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="$match2[position() < 5]">
<xsl:sort select="date/@year"
order="descending" data-type="number"/>
<xsl:sort select="date/@month"
order="descending" data-type="number"/>
<xsl:sort select="date/@day"
order="descending" data-type="number"/>
<xsl:call-template name="dump-item">
<xsl:with-param name="item"
select="."/>
</xsl:call-template>
</xsl:for-each>
<xsl:for-each select="$match1all[position() <
(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() <
(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(_at_)objectiva(_dot_)fr]
Sent: Thursday, January 16, 2003 7:13 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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
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XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list