RE: generic sort based on attribute names

Hey Joris, 
Thanks for your help. But I am still not getting the desired output. 
This is what I get: 
Input:
<root>
        <e b="bb" y="yy">
                <node x="123" a="123"/>
                <node x="999" a="999" b="999"/>
        </e>
        <e z="zz" a="aa"/>
</root>

Output:
<root>
    <e b="bb" y="yy">
        <node a="123" x="123"/>  
        <node a="999" b="999" x="999"/>
        </e>
    <e a="aa" z="zz"/>
</root>
Basically it is only sorting attributes within an element, not elements within 
a set of siblings. 
Thanks.


-----Original Message-----
From: Joris Gillis [mailto:roac(_at_)pandora(_dot_)be] 
Sent: Tuesday, January 25, 2005 1:33 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] generic sort based on attribute names

Tempore 21:48:58, die 01/25/2005 AD, hinc in  
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com scripsit Chaitanya Desai  
<cdesai(_at_)syncata(_dot_)com>:

> OK,
> This is the recursive stylesheet I have, to handle a generic XML.
> How does the thing you explained with the 'e' fit here.
This is an example of how it could be used;


<xsl:template match="*">
        <xsl:variable name="sortkey">
                <xsl:for-each select="*">
                        <e>
                                <xsl:for-each select="@*">
                                        <xsl:sort select="name(.)"/>
                                        <xsl:value-of select="name()"/>
                                </xsl:for-each>
                        </e>
                </xsl:for-each>
        </xsl:variable>
        <xsl:copy>
                <xsl:for-each select="@*">
                        <xsl:sort select="name(.)"/>
                        <xsl:copy-of select="."/>
                </xsl:for-each> 
                <xsl:apply-templates>
                                        <!-- the select for this sort needs to 
be the concatenated name-value  
pairs of
        all the attributes sorted in alphabetical order -->
                        <xsl:sort  
select="xx:node-set($sortkey)//e[position()=count(current()/preceding-sibling::*)]"/>
                </xsl:apply-templates>
        </xsl:copy>
</xsl:template>



It will convert an input like this:
<root>
        <e b="bb" y="yy">
                <node x="123" a="123"/>
                <node x="999" a="999" b="999"/>
        </e>
        <e z="zz" a="aa"/>
</root>

to this:
<root>
        <e a="aa" z="zz"/>
        <e b="bb" y="yy">
                <node a="999" b="999" x="999"/>
                <node a="123" x="123"/>
        </e>
</root>


For each element in the source doument, it creates a table of the  
attributes of all child elements and uses this table (stored in the  
'sortkey' variable) to perform the sort.

e.g. when the 'root' element is matched, the 'sortkey' variable contains:
<e>by</e>
<e>az</e>

because these entries follow the document order, you can use them as  
sortkey for the child elements by matching the position:
xx:node-set($sortkey)//e[position()=count(current()/preceding-sibling::*)]

(please keep in mind that this is an XSLT1.0 solution. The XSLT2.0  
solutions of Michael Kay and Dimitre Novatchev are of course better if  
your XSLT processor can handle them)

regards,
-- 
Joris Gillis (http://www.ticalc.org/cgi-bin/acct-view.cgi?userid=38041)
"Φιλήκοον ειναι μαλλον η φιλόλαλον"  -  
Κλεόβουλος

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