/items/item[key('weight',@category)=0]"/>
	<xsl:variable name="missing"
select="/root/items/item[not(key('weight',@category))]"/>
	<xsl:variable name="match1all" select="$match1|$missing"/>

	<xsl:template match="/">
		<table border="1">
			<tr>
				<td><b>data</b></td>
				<td><b>weight</b></td>
				<td><b>date</b></td>
			</tr>
			<xsl:for-each select="$match2[position() &lt; 5]">
				<xsl:sort select="date/@year"
order="descending" data-type="number"/>
				<xsl:sort select="date/@month"
order="descending" data-type="number"/>
				<xsl:sort select="date/@day"
order="descending" data-type="number"/>
				<xsl:call-template name="dump-item">
					<xsl:with-param name="item"
select="."/>
				</xsl:call-template>
			</xsl:for-each>
			<xsl:for-each select="$match1all[position() &lt;
(5-count($match2))]">
				<xsl:sort select="date/@year"
order="descending" data-type="number"/>
				<xsl:sort select="date/@month"
order="descending" data-type="number"/>
				<xsl:sort select="date/@day"
order="descending" data-type="number"/>
				<xsl:call-template name="dump-item">
					<xsl:with-param name="item"
select="."/>
				</xsl:call-template>
			</xsl:for-each>
			<xsl:for-each select="$match0[position() &lt;
(5-count($match2)-count(match1all))]">
				<xsl:sort select="date/@year"
order="descending" data-type="number"/>
				<xsl:sort select="date/@month"
order="descending" data-type="number"/>
				<xsl:sort select="date/@day"
order="descending" data-type="number"/>
				<xsl:call-template name="dump-item">
					<xsl:with-param name="item"
select="."/>
				</xsl:call-template>
			</xsl:for-each>
		</table>
		
	</xsl:template>
	
	<xsl:template name="dump-item">
		<xsl:param name="item"/>
		<tr>
			<td>
				<xsl:value-of select="$item/@data"/>
			</td>
			<td>
				<xsl:value-of
select="key('weight',$item/@category)"/>
			</td>		
			<td>
				<xsl:value-of select="date/@year"/>
				<xsl:text>/</xsl:text>
				<xsl:value-of select="date/@month"/>
				<xsl:text>/</xsl:text>
				<xsl:value-of select="date/@day"/>
			</td>
		</tr>	
	</xsl:template>

</xsl:stylesheet>

:-)

> -----Original Message-----
> From: Philippe Drix [mailto:pdrix@xxxxxxxxxxxx]
> Sent: Thursday, January 16, 2003 7:13 PM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] Complex sorting problem (looking for an XSLT outer
> join?)
> 
> 
> At 18:43 16/01/2003 +0100, you wrote:
> >Hi,
> >
> >My source xml has the following format:
> >
> ><root>
> >     <items>
> >         <item category="1" data="ggg">
> >             <date year="1995" month="4" day="13"/>
> >         </item>
> >         <item category="2" data="hhh">
> >             <date year="1984" month="7" day="22"/>
> >         </item>
> >         <item category="3" data="www">
> >             <date year="1991" month="3" day="12"/>
> >         </item>
> >         <item category="3" data="rrr">
> >             <date year="1999" month="6" day="19"/>
> >         </item>
> >         <item category="4" data="xxx">
> >             <date year="1982" month="2" day="17"/>
> >         </item>
> >         <item category="5" data="kkk">
> >             <date year="2000" month="12" day="11"/>
> >         </item>
> >     </items>
> >
> >     <categories>
> >         <category id="1" weight="0">
> >         <category id="3" weight="2">
> >         <category id="4" weight="1">
> >     </categories>
> ></root>
> >
> >1) The source xml contains a list of items.
> >    Each item carries a 'data' attribute, which is the 
> actual content of the
> >item.
> >    Apart from that, each item contains:
> >    a) a 'catagory' attribute, identifying the catagory that 
> the item belogs
> >to.
> >    b) a 'date' child element, representing the date of the item
> >    Each item has an implied weight, implied by the category that it
> >references. (but see [2] and [3])
> >
> >2) The source xml also contains a list of categories.
> >    Each category contains a 'weight' attribute, which carries the
> >'importance' of the category.
> >    Weight is from 0 (least important) to 2 (most 
> important), in other words
> >[0,2].
> >
> >3) Some of the categories referred to by the item elements 
> are not present
> >in the source xml.
> >    A default weight of 1 should be assumed in that case.
> >    In the example xml, the categories (2,5) are absent.
> >
> >I would like to sort the item elements using the following criteria:
> >    1) first, by (implied) weight
> >    2) second, by date
> >
> >It seems to me that the problem would be easy if all 
> referenced categories
> >were present in the source xml.
> >In that case, I could use xsl:key to retreive the matching 
> weight for each
> >item.
> >
> >However, that is not the case. It seems that what I want is to grab a
> >default weight of 1 wherever the references category is not 
> present in the
> >source xml.
> >Using SQL, for instance, I could use an outer join, 
> specifying a default
> >value.
> >
> >Is there any solution for this problem in XSLT?
> >
> >Thanks in advance for any feedback,
> >
> >tt
> >
> >
> >
> >
> >
> >
> >
> >  XSL-List info and archive:  
http://www.mulberrytech.com/xsl/xsl-list


You can copy your XML into a variable, but with a copy that will not be a 
true copy, because you will have to instanciate attribute "weight", 
defaulted to value 1 if absent.
Then you put the current node on the root of this variable (use Saxon with 
version="1.1", or use a function like node-set() to convert RTF to 
nodes-set ):
<xsl:for-each select="$myNewXMLtreeWithAttributeWeightAlwaysPresent">
         <xsl:for-each select="what you want">
                 <xsl:sort what you want>
         </xsl:for-each>
</xsl:for-each>
Regards -- Ph D

==
Philippe Drix
      ___________
__| OBJECTIVA |___________________
http://www.objectiva.fr
21-23, rue Aristide Briand - 92170 Vanves
tel : +33 1 47 36 60 30
fax : +33 1 47 36 61 93



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