Looping Element to Single Element in XSLT - xslt

Question

I need to convert the following content
<QTLS_ITEM>
<ID>123</ID>
<ID1>1345</ID1>
<SERAIL_NUMBER>1026977­04257</SERAIL_NUMBER>
<PROD_NAME>upgrade</PROD_NAME>
</QTLS_ITEM>
<QTLS_ITEM>
<ID>123</ID>
<ID1>1345</ID1>
<SERAIL_NUMBER>1026977­04257</SERAIL_NUMBER>
<PROD_NAME>Plug­in</PROD_NAME>
</QTLS_ITEM>
<QTLS_ITEM>
<ID>123</ID>
<ID1>1345</ID1>
<SERAIL_NUMBER>1026977­04257</SERAIL_NUMBER>
<PROD_NAME>License</PROD_NAME>
</QTLS_ITEM>
This is a looping element type.<QTLS_ITEM> is a repeating element. For each element I have to concatenate those two fields and get the value as below.
I want to transform it into a single Element like
<Item_description>
1026977­04257 upgrade
1026977­04257 Plug­in
1026977­04257 License
<Item_Description>
Which means I need to concatenate both the SERAIL_NUMBER and PROD_NAME.
Can anyone help on this?
<xsl:template name="string-join">
<xsl:param name="nodes"/>
<xsl:param name="delimiter"/>
<xsl:for-each select="$nodes">
<xsl:value-of select="."/>
<xsl:if test="$nodes[position()!=last()-1]">
<xsl:value-of select="$delimiter"/>
</xsl:if>
</xsl:for-each>
</xsl:template>
I am using this to get the Serial numbers separated by comma. But I want to concatenate and get the result in the above format.
The answer for my case is
<xsl:template name="join">
<xsl:param name="list"/>
<xsl:param name="separator"/>
<xsl:for-each select="db:SERAIL_NUMBER | db:PROD_NAME">($list value)
<xsl:value-of select="."/>
<xsl:if test="position() != last()">
<xsl:value-of select="$separator"/>
</xsl:if>
</xsl:for-each>
</xsl:template>

Answer 1

The input that you show us is not well-formed XML, because it does not have a single root element. Given a well-formed XML input such as:
XML
<root>
<QTLS_ITEM>
<SERAIL_NUMBER>102697704257</SERAIL_NUMBER>
<PROD_NAME>upgrade</PROD_NAME>
</QTLS_ITEM>
<QTLS_ITEM>
<SERAIL_NUMBER>102697704257</SERAIL_NUMBER>
<PROD_NAME>Plugin</PROD_NAME>
</QTLS_ITEM>
<QTLS_ITEM>
<SERAIL_NUMBER>102697704257</SERAIL_NUMBER>
<PROD_NAME>License</PROD_NAME>
</QTLS_ITEM>
</root>
you can do simply:
XSLT 2.0
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/root">
<Item_description>
<xsl:for-each select="QTLS_ITEM">
<xsl:value-of select="SERAIL_NUMBER, PROD_NAME" />
<xsl:text>
</xsl:text>
</xsl:for-each>
</Item_description>
</xsl:template>
</xsl:stylesheet>
to get:
Result
<?xml version="1.0" encoding="UTF-8"?>
<Item_description>102697704257 upgrade
102697704257 Plugin
102697704257 License
</Item_description>

Answer 2

This has worked out in my case
<xsl:template name="join">
<xsl:param name="list"/>
<xsl:param name="separator"/>
<xsl:for-each select="db:SERAIL_NUMBER|db:PROD_NAME">($list value)
<xsl:value-of select="."/>
<xsl:if test="position() != last()">
<xsl:value-of select="$separator"/>
</xsl:if>
</xsl:for-each>
</xsl:template>