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Grouping XML using XSLT

, 17 Feb 2002
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Using XSLT to group XML elements based on unique ID values in the XML. The XML is transformed into an HTML table.
<!-- Download Links --> <!-- Add the rest of your HTML here -->


Processing a list of XML elements using XSLT is fairly simple if you want to process each element. But what if you want to group the XML elements, to show a summary? Consider the following XML:

<?xml version="1.0" ?>

Suppose that you need to show a summary of Employee hours, grouped by Team. Something like this:

<!-- Article image -->

Sample Image - groupxml.jpg

The unwieldy approach

One way to do this is to loop through the list of <Employee> elements, and only show a row whenever the EmployeeID changes. While this would work, this approach is unwieldy and inefficient, because for each <Employee> being processed, you would be required to keep track of the IDs of the previous <Employee> element. This is not a pretty sight.

The efficient approach

A cleaner, more efficient way to do this is to build a list of unique keys, then use these keys to group the results. (This is called the Muenchian Method.)

First, you must define the keys required to group the <Employee> elements. You will need one for the TeamID, and one for the EmployeeID.

<xsl:key <code>name="keyTeamID"</code> match="Employee" <code>use="TeamID"</code> />
<xsl:key <code>name="keyEmployeeID"</code> match="Employee" <code>use="EmployeeID"</code> />

Select the first element of each group of elements for each unique TeamID.

<xsl:for-each select="//<code>Employee[generate-id(.) = generate-id(key('keyTeamID', TeamID)[1]</code>)]">

Get all the <Employee> elements that belong to that Team, into a variable.

<!-- Save the ID of the Team to a variable -->
<xsl:variable name="lngTeamID"><xsl:value-of select="TeamID" /></xsl:variable>
<!-- Select all the Employees belonging to the Team -->
<xsl:variable name="lstEmployee" select="<code>//Employee[TeamID=$lngTeamID]</code>" />

The <Employee> elements in this list must now be grouped by EmployeeID. This is similar to grouping by TeamID, except that in this case you only need to select elements in the list contained in the variable; you do not need to select elements from the entire result set.

<xsl:for-each select="<code>$lstEmployee</code>[generate-id(.) = generate-id(key(<code>'keyEmployeeID', EmployeeID</code>)[1])]">

It is now fairly simple to show the total Hours for each Employee.

<xsl:value-of select="sum(<code>$lstEmployee[EmployeeID=$lngEmployeeID]/Hours</code>)" />

The full source

This is the entire XSLT used to render the table in the image:

<?xml version="1.0" ?>
<xsl:stylesheet version="1.0" 

<!-- Define keys used to group elements -->
<code><xsl:key name="keyTeamID" match="Employee" use="TeamID" />
<xsl:key name="keyEmployeeID" match="Employee" use="EmployeeID" /></code>

<xsl:template match="/">
   <title>Employee Hours By Team</title>
   <link type="text/css" rel="stylesheet" href="groupxml.css" />
   <h3>Employee Hours By Team</h3>
    <!-- Process each Team -->
    <code><xsl:for-each select="//Employee[generate-id(.) = generate-id(key('keyTeamID', TeamID)[1])]"></code>
     <xsl:variable name="lngTeamID"><xsl:value-of select="TeamID" /></xsl:variable>
     <!-- Select all the Employees belonging to the Team -->
     <code><xsl:variable name="lstEmployee" select="//Employee[TeamID=$lngTeamID]" /></code>
     <!-- Show details for Employees in Team -->
     <xsl:call-template name="ShowEmployeesInTeam">
      <code><xsl:with-param name="lstEmployee" select="$lstEmployee" /></code>
     <td colspan="4" class="RightJustified DarkBack">Grand Total</td>
     <td colspan="1" class="RightJustified DarkBack">
      <!-- Show Grand Total of hours for all Employees -->
      <xsl:value-of select="sum(//Employee/Hours)" />

<xsl:template name="ShowEmployeesInTeam">
 <code><xsl:param name="lstEmployee" /></code>
 <!-- Show the name of the Team currently being processed -->
  <td colspan="4" class="DarkBack">TEAM: <xsl:value-of select="<code>$lstEmployee[1]/TeamName</code>" /></td>
  <td colspan="1" class="DarkBack RightJustified">HOURS</td>
 <!-- Show the total hours for each Employee in the Team -->
 <code><xsl:for-each select="$lstEmployee[generate-id(.) = generate-id(key('keyEmployeeID', EmployeeID)[1])]"></code>
  <xsl:variable name="lngEmployeeID" select="EmployeeID" />
  <!-- Show details of each Employee -->
   <td colspan="4">
    <xsl:value-of select="$lstEmployee[EmployeeID=$lngEmployeeID]/Name" />
     <xsl:value-of select="$lstEmployee[EmployeeID=$lngEmployeeID]/Surname" />
   <td colspan="1" class="RightJustified">
    <!-- Show the total hours for the current Employee -->
    <code><xsl:value-of select="sum($lstEmployee[EmployeeID=$lngEmployeeID]/Hours)" /></code>
  <td colspan="4" class="LightBack RightJustified">Sub-Total</td>
  <td colspan="1" class="LightBack RightJustified">
   <!-- Show the total hours for all Employees in the Team -->
   <xsl:value-of select="sum($lstEmployee/Hours)" />


The CSS used to render the table in the image:

{  border-collapse: collapse;
   width: 30%;
   table-layout: fixed;
   border-style: solid;
table, td
{  border-width: 1px;
{  color: black;
   font-family: Arial;
   font-size: x-small;
   border-right-style: none;
   border-left-style: none;
   border-top-style: solid;
   border-bottom-style: solid;
{  background-color: #0066FF;
   background-color: blue;
   color: white;
   font-weight: bold;
{  background-color: #99CCFF;
   color: black;
{  text-align: right;


This article has no explicit license attached to it but may contain usage terms in the article text or the download files themselves. If in doubt please contact the author via the discussion board below.

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About the Author

MS le Roux
Web Developer
South Africa South Africa
I live in the Northern Suburbs of Cape Town (South Africa).

Comments and Discussions

QuestionXSLT grouping Pinmemberavdhoute6-Jul-14 18:54 
QuestionXSLT grouping Pinmemberavdhoute30-Jun-14 1:15 
GeneralMy vote of 4 PinmemberVarunKumarGB14-Dec-10 22:23 
GeneralMy vote of 3 Pinmemberlijo011@yahoo.com15-Aug-10 23:57 
GeneralGrouping XML data Pinmemberblindcapt31-Mar-09 18:15 
GeneralPerformance issue PinmemberQing Jiang30-Mar-06 3:59 
QuestionHow can I describe binary data in XML? Pinsusslei cheng28-Sep-02 20:29 
AnswerRe: How can I describe binary data in XML? PinmemberMS le Roux29-Sep-02 20:06 
AnswerRe: How can I describe binary data in XML? PinmemberHeath Stewart9-Jan-03 11:26 
The common way is with base64 encoding. There are many classes to help you accomplish that in .NET. This is the common approach for MIME and SOAP when dealing with binary data (although you can use a new protocol named DIME that is separate from SOAP but part of the same messages).
So, if you had binary data in byte[] buffer, you could get a string value using string base64 = Convert.ToBase64String(buffer) and put that in an element. For a good schema, you should also specify the MIME type, which is "application/octet-stream" using a transfer encoding method of "base64". A MIME header would look like this:
Content-Type: application/octet-stream
Content-Transfer-Encoding: bas64
Content-Disposition: attachment
A typical XML fragment (while the element name is completely arbitrary, it's good practice to remain consistent with the attribute names used) would look like:
<attachment type="application/octet-stream" encoding="base64">SVRTRgA...AAA=</attachment>

"Well, I wouldn't say I've been missing it, Bob." - Peter Gibbons
GeneralRendering an xml document with MSIE 5.5 PinmemberBalteo5-Sep-02 22:13 
GeneralRe: Rendering an xml document with MSIE 5.5 PinmemberMS le Roux6-Sep-02 0:19 
GeneralRe: Rendering an xml document with MSIE 5.5 PinmemberBalteo6-Sep-02 4:19 
Generalgrouping and xslt PinmemberAbsynthE22-Aug-02 8:34 
QuestionWhy the data are gone? Pinmembersshhz8-May-02 0:48 
AnswerRe: Why the data are gone? PinmemberMarSCoZa8-May-02 1:06 
GeneralRe: Why the data are gone? Pinmembersshhz8-May-02 15:50 
GeneralRe: Why the data are gone? PinmemberMS le Roux8-May-02 20:17 
GeneralRe: Why the data are gone? Pinmembergdltlxb5-Jun-02 2:45 
GeneralNice going, one thing though... PinmemberPaul Watson17-Feb-02 21:15 
GeneralRe: Nice going, one thing though... PinmemberMarSCoZa17-Feb-02 22:00 
GeneralRe: Nice going, one thing though... PinmemberMatt Berther20-Feb-02 7:11 
GeneralRe: Nice going, one thing though... PinmemberPaul Watson20-Feb-02 8:36 

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