rearrange xml nodes including sub-nodes by xslt

Question

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 ?

Answer 1

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.

Answer 2

How can i make the order of all nodes as my expected ?

The short answer: By using <xsl:apply-templates/> and <xsl:template> instead of <xsl:copy-of>

---

Here is a complete transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="node()|@*">
  <xsl:copy>
   <xsl:apply-templates select="node()|@*"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="order">
   <xsl:copy>
    <xsl:apply-templates select="customer"/>
    <xsl:apply-templates select="*[not(self::customer)]"/>
   </xsl:copy>
 </xsl:template>

 <xsl:template match="ship">
  <xsl:copy>
   <xsl:apply-templates select="street"/>
   <xsl:apply-templates select="city"/>
   <xsl:apply-templates select="zipcode"/>
   <xsl:apply-templates select="country"/>
  </xsl:copy>
 </xsl:template>

 <xsl:template match="items">
  <xsl:copy>
   <xsl:apply-templates select="itemno"/>
   <xsl:apply-templates select="quantity"/>
  </xsl:copy>
 </xsl:template>
</xsl:stylesheet>

when this transformation is applied on the provided XML document:

<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 wanted, correct result is produced:

<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>

Explanation:

<xsl:copy-of select="someElement"/>

copies the whole subtree rooted by someElement exactly as is (and if we had an instruction that rearranges the descendants, how would this instruction know the order that we want???).

In order to change the order of any siblings - elements, we must specify the new, wanted order.

This can be done by writing a sequence of <xsl:apply-templates> instructions, each selecting the desired element -- in the desired order. We could write <xsl:copy-of> instructions, but only for copying elements, whose descendents we want to remain in their original order.