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Everett TSubject: XML-to-XML Transformation
Author: Everett T
Date: 21 Dec 2005 10:44 AM
I am a newbie to XSLT and I'm trying to do something that is somewhat non-standard.

My original XML is formatted in the following manner:
<?xml version= '1.0' ?>
<Bills>
<Bill No="10030" Name="John Doe">
.
. some other XML
.
</Bill>
<Bill No="10009" Name="Jane Jones">
.
. some other XML
.
</Bill>
<Bill No="10100" Name="Bob Smith">
.
. some other XML
.
</Bill>
<Bill No="10030" Name="John Doe">
.
. some other XML
.
</Bill>
</Bills>


I would like to transform it into XML that looks like the following:
<?xml version= '1.0' ?>
<Bills>
<Bill No="10009" Name="Jane Jones" NewBill="1">
.
. some other XML
.
</Bill>
<Bill No="10030" Name="John Doe" NewBill="1">
.
. some other XML
.
</Bill>
<Bill No="10030" Name="John Doe" NewBill="0">
.
. some other XML
.
</Bill>
<Bill No="10100" Name="Bob Smith" NewBill="1">
.
. some other XML
.
</Bill>
</Bills>


Basically, I'm trying to, first, sort all of the Bill tags by the @No attribute. After sorting, I want to go through and add a @NewBill attribute that distinguished whether or not the @No attribute has changed. In the example output above I use a "1" when it is a new bill and a "0" if the @No attribute is the same as the previous bill.

Any help would be greatly appreciated...
Postnext
Minollo I.Subject: XML-to-XML Transformation
Author: Minollo I.
Date: 21 Dec 2005 01:26 PM
There is no automatic mapping for this.

Hand coding it, you can write something like:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
<xsl:variable name="orderedBills">
<Bills>
<xsl:for-each select="Bills/Bill">
<xsl:sort select="@No"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</Bills>
</xsl:variable>
<Bills>
<xsl:for-each select="$orderedBills/Bills/Bill">
<Bill>
<xsl:for-each select="@*">
<xsl:copy-of select="."/>
</xsl:for-each>
<xsl:attribute name="NewBill">
<xsl:variable name="previousBill" select="preceding-sibling::Bill[position()=1]"/>
<xsl:choose>
<xsl:when test="not($previousBill) or $previousBill/@No != @No">1</xsl:when>
<xsl:otherwise>0</xsl:otherwise>
</xsl:choose>
</xsl:attribute>
<xsl:apply-templates/>
</Bill>
</xsl:for-each>
</Bills>
</xsl:template>
</xsl:stylesheet>

Postnext
Everett TSubject: XML-to-XML Transformation
Author: Everett T
Date: 22 Dec 2005 10:59 AM
This seems to me like it would work just fine but I keep receiving the following error message during the transformation process.

Error during XSLT transformation: An XPath expression was expected to return a NodeSet.

I assume that it is referring to line where the for-each loop is selecting the $orderedBills variable.

Any idea how to get around this error message?

Thanks again.
Posttop
Minollo I.Subject: XML-to-XML Transformation
Author: Minollo I.
Date: 22 Dec 2005 11:13 AM
What processor are you using?

If you use MSXML you will have to define the xmlns:msxsl="urn:schemas-microsoft-com:xslt" namespace and then use msxsl:node-set($orderedBills).

If you use Saxon 6.x you will have to define the xmlns:exslt="http://exslt.org/common" namespace and then use exslt:node-set($orderedBills).

In Xalan-J you will have to define the xmlns:xalan="http://xml.apache.org/xalan" namespace and then use xalan:nodest($orderedBills).

Minollo

   
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