Transform the Input xml into Output xml using xslt

Question

Input XML structure:

<Customer>
   <Order>
      <item>
        <name>ID</name>
        <value>11111</value>
      </item>
   </Order>
   <Order>
      <item>
        <name>ID</name>
        <value>11111</value>
      </item>
   </Order>
    <Order>
      <item>
        <name>ID</name>
        <value>22222</value>
      </item>
   </Order>
   <Order>
      <item>
        <name>ID</name>
        <value>33333</value>
      </item>
   </Order>
</Customer>

Output should be :

<Customer>
   <Order>
      <item>
        <name>ID</name>
        <value>11111</value>
      <item>
   </Order>
   <Order>
      <item>
        <name>ID</name>
        <value>11111</value>
      </item>
   </Order>
</Customer>
<Customer>
    <Order>
      <item>
        <name>ID</name>
        <value>22222</value>
      </item>
   </Order>
</Customer>
<Customer>
    <Order>
      <item>
        <name>ID</name>
        <value>33333</value>
      </item>
   </Order>
</Customer>

Here the /Customer/<Order/item/value will come dynamically. Please anyone give a solution for this transformation.

You do realize that the requested output is invalid XML (you need a single document element)? Is there an element that is supposed to wrap that content(e.g. <doc><Customer>...</Customer><Customer>...</Customer></doc>)? — Mads Hansen
– Mads Hansen, Commented Mar 15, 2011 at 11:13
@Mads Hansen: About output validation, do note that for xml serialization method there are two options, from w3.org/TR/xslt#section-XML-Output-Method: "The xml output method outputs the result tree as a well-formed XML external general parsed entity. If the root node of the result tree has a single element node child and no text node children, then the entity should also be a well-formed XML document entity." — user357812
– user357812, Commented Mar 16, 2011 at 0:42

user357812 · Accepted Answer · 2011-03-15 15:42:00Z

1

This stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:key name="kOrderByValue" match="Order" use="item/value"/>
    <xsl:template match="Customer">
        <xsl:for-each select="Order[count(.|key('kOrderByValue',
                                                item/value
                                            )[1]
                                    ) = 1]">
            <Customer>
                <xsl:apply-templates select="key('kOrderByValue',
                                                 item/value
                                             )"/>
            </Customer>
        </xsl:for-each>
    </xsl:template>
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

Output:

<Customer>
    <Order>
        <item>
            <name>ID</name>
            <value>11111</value>
        </item>
    </Order>
    <Order>
        <item>
            <name>ID</name>
            <value>11111</value>
        </item>
    </Order>
</Customer>
<Customer>
    <Order>
        <item>
            <name>ID</name>
            <value>22222</value>
        </item>
    </Order>
</Customer>
<Customer>
    <Order>
        <item>
            <name>ID</name>
            <value>33333</value>
        </item>
    </Order>
</Customer>

Note: grouping Customer's Order children by value.

answered Mar 15, 2011 at 15:42

user357812

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2 Comments

Mads Hansen Over a year ago

+1 Good solution. I hadn't noticed the grouping of orders when I posted mine. Deleting my answer.

user357812 Over a year ago

@Mads Hansen: Thanks. Reposting comment above because I think it has some value.

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