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Home/ Questions/Q 6864563
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T02:55:38+00:00 2026-05-27T02:55:38+00:00

I have a xml document, now i want to translate it to another xml

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I have a xml document, now i want to translate it to another xml document with same content but different element orders.

The original xml document like:

<?xml version = "1.0" encoding = "UTF-8"?>  
<order xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance" >  
 <ship>  
    <zipcode>78712</zipcode>  
    <street>1234 Main Street</street>  
    <country>CN</country>    
    <city>Beijing</city>  
 </ship>   
 <items>     
    <quantity>1</quantity>     
    <itemno>1234</itemno>  
 </items>     
 <items>     
    <quantity>3</quantity>    
    <itemno>1235</itemno>    
 </items>    
 <price>456</price>  
 <customer>Tom Hill</customer>    
</order>  

The expected output xml document like:

<?xml version = "1.0" encoding = "UTF-8"?>  
<order xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance" >  
 <customer>Tom Hill</customer>    
 <ship>  
    <street>1234 Main Street</street>  
    <city>Beijing</city>  
    <zipcode>78712</zipcode>  
    <country>CN</country>    
 </ship>    
 <items>     
    <itemno>1234</itemno>    
    <quantity>1</quantity>     
 </items>     
 <items>     
    <itemno>1235</itemno>    
    <quantity>3</quantity>    
 </items>    
 <price>456</price>  
</order> 

I used following xslt document to translate it.

<?xml version="1.0"?>  
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">  
<xsl:template match="/order">  
 <xsl:copy>  
  <xsl:copy-of select="customer" />  
  <xsl:copy-of select="ship" >  
  <xsl:call-template name="TShip" />  
  </xsl:copy-of>  
  <xsl:copy-of select="items">  
  <xsl:call-template name="TItems" />  
  </xsl:copy-of>  
 <xsl:copy-of select="price" />  
 </xsl:copy>  
</xsl:template>  

<xsl:template name="TShip">  
 <xsl:copy>  
  <xsl:copy-of select="street" />  
  <xsl:copy-of select="city" />  
  <xsl:copy-of select="zipcode" />  
  <xsl:copy-of select="country" />  
 </xsl:copy>  
</xsl:template>  

<xsl:template name="TItems">  
 <xsl:for-each select="items">  
  <xsl:copy>  
   <xsl:copy-of select="itemno" />  
   <xsl:copy-of select="quantity" />  
  </xsl:copy>  
 </xsl:for-each>  
</xsl:template>  

</xsl:stylesheet>  

However, the translated result is not my expected.
Translated result xml:

<?xml version = "1.0" encoding = "UTF-8"?>  
<order xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance" >  
 <customer>Tom Hill</customer>    
 <ship>  
    <zipcode>78712</zipcode>  
    <street>1234 Main Street</street>  
    <country>CN</country>    
    <city>Beijing</city>    
 </ship>    
 <items>     
    <quantity>1</quantity>     
    <itemno>1234</itemno>    
 </items>     
 <items>     
    <quantity>3</quantity>    
    <itemno>1235</itemno>   
 </items>    
 <price>456</price>  
</order>  

It just made the first level nodes in expected order. All sub-nodes are kept in original order. How can i make the order of all nodes as my expected ?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T02:55:39+00:00Added an answer on May 27, 2026 at 2:55 am

    xsl:copy-of copies all child nodes as well and child nodes of it are not evaluated.

    So your TShip and TItems templates are never even being evaluated. <xsl:copy-of select="ship"> copies all of <ship>...</ship>.

    This modification to your template will demonstrate that your TShip and TItems templates are not being called.

    <?xml version="1.0"?>  
    <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">  
    <xsl:template match="/order">  
     <xsl:copy>  
      <xsl:copy-of select="customer" />
        <xsl:copy-of select="ship">
      <xsl:call-template name="TShip" />  
    </xsl:copy-of>
      <xsl:copy-of select="items">  
      <xsl:call-template name="TItems" />  
      </xsl:copy-of>  
     <xsl:copy-of select="price" />  
     </xsl:copy>  
    </xsl:template>  
    
    <xsl:template name="TShip">  
     <xsl:copy>  
      <test>TShip called</test>
      <xsl:copy-of select="street" />  
      <xsl:copy-of select="city" />  
      <xsl:copy-of select="zipcode" />  
      <xsl:copy-of select="country" />  
     </xsl:copy>  
    </xsl:template>  
    
    <xsl:template name="TItems">  
     <xsl:for-each select="items">  
      <xsl:copy> 
      <test>TItems called</test>
       <xsl:copy-of select="itemno" />  
       <xsl:copy-of select="quantity" />  
      </xsl:copy>  
     </xsl:for-each>  
    </xsl:template>  
    
    </xsl:stylesheet>
    

    Notice that the output does not contain the <test> elements I added.

    What you need to do instead is recursive implicit copying. Usually xsl:copy, xsl:copy-of and xsl:for-each are a sign of bad xsl template design–there are very few problems which xsl:template and xsl:apply-template with an identity transform do not handle better.

    This is how I would do it:

    <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    
        <xsl:output encoding="UTF-8" indent="yes" method="xml" />
    
        <xsl:template match="order">
            <xsl:copy>
                <!-- copy all attributes; maybe you don't want this -->
                <xsl:apply-templates select="@*" />
                <!-- copy some elements in a specific order  -->
                <xsl:apply-templates select="customer" />
                <xsl:apply-templates select="ship" />
                <xsl:apply-templates select="items" />
                <xsl:apply-templates select="price" />
                <!-- now copy any other children that we haven't explicitly reordered; again, possibly this is not what you want -->
                <xsl:apply-templates select="*[not(self::customer or self::ship or self::items or self::price)]"/>
            </xsl:copy>
        </xsl:template>
    
        <xsl:template match="ship">
            <xsl:copy>
                <xsl:apply-templates select="@*" />
                <xsl:apply-templates select="street" />
                <xsl:apply-templates select="city" />
                <xsl:apply-templates select="zipcode" />
                <xsl:apply-templates select="country" />
                <xsl:apply-templates select="*[not(self::street or self::city or self::zipcode or self::country)]"/>
            </xsl:copy>
        </xsl:template>
    
        <xsl:template match="items">
            <xsl:copy>
                <xsl:apply-templates select="@*" />
                <xsl:apply-templates select="itemno" />
                <xsl:apply-templates select="quantity" />
                <xsl:apply-templates select="*[not(self::itemno or self::quantity)]"/>
            </xsl:copy>
        </xsl:template>
    
        <!-- this is the identity transform: it copies everything that isn't matched by a more specific template -->
        <xsl:template match="@*|node()">
            <xsl:copy>
                <xsl:apply-templates select="@*|node()"/>
            </xsl:copy>
        </xsl:template>
    
    </xsl:stylesheet>
    

    Notice how many fewer assumptions this template design makes about the structure of your source XML. It is also much easier to change: for example, if you want to silence or rename a particular element that may itself have children, you just add a new xsl:template that matches that element, do whatever you need to do, and xsl:apply-templates on the children.

    You should learn more about this XSLT pattern because it is very versatile and makes template authoring much less tedious and your templates much less brittle.

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