Original Input XML:

<?xml version="1.0" encoding="UTF-8"?>
<bb3:Reservations xmlns:bb3="http://www.ourbedandbreakfast.com/sample1">
    <bb3:Reservation RoomNum="106" InvoiceNum="BB3002">
        <bb3:Customer FirstName="Tom" LastName="Cruise" />
        <bb3:Customer FirstName="Nicole" LastName="Kidman" />
    </bb3:Reservation>
    <bb3:Reservation RoomNum="206" InvoiceNum="BB5010">
        <bb3:Customer FirstName="Tom" LastName="Cruise" />
    </bb3:Reservation>
    <bb3:Reservation RoomNum="506" InvoiceNum="BB6020">
        <bb3:Customer LastName="Madonna" />
    </bb3:Reservation>
</bb3:Reservations>


I want to get a list of all invoice numbers per customer, like below.

Example of Output XML:

<bb3:CustomerVisits>
    <bb3:Booking>
        <bb3:Customer FirstName="Tom" LastName="Cruise" />
        <InvoiceNum>BB3002</InvoiceNum>
        <InvoiceNum>BB5010</InvoiceNum>
    </bb3:Booking>
    <bb3:Booking>
        <bb3:Customer FirstName="Nicole" LastName="Kidman" />
        <InvoiceNum>BB3002</InvoiceNum>
    </bb3:Booking>
    <bb3:Booking>
        <bb3:Customer LastName="Madonna" />
        <InvoiceNum>BB6020</InvoiceNum>
    </bb3:Booking>
</bb3:CustomerVisits>

Not sure where to start. I read something about Grouping by 2 levels, but was confused since I am new to XSLT.
Grouping by 2 levels: http://www.dpawson.co.uk/xsl/sect2/N4486.html#d5891e235

This is a straightforward single-level grouping problem, except that your grouping key is a compound key, so you need an expression such as concat(@FirstName, '~', @LastName).

Grouping in XSLT 2.0 is easy using the xsl:for-each-group construct.

In XSLT 1.0 it's surprisingly difficult, but there are coding patterns that you can reuse. See http://www.jenitennison.com/xslt/grouping. For a compound grouping key, I think you'll need to use the Muenchian grouping approach.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Thanks Michael, I will look at the resources you suggested and give it a go.

Also FirstName is an optional will the concat key you suggest still work then?

concat() treats an empty sequence the same as a zero-length string, so @FirstName="" will be treated the same as an element with no @FirstName. If you want to treat them differently, you'll have to refine the grouping key, for example include count(@FirstName) as a component.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

I tried the below in oXygen's XSLT debugger and it is spitting out the names. But how do I get it to output as expected with InvoiceNum as well?

<?xml version="1.0"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:bb3="http://www.ourbedandbreakfast.com/sample1">
<xsl:output method="text" indent="yes" encoding="UTF-8"/>
<xsl:template match="bb3:Reservations">
  <xsl:for-each-group select="bb3:Reservation/bb3:Customer" group-by="concat(@FirstName, ' ', @LastName)">
    <xsl:value-of select="current-grouping-key()"/>
    <xsl:for-each select="current-group()">
      <xsl:value-of select="@InvoiceNum"/>
    </xsl:for-each>
  </xsl:for-each-group>
</xsl:template>
</xsl:stylesheet>

I am trying to get my output to look like this:

Tom Cruise
BB3002
BB5010
Nicole Kidman
BB3002
Madonna
BB6020

Did you add the declaration to map bb3 to the appropriate URI?

--

Joe (Microsoft MVP - XML)

Joe I figured that out and edited my message right before your post. I am getting closer, but still not there yet.

Can you look at where I am at now and assist. Thanks.

Okay making this change:
<xsl:value-of select="../@InvoiceNum"/>

Seems to get what I am expecting.

I am very new to this and I am wondering if this code looks efficient enough, correct, and resistant to change. Any improvements or suggestions appreciated.

Now onto getting it into the XML format I need. Thanks all!

I might run into some problems getting it formatted into XML like needed so if those who helped and others can check back here in this post, I would appreciate the assistance while I learn the ropes.

I have this so far, but I need to figure out how to get
<bb3:Customer FirstName="Tom" LastName="Cruise" />
into each Booking
I am only getting: FirstName="Tom" LastName="Cruise">

Obviously this:
<xsl:copy-of select="@*"/>
is not correct, since its missing the <bb3:Customer />.

Can someone help.

Okay this seems to work (see below). I think the XSL is complete, experts let me know if there are any mistakes as this is my first attempt at learning XSLT.

<?xml version="1.0"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  xmlns:bb3="http://www.ourbedandbreakfast.com/sample1">
<xsl:output method="xml" indent="yes" encoding="UTF-8"/>
<xsl:template match="bb3:Reservations">
  <bb3:CustomerVisits>
  <xsl:for-each-group select="bb3:Reservation/bb3:Customer" group-by="concat(@FirstName, '~', @LastName)">
         <bb3:Booking>
             <xsl:copy>
               <xsl:copy-of select="@*"/>
               <xsl:apply-templates mode="copy"/>
             </xsl:copy>
    <xsl:for-each select="current-group()">
      <InvoiceNum><xsl:value-of select="../@InvoiceNum"/></InvoiceNum>
    </xsl:for-each>
    </bb3:Booking>
  </xsl:for-each-group>
   </bb3:CustomerVisits>
</xsl:template>
</xsl:stylesheet>

Final Output:

<?xml version="1.0" encoding="UTF-8"?>
<bb3:CustomerVisits xmlns:bb3="http://www.ourbedandbreakfast.com/sample1">
 <bb3:Booking>
  <bb3:Customer FirstName="Tom" LastName="Cruise"/>
      <InvoiceNum>BB3002</InvoiceNum>
      <InvoiceNum>BB5010</InvoiceNum>
 </bb3:Booking>
 <bb3:Booking>
  <bb3:Customer FirstName="Nicole" LastName="Kidman"/>
      <InvoiceNum>BB3002</InvoiceNum>
 </bb3:Booking>
 <bb3:Booking>
  <bb3:Customer LastName="Madonna"/>
      <InvoiceNum>BB6020</InvoiceNum>
 </bb3:Booking>
</bb3:CustomerVisits>