Looping Element to Single Element in XSLT

问题

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> 

回答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>&#10;</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>

回答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> 

来源：https://stackoverflow.com/questions/40935863/looping-element-to-single-element-in-xslt