I am getting error while tranforming xml by XSLT when xml file having rowset with no rows.

And I suppose you expect me to tell you that your mistake is a missing closing bracket on line 126 of your stylesheet.

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

my xml source is like :

 <?xml version="1.0" encoding="WINDOWS-1252" ?>
- <page>
  <rowset />
  </page>

and XSL is like :

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

<xsl:key name="cd-by-network" match="ROW" use="NETWORK_NAME" />
<xsl:key name="cd-by-show" match="ROW" use="BASIS_NUMBER" />
<xsl:key name="cd-by-seg" match="ROW" use="SEG" />
<xsl:template match="ROWSET/ROW">
<xsl:for-each select="current()[generate-id()=generate-id(key('cd-by-network',NETWORK_NAME)[1])]">
   <ws_structure_import network_name="{NETWORK_NAME}" broadcast_Date="{BROADCAST_DATE}" schedule_name="{SCHEDULE_NAME}" >
   <xsl:for-each select="key('cd-by-network', NETWORK_NAME)">

      <xsl:for-each select="current()[generate-id()=generate-id(key('cd-by-show',BASIS_NUMBER)[1])]">

        <show basis_number="{BASIS_NUMBER}" program_title="{PROGRAM_TITLE}" start_time="{START_TIME}">
            <xsl:for-each select="key('cd-by-show', BASIS_NUMBER)">

              <xsl:for-each select="current()[generate-id()=generate-id(key('cd-by-seg',SEG)[1])]">

               <segment sort_col="{SORT_COL}" segment_no="{SEGMENT_NO}">
                  <xsl:for-each select="key('cd-by-seg', SEG)">
                    <avail pos_in_seg="{POS_IN_SEG}" record_type="{RECORD_TYPE}" avail_type="{AVAIL_TYPE}" duration="{DURATION}" />
                </xsl:for-each>
               </segment>
            </xsl:for-each>
        </xsl:for-each>
       </show>
     </xsl:for-each>
    </xsl:for-each>
   </ws_structure_import>
  </xsl:for-each>
</xsl:template>
</xsl:stylesheet>


   Suppose I am having rows under rowset then I get proper result. But in case of zero rows getting following error:



The XML page cannot be displayed
Cannot view XML input using style sheet. Please correct the error and then click the Refresh button, or try again later.


--------------------------------------------------------------------------------

Invalid at the top level of the document. Error processing resource 'http://localhost:8080/ASM/getFormat.xsql?bdate=2007111...

<?xml version = '1.0'?>

When you use XSLT to generate XML, there is no guarantee that the resulting XML will be well-formed, that is, that it will have a single element node as the document element. You can generate XML with no top-level elements or with multiple top-level elements. If you do that (as you have done) then IE will not be able to display it. It's best always to generate well-formed XML, which you can do by adding the template rule:

<xsl:template match="ROWSET">
<doc>
  <xsl:apply-templates/>
</doc>
</xsl:template>

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

thanks a lot Michael .............
its working now.......