Newbie question on Recusrion using XLST 1.0 Part 2

Hi All,
       Following from my previous email. I have updated my code to no avail.
Again I won't to display each EID attribute for each employee grouped by
division, using XSLT version 1.0.

Any suggestions most appreciated.

Thanks Marco Mastrocinque 

 
<xsl:variable name="result">
   <xsl:call-template name="DisplayEmployees">
	<xsl:with-param name="employeeList" select="company/division"/>
   </xsl:call-template>
</xsl:variable>								
<br/>
<h3>The value of result is <xsl:value-of select="$result"/></h3>	
	

</xsl:template>
    <xsl:template name="DisplayEmployees">
    <!--This is a recursive routine-->
    <xsl:param name="employeeList"/>
	 <xsl:choose>
	    <xsl:when test="not($employeeList)">
	    <!--If the nodelist is empty return true-->
		 <xsl:text>true</xsl:text>
	    </xsl:when>
	    <xsl:otherwise>
		<xsl:choose>
		<!--Get the first employee EID attribute -->
		<xsl:when test="$employeeList/employees/employee[1]">
		  <xsl:value-of select="@EID"/>	
		</xsl:when>
		<xsl:otherwise>
		<!--Get the remaianing EID attributes -->
		<xsl:choose>
		  <xsl:when
test="$employeeList/employees/employee[position() &gt; 1]"/>
		  </xsl:choose>
		  <!--Get the remaining employee EID attributes -->
		  <xsl:call-template name="DisplayEmployees">
			<xsl:with-param name="employeeList"
select="$employeeList[position() &gt; 1]"/>
		  </xsl:call-template>
	     </xsl:otherwise>
	  </xsl:choose>
	 </xsl:otherwise>
      </xsl:choose>
    </xsl:template>       		
</xsl:stylesheet>


-----Original Message-----
From: Marco Mastrocinque [mailto:mmfive@xxxxxxxxxxxxxxx] 
Sent: Saturday, 30 April 2005 5:16 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject:  Newbie question on Recusrion using XLST 1.0

Hi All,
       I have an example file with the following structure:

<company>
   <division DID="Sal">
      <employees>
		<employee EID="EMP001">
		   <ENAME>Joe Blow</ENAME>
               <OFFICE>1.2.27</OFFICE>
               <BIRTHDATE>1963-08-01</BIRTHDATE>
		   <SALARY>28790</SALARY>	  	
		</employee>	
		<employee EID="EMP002">
		   <ENAME>John Doe</ENAME>
               <OFFICE>1.2.2</OFFICE>
               <BIRTHDATE>1970-08-01</BIRTHDATE>
		   <SALARY>29855</SALARY>	  	
		</employee>	
	</employees>	
   </division>
   <division DID="SOF">
      <employees>
		<employee EID="EMM001">
		   <ENAME>Mark Smith</ENAME>
               <OFFICE>1.2.27</OFFICE>
               <BIRTHDATE>1966-08-01</BIRTHDATE>
		   <SALARY>35000</SALARY>	  	
		</employee>	
		<employee EID="EMM002">
		   <ENAME>John Doe</ENAME>
               <OFFICE>1.2.2</OFFICE>
               <BIRTHDATE>1977-08-01</BIRTHDATE>
		   <SALARY>40000</SALARY>	  	
		</employee>	
	</employees>	
   </division>	
</company>



I'm having using a recursive routine (see below) to display the following
elements and attributes:
1) The employee EID attribute such as 'EMP001', followed by
2) The element text of ENAME, such as 'Joe Blow',
3) The element text of OFFICE, such as '1.2.27'
4) The output must be formatted such that the employees of each division are
grouped together.

Here is my code so far, for the above snippet of XML code which has numerous
errors:

<xsl:variable name="result">
  <xsl:call-template name="DisplayEmployees">
	<xsl:with-param name="employeeList" select="company/division"/>
  </xsl:call-template>
</xsl:variable>	
<xsl:value-of select="$result"/>		


<xsl:template name="DisplayEmployees">
    <!--This is a recursive routine-->
		<xsl:param name="employeeList"/>
		<xsl:choose>
			<xsl:when test="not($employeeList)">
			    <!--If the nodelist is empty return true-->
				<xsl:text>true</xsl:text>
			</xsl:when>
			<xsl:otherwise>
				<xsl:choose>
					<xsl:when
test="employees/employee[@EID]">
						<xsl:value-of
select="@EID"/>	
					</xsl:when>
				</xsl:choose>
			</xsl:otherwise>
		</xsl:choose>
    </xsl:template>

I can't seem to get any output. Trying to put it in a html file. Please note
I'm new to XSLT and XPATH so please excuse the numerous errors.

Thanks Marco Mastrocinque