Normalization in DOM parsing with java - how does it work?

JavaXmlDom

Java Problem Overview


I saw the line below in code for a DOM parser at this tutorial.

doc.getDocumentElement().normalize();

Why do we do this normalization ?
I read the docs but I could not understand a word.

> Puts all Text nodes in the full depth of the sub-tree underneath this Node

Okay, then can someone show me (preferably with a picture) what this tree looks like ?

Can anyone explain me why normalization is needed?
What happens if we don't normalize ?

Java Solutions


Solution 1 - Java

The rest of the sentence is:

> where only structure (e.g., elements, comments, processing instructions, CDATA sections, and entity references) separates Text nodes, i.e., there are neither adjacent Text nodes nor empty Text nodes.

This basically means that the following XML element

<foo>hello 
wor
ld</foo>

could be represented like this in a denormalized node:

Element foo
    Text node: ""
    Text node: "Hello "
    Text node: "wor"
    Text node: "ld"
    

When normalized, the node will look like this

Element foo
    Text node: "Hello world"

And the same goes for attributes: <foo bar="Hello world"/>, comments, etc.

Solution 2 - Java

In simple, Normalisation is Reduction of Redundancies.
Examples of Redundancies:
a) white spaces outside of the root/document tags(...<document></document>...)
b) white spaces within start tag (<...>) and end tag (</...>)
c) white spaces between attributes and their values (ie. spaces between key name and =")
d) superfluous namespace declarations
e) line breaks/white spaces in texts of attributes and tags
f) comments etc...

Solution 3 - Java

As an extension to @JBNizet's answer for more technical users here's what implementation of org.w3c.dom.Node interface in com.sun.org.apache.xerces.internal.dom.ParentNode looks like, gives you the idea how it actually works.

public void normalize() {
    // No need to normalize if already normalized.
    if (isNormalized()) {
        return;
    }
    if (needsSyncChildren()) {
        synchronizeChildren();
    }
    ChildNode kid;
    for (kid = firstChild; kid != null; kid = kid.nextSibling) {
         kid.normalize();
    }
    isNormalized(true);
}

It traverses all the nodes recursively and calls kid.normalize()
This mechanism is overridden in org.apache.xerces.dom.ElementImpl

public void normalize() {
     // No need to normalize if already normalized.
     if (isNormalized()) {
         return;
     }
     if (needsSyncChildren()) {
         synchronizeChildren();
     }
     ChildNode kid, next;
     for (kid = firstChild; kid != null; kid = next) {
         next = kid.nextSibling;

         // If kid is a text node, we need to check for one of two
         // conditions:
         //   1) There is an adjacent text node
         //   2) There is no adjacent text node, but kid is
         //      an empty text node.
         if ( kid.getNodeType() == Node.TEXT_NODE )
         {
             // If an adjacent text node, merge it with kid
             if ( next!=null && next.getNodeType() == Node.TEXT_NODE )
             {
                 ((Text)kid).appendData(next.getNodeValue());
                 removeChild( next );
                 next = kid; // Don't advance; there might be another.
             }
             else
             {
                 // If kid is empty, remove it
                 if ( kid.getNodeValue() == null || kid.getNodeValue().length() == 0 ) {
                     removeChild( kid );
                 }
             }
         }

         // Otherwise it might be an Element, which is handled recursively
         else if (kid.getNodeType() == Node.ELEMENT_NODE) {
             kid.normalize();
         }
     }

     // We must also normalize all of the attributes
     if ( attributes!=null )
     {
         for( int i=0; i<attributes.getLength(); ++i )
         {
             Node attr = attributes.item(i);
             attr.normalize();
         }
     }

 	// changed() will have occurred when the removeChild() was done,
 	// so does not have to be reissued.

     isNormalized(true);
 } 

Hope this saves you some time.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionApple GrinderView Question on Stackoverflow
Solution 1 - JavaJB NizetView Answer on Stackoverflow
Solution 2 - JavaAVAView Answer on Stackoverflow
Solution 3 - JavaMatas VaitkeviciusView Answer on Stackoverflow