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CaptainHastings
CaptainHastings

Reputation: 1597

Digester: Extracting node name

Is it possible to extract the node name using apache digester?

So, if the xml looks like 

   <furniture>
     <sofa>
       .....
     </sofa>
     <coffeeTable>
       .....
     </coffeeTable>
   </furniture>

is it possible to extract the node name "sofa", "coffeeTable"?

I know it is possible using xpath but is it possible using digester?

Cheers

Upvotes: 1

Views: 1168

Answers (2)

polygenelubricants
polygenelubricants

Reputation: 383876

This is the kind of matching that ExtendedBaseRules afford.

Let's say this is the content of furniture.xml:

<furniture>
   <sofa>
      <blah/>
   </sofa>
   <coffeeTable>
      <bleh/>
   </coffeeTable>
</furniture>

Let's say you want to get the element names of the direct children of furniture element. This is what furniture/? matches in the ExtendedBaseRules.

import java.io.*;
import java.util.*;
import org.apache.commons.digester.*;

public class FurnitureDigest {
    public static void main(String[] args) throws Exception {
        File file = new File("furniture.xml");
        Digester digester = new Digester();
        digester.setRules(new ExtendedBaseRules());
        final List<String> furnitures = new ArrayList<String>();
        digester.addRule("furniture/?", new Rule() {
            @Override public void end(String nspace, String name) {
                furnitures.add(name);
            }
        });
        digester.parse(file);
        System.out.println(furnitures); // prints "[sofa, coffeeTable]"
    }
}

API links

Upvotes: 0

Barend
Barend

Reputation: 17419

(original answer)

Create a Digester for pattern "furniture/*" with a simple Rule that takes the second parameter to each call to the begin method and sticks it in a collection of your choice (a list to get all of them, a set to get only all unique names).

(edit)

Scratch that, it's a bit more complicated.

This works:

public class App 
{
    final static Rule printRule = new Rule() {
        public void begin(String namespace, String name,
                Attributes attributes) throws Exception {
            System.out.println(name);
        }
    }; 
    public static void main( String[] args ) throws IOException, SAXException
    {
        InputStream instr = App.class.getResourceAsStream("/sample.xml");
        Digester dig = new Digester();
        dig.setRules(new RulesBase(){
            public List<Rule> match(String namespaceURI, String pattern) {
                return Arrays.asList(printRule);
            }
        });
        dig.parse(instr);
    }
}

This particular sample will print all element names including the root furniture element. I'll leave it to you to adjust the match() method to your needs.

Upvotes: 1

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