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Home/ Questions/Q 770165
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T18:27:17+00:00 2026-05-14T18:27:17+00:00

I am trying to learn XSLT but I work best by example. I want

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I am trying to learn XSLT but I work best by example. I want to perform a trivial schema to schema transformation. How do I perform this transformation in only one pass (my current solution uses two passes and loses the original order of customers)?

From:

<?xml version="1.0" encoding="UTF-8"?>
<sampleroot>

<badcustomer>
    <name>Donald</name>
    <address>Hong Kong</address>
    <age>72</age>
</badcustomer>

<goodcustomer>
    <name>Jim</name>
    <address>Wales</address>
    <age>22</age>
</goodcustomer>

<goodcustomer>
    <name>Albert</name>
    <address>France</address>
    <age>51</age>
</goodcustomer>

</sampleroot>

To :

<?xml version="1.0" encoding="UTF-8"?>
<records>

<record id="customer">
    <name>Donald</name>
    <address>Hong Kong</address>
    <age>72</age>
    <customertype>bad</customertype>
</record>

<record id="customer">
    <name>Jim</name>
    <address>Wales</address>
    <age>22</age>
    <customertype>good</customertype>
</record>

<record id="customer">
    <name>Albert</name>
    <address>France</address>
    <age>51</age>
    <customertype>good</customertype>
</record>

</records>

I already solved this a bad way (I lose the order of customers and I think that I have to parse the file multiple times:

<?xml version='1.0'?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="/sampleroot">

    <records>

        <xsl:for-each select="goodcustomer">
            <record id="customer">
                <name><xsl:value-of select="name" /></name>
                <address><xsl:value-of select="address" /></address>
                <age><xsl:value-of select="age" /></age>
                <customertype>good</customertype>
            </record>
        </xsl:for-each>

        <xsl:for-each select="badcustomer">
            <record id="customer">
                <name><xsl:value-of select="name" /></name>
                <address><xsl:value-of select="address" /></address>
                <age><xsl:value-of select="age" /></age>
                <customertype>bad</customertype>
            </record>
        </xsl:for-each>

    </records>
    </xsl:template>
</xsl:stylesheet>

Please can someone help me out with the correct XSLT construct where I only have to use a single parse (only one for-each)?

Thanks,

Chris

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T18:27:18+00:00Added an answer on May 14, 2026 at 6:27 pm

    It is a good XSLT practice to avoid using <xsl:for-each> as much as possible.

    Here is a simple solution, using this principle:

    <xsl:stylesheet version="1.0"
     xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
     <xsl:output omit-xml-declaration="yes" indent="yes"/>
    
     <xsl:template match="node()|@*">
      <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
      </xsl:copy>
     </xsl:template>
    
     <xsl:template match="/*">
      <records>
        <xsl:apply-templates/>
      </records>
     </xsl:template>
    
     <xsl:template match="badcustomer | goodcustomer">
      <record>
       <xsl:apply-templates/>
       <customertype>
         <xsl:value-of select="substring-before(name(), 'customer')"/>
       </customertype>
      </record>
     </xsl:template>
    </xsl:stylesheet>
    

    Do note:

    1. Only templates and <xsl:apply-templates> are used.

    2. The use of the identity rule and its overriding wherever necessary. This is one of the most fundamental XSLT design pattern.

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