Wendell,
Thanks a lot for the pointer.... I have now progressed some what, but I
still can see to get subtotals...
I stripped the XSL down so I can concentrate on the problem. So now I
have....
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:key name="cards" match="ITEM" use="CARD/@number" />
<xsl:template match="/">
<xsl:for-each select="/ITEMS" >
<xsl:call-template name="create-body" />
</xsl:for-each>
</xsl:template>
<xsl:template name="create-body" >
<BR/>
SELECTING UNIQUE USING A FOR-EACH
<xsl:for-each select="./ITEM[CARD/@number and
generate-id(.) = generate-id(key('cards', CARD/@number))]" >
<BR/>
CARD: <xsl:value-of select="./CARD/@number"/>
NUMBER: <xsl:value-of select="count(key('cards', ./CARD/@number))" />
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
The Raw XML is...
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="test.xsl"?>
<ITEMS>
<ITEM item_id='1'> <CARD number='123456789'/> <AMOUNT amt='100'/>
</ITEM>
<ITEM item_id='2'> <CARD number='123456789'/> <AMOUNT amt='200'/>
</ITEM>
<ITEM item_id='3'> <CARD number='987654321'/> <AMOUNT amt='30'/>
</ITEM>
. etc. etc.
</ITEMS>
I don't understand!!!
If I can count the nodes referenced by the key, why can't I sum the values
in those nodes.
I am going to read up on this tonight and may be I will crack it, but any
more pointers would be much appreciated...
Roger
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Roger Wyatt
Principal Consultant
Nomad Software Ltd.
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Wendell Piez
<wapiez(_at_)mulberryt To:
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
ech.com> cc:
Subject: Re: [xsl] Sorting
unique and sub-total problem
02/06/2004 16:48
Please respond to
xsl-list
Roger,
At 11:27 AM 6/2/2004, you wrote:
The <ITEM> nodes can be in any order. What I would like to get at the end
is the following out put.
Card: 123456789 Total:350
Card: 987654321 Total: 30
At the moment I can't even sort unique! I have tried to sort the nodes and
then compare the current node to the following sibling, using...
<xsl:for-each select=".//ITEM" >
<xsl:sort select="CARD/@number" order="ascending" />
<xsl:if test="./CARD/@number[ not
(.=following-sibling::ITEM/CARD/@number) ]">
....print number etc. etc....
But what I have found is that the following sibling is as it would be if
the nodes have not been ordered.
Yes: the axes provide access to the document in its original order.
You have a fairly standard grouping problem. Grouping is a very common
requirement that is not addressed well by XSLT 1.0 (it is a species of
up-conversion), but for which a multitude of sophisticated techniques have
nonetheless been devised.
Start researching them on the FAQ at
http://www.dpawson.co.uk/xsl/sect2/sect21.html, and at Jeni's pages at
http://www.jenitennison.com/xslt/grouping.
I fear that the only way to deal with this is to use XSLT to preorder the
nodes and then use a 2nd script using the same approach as described.
You could do this, but your problem does not look so difficult that you
couldn't do it in one pass.
Alternatively I could go right back to the process that originally created
the XML file and order / sub total from there... something I really want
to
avoid if at all possible.
This, however, is sometimes the best solution: sometimes the source comes
out of a relational database, and relational technology is very good at
this kind of operation.
Is there another approach I can take ? Am I expecting too much from XSLT
with all this... ?
Nope. Just dig a little deeper under "grouping".
Good luck,
Wendell
======================================================================
Wendell Piez
mailto:wapiez(_at_)mulberrytech(_dot_)com
Mulberry Technologies, Inc. http://www.mulberrytech.com
17 West Jefferson Street Direct Phone: 301/315-9635
Suite 207 Phone: 301/315-9631
Rockville, MD 20850 Fax: 301/315-8285
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