Question

I have been trying to write a style sheet for the following documents which can check for all the users their awards and add the name of that user as subelement to the /awards/award element

<document>
    <users>
        <user identity="1">
            <name>
                <first>abc</first>
                <last>xyz</last>
            </name>
            <achievements>
                <achievement>Award A</achievement>
                <achievement>Award B</achievement>
            </achievements>
        </user>
        <user identity="2">
            <name>
                <first>123</first>
                <last>DEF</last>
            </name>
            <achievements>
                <received>Award A</received>
            </achievements>
        </user>
        <user identity = "3">
            <name>
                <first>aaa</first>
                <last>bbb</last>
            </name>
            <achievements>
                <received>Award B</received>
            </achievements>
        </user>
    </users>
    <!--awards-->
    <awards>
        <award>
            <name>Award A</name>
            <!--here the list of the users who recieved this award has to be included-->
        </award>
        <award>
            <name>Award B</name>
        </award>
    </awards>
</document>

I wrote down the following incomplete stylesheet

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="awards/award">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
                <!--how can i create a list of all users who have this award-->
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

I can not figure out the way of writing a query in the template “awards/award” like “add all those users who have received this award as sub elements called user and the value of added element to be the name of the users” and gives following output. I will much appreciate if someone can guide me to the correct approach.

<document>
    <users>
        <user identity="1">
            <name>
                <first>abc</first>
                <last>xyz</last>
            </name>
            <achievements>
                <achievement>Award A</achievement>
                <achievement>Award B</achievement>
            </achievements>
        </user>
        <user identity="2">
            <name>
                <first>123</first>
                <last>DEF</last>
            </name>
            <achievements>
                <received>Award A</received>
            </achievements>
        </user>
        <user identity="3">
            <name>
                <first>aaa</first>
                <last>bbb</last>
            </name>
            <achievements>
                <received>Award B</received>
            </achievements>
        </user>
    </users>
    <!--awards-->
    <awards>
        <award>
            <name>Award A</name>
            <user identity="1">abc xyz</user>
            <user identity="2">123 DEF</user>    
        </award>
        <award>
            <name>Award B</name>
            <user identity="1">abc xyz</user>
            <user identity="3">aaa bbb</user>
        </award>
    </awards>
</document>

thanks in advance

Answer 1

This transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:key name="kUserByAward" match="user"
          use="achievements/*[self::achievement or self::received]"/>

 <xsl:template match="node()|@*">
     <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
     </xsl:copy>
 </xsl:template>

 <xsl:template match="award">
   <award>
    <xsl:apply-templates select="node()"/>
    <xsl:apply-templates select="key('kUserByAward', name)" mode="inclusion"/>
   </award>
 </xsl:template>

 <xsl:template match="user" mode="inclusion">
  <user>
    <xsl:copy-of select="name"/>
  </user>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document:

<document>
    <users>
        <user>
            <name>
                <first>abc</first>
                <last>xyz</last>
            </name>
            <achievements>
                <achievement>Award A</achievement>
                <achievement>Award B</achievement>
            </achievements>
        </user>
        <user>
            <name>
                <first>dsd</first>
                <last>sdasdsadsa</last>
            </name>
            <achievements>
                <received>Award A</received>
            </achievements>
        </user>
        <user>
            <name>
                <first>aaa</first>
                <last>bbb</last>
            </name>
            <achievements>
                <received>Award B</received>
            </achievements>
        </user>
    </users>
    <!--awards-->
    <awards>
        <award>
            <name>Award A</name>
            <!--here the list of the users who recieved this award has to be included-->
        </award>
        <award>
            <name>Award B</name>
        </award>
    </awards>
</document>

produces (what I guess should be) the wanted, correct result:

<document>
   <users>
      <user>
         <name>
            <first>abc</first>
            <last>xyz</last>
         </name>
         <achievements>
            <achievement>Award A</achievement>
            <achievement>Award B</achievement>
         </achievements>
      </user>
      <user>
         <name>
            <first>dsd</first>
            <last>sdasdsadsa</last>
         </name>
         <achievements>
            <received>Award A</received>
         </achievements>
      </user>
      <user>
         <name>
            <first>aaa</first>
            <last>bbb</last>
         </name>
         <achievements>
            <received>Award B</received>
         </achievements>
      </user>
   </users><!--awards-->
   <awards>
      <award>
         <name>Award A</name><!--here the list of the users who recieved this award has to be included-->
         <user>
            <name>
               <first>abc</first>
               <last>xyz</last>
            </name>
         </user>
         <user>
            <name>
               <first>dsd</first>
               <last>sdasdsadsa</last>
            </name>
         </user>
      </award>
      <award>
         <name>Award B</name>
         <user>
            <name>
               <first>abc</first>
               <last>xyz</last>
            </name>
         </user>
         <user>
            <name>
               <first>aaa</first>
               <last>bbb</last>
            </name>
         </user>
      </award>
   </awards>
</document>

Explanation: Proper use of keys and overriding the identity rule.

UPDATE: The OP has shown the actually wanted output format. This is very close to what I guessed. The only modification is to use this template, insted the one in the initial solution:

 <xsl:template match="user" mode="inclusion">
  <user identity="{@identity}">
    <xsl:copy-of select="concat(name/first, ' ', name/last)"/>
  </user>
 </xsl:template>