My xml contains the reords of an employee shift and it's task. so it might be possible that one employee in one has to do many task. so my xml contains the reords as:
 <Detail>
    <idShiftNo>72</idShiftNo>
    <idEmployee>221265</idEmployee>
    <NameEmployee>Biswal, Antje</NameEmployee>
    <Date>2007-01-05T00:00:00+05:30</Date>
    <Department>1</Department>
    <ShiftName>F11</ShiftName>
    <ShiftStart>2007-01-05T02:30:00+05:30</ShiftStart>
    <ShiftEnd>2007-01-05T11:00:00+05:30</ShiftEnd>
    <idTaskNo>324</idTaskNo>
    <strNameJob>C-DE</strNameJob>
    <idTaskNo1>324</idTaskNo1>
    <TaskStart>2007-01-05T02:30:00+05:30</TaskStart>
    <TaskEnd>2007-01-05T04:15:00+05:30</TaskEnd>
    <TaskName>C-DE5824</TaskName>
    #000000
    <backColorHTML>#C0C0C0</backColorHTML>
    <ToolTip>Flight: Job: C-DE Task:C-DE5824 STA: STD: 03:45 AcOP: DE AcType: 320 Dest: TFS

PaxMax: 174 PaxBkd: 176</ToolTip>
  </Detail>
  <Detail>
    <idShiftNo>72</idShiftNo>
    <idEmployee>221265</idEmployee>
    <NameEmployee>Biswal, Antje</NameEmployee>
    <Date>2007-01-05T00:00:00+05:30</Date>
    <Department>1</Department>
    <ShiftName>F11</ShiftName>
    <ShiftStart>2007-01-05T02:30:00+05:30</ShiftStart>
    <ShiftEnd>2007-01-05T11:00:00+05:30</ShiftEnd>
    <idTaskNo>292</idTaskNo>
    <strNameJob>BO</strNameJob>
    <idTaskNo1>292</idTaskNo1>
    <TaskStart>2007-01-05T04:15:00+05:30</TaskStart>
    <TaskEnd>2007-01-05T04:55:00+05:30</TaskEnd>
    <TaskName>DE5824</TaskName>
    #000000
    <backColorHTML>#C0C0C0</backColorHTML>
    <ToolTip>Flight: Job: BO Task:DE5824 STA: STD: 03:45 AcOP: DE AcType: 320 Dest: TFS PaxMax:

174 PaxBkd: 176</ToolTip>
  </Detail>

The problem is that as i have to show this on a chart then there it shows as

72 Biswal, Antje shiftstarttime shiftendtime task1
72 Biswal, Antje shiftstarttime shiftendtime task2

but i wan't to show it like
72 Biswal, Antje shiftstarttime shiftendtime task1 task2


i have done to display the id,name,shiftstart and end as thay are same in all by just using distinct
72 Biswal, Antje shiftstarttime shiftendtime

but how can i merge the two task records .


please help me in this
m waiting for ur replies :(

This problem is known as grouping, and if you google for "XSLT Grouping" (or preferably, look it up in your favourite XSLT textbook) you will find lots of information on the subject. It's very easy in XSLT 2.0 using the <xsl:for-each-group> instruction. It'd rather harder in XSLT 1.0, but there are well-known techniques.

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

Thanx for ur reply but as m new to xslt can u solve my problem as how can we merge the two records here.

Sorry, I don't solve problems for people, I try to help people solve them for themselves.

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

Sorry sir but u r getting me wrong as m searching on xslt grouping and able to find the distinct records but not able to get merge two records

what m doing is as:
xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:foo="http://whatever">



  <xsl:key name="distinctShift" match="NewDataSet/Detail" use="./idShiftNo"/>

 <xsl:for-each select="NewDataSet/Detail[generate-id() = generate-id(key('distinctShift', ./idShiftNo))]">
<xsl:template match="/">
<table>
<tr>
                <td>
                  <xsl:value-of select="idShiftNo"/>
                </td>
                <td>
                  <xsl:value-of select="ShiftName"/>
                </td>
</tr>
</table>
</xsl:template>
</xsl:stylesheet>
not able to get record combine

Looks like you are close, but your for-each loop is in the wrong place.

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



<xsl:key name="distinctShift" match="NewDataSet/Detail" use="./idShiftNo"/>

<xsl:template match="/">

<table>

<xsl:for-each select="NewDataSet/Detail[generate-id() = generate-id(key('distinctShift', ./idShiftNo)[1])]">
  <tr>
    <td>
      <xsl:value-of select="idShiftNo"/>
    </td>
    <td>
      <xsl:value-of select="ShiftName"/>
    </td>
    <td>
      <xsl:for-each select="/NewDataSet/Detail[idShiftNo = current()/idShiftNo]">
        <xsl:if test="position() != 1">,</xsl:if>
        <xsl:value-of select="idTaskNo"/>
      </xsl:for-each>
    </td>
  </tr>
</xsl:for-each>
</table>
</xsl:template>

</xsl:stylesheet>

I think that might do what you want.

/- Sam Judson : Wrox Technical Editor -/

In Muenchian grouping there is always an outer loop, which selects one element for each group, and an inner loop, which processes all the elements of the group. Your code has the outer loop, so it executes once for each group, but is missing the inner loop, which you will need to get the values (such as task1, task2) that are different between different elements in the group.

(The code also seems to have got mangled somehow, it has the for-each outside an xsl:template. I assume that's just carelessness, of the same kind that causes you to write in text shorthand. You need to cure yourself of this - careless people never make good programmers.)

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