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

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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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