Hi!
I have seen "tranforming xml to xml" topic but my problem is little complicated.The X markup needs to be converted into Y markup.
The X markup is as follows:


<class classscheme="SAT">
   <classitem>
     <identifier>
        <name>Sales </name>
               <code>1234</code>
     </identifier>
     <classitem>
        <identifier>
          <name> Tax</name>
          <code>5678</code>
       </identifier>
     <classitem>
       <identifier>
         <name>Items </name>
         <code>9101</code>
       </identifier>
      <classitem>
         <identifier>
        <name>Manufacturing </name>
        <code>1121</code>
         </identifier>
           </classitem>
    </classitem>
     </classitem>
    </classitem>
</class>

The Y Markup is has follows:

     <hier>
          <hierlev role="ancestor">
             <heading>
           <title>Sales</title>
             </heading>
             <hierlev role="ancestor">
                <heading>
                <title> Tax</title>
                </heading>
                <hierlev role="ancestor">
              <heading>
                     <title>Items </title>
               </heading>
                   <hierlev role="ancestor">
                  <heading>
                 <title>Manufacturing </title>
                  </heading>
                      <hierlev role="ancestor">
                  <heading>
                  <title>Revised Code </title>
                   </heading>
               </hierlev>
              <hierlev role="me">
                       <heading>
                    <title> Title Value</title>
                   </heading>
               </hierlev>
                  </hierlev>
               </hierlev>
            </hierlev>
        </hierlev>
    </hier>

The preview screen showing little different format of the Y markup, so the tree structure is as follows:
  Hier
    -Hierlev
        -Heading
           -Title
        -Hierlev
           -Heading
              -Title
           -Hierlev
              -Heading
                 -Title
         so and so

Now two rules for the above conversion:
1)In X markup--The classitem depth is arbitary.
2)In Y Markup--- always the last hierlev element contains two hierlev as children.

I have tried to convert the above conversion but always I am getting following output:
   Hier
     -Hierlev
        -Heading
          -Title
     -Hierlev
        -Heading
          -Title
      so and so
At last I got the answer if the depth is fixed. But I need the way when the classitem depth is arbitary.
Thanks in advance for any suggestion/idea.

Please give me a suggestion or idea for my question....because another task also needs same needs.
Thanks in advance.

You asked the same question on another forum and you have received responses there telling you that poeple can't work out what your requirements are (which is actually the most common reason for not getting a response - most people don't bother replying if they can't work out what the question is).

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

I will try to explain my question briefly with small sample markup and small xslt code. At the end I will explain what is needed.
Question: Converting an XML markup to another XML Markup Using XSLT.
Sample XML Markup:
<class >
   <classitem>
     <identifier>
        <name>Sales </name>
               <code>1234</code>
     </identifier>
     <classitem>
        <identifier>
          <name> Tax</name>
          <code>5678</code>
       </identifier>
     </classitem>
    </classitem>
</class>

After XSL Transformation, the output markup is as follows:
  <hier>
          <hierlev >
             <heading>
                <title>Sales</title>
             </heading>
             <hierlev>
                <heading>
                    <title> Tax</title>
                </heading>
               </hierlev>
            </hierlev>
    </hier>

The mapping between Input and output markup elements as follows:

class - hier
Classitem - heirlev
identifier - heading
name - title
code has no mapping i.e. the element can be ignored in XSLT Code

I have written XSLT code which works for fixed depth of class.Actually I have written XSLT code for classItem depth of 4, but for simplicity... the following code is for classItem depth of 2 only.
<xsl:template match="class">
        <xsl:param name="count"/>
        <xsl:param name="v1" select="descendant::name[1]/text()"/>
        <xsl:param name="v2" select="descendant::name[2]/text()"/>
        <xsl:call-template name="createHierLevElement">
          <xsl:with-param name="hierLevContent">
           <xsl:element name="heading">
              <xsl:element name="title">
              <xsl:copy-of select="$v1"/>
               </xsl:element>
            </xsl:element>

                <xsl:if test=" $count = 2 ">
          <xsl:call-template name="createHierLevElement">
                      <xsl:with-param name="hierLevContent">
              <xsl:element name="heading">
                      <xsl:element name="title">
                                  <xsl:copy-of select="$v2"/>
                 </xsl:element>
              </xsl:element>
                       </xsl:with-param>
                   </xsl:call-template>
                </xsl:if>
             </xsl:with-param>
        </xsl:call-template>
  </xsl:template>


  <xsl:template name="createHierLevElement">
      <xsl:param name="hierLevContent" />

        <xsl:element name="hierlev">
           <xsl:copy-of select="$hierLevContent" />
        </xsl:element>
    </xsl:template>

If my input markup has arbitary depth, the above XSLT code will not work, so I need code which accomodates any depth of ClassItem.



Thanks.
Sridhar

You're raising the same questions on xsl-list. Please stick to one forum at a time.

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