how to sort xml code in ascending order in java

Question

Below is my XML code i need to display the 'node name' in ascending order.

XML :

<node label="Tree Data">
<node name="View" type="Page">
<node name="Organisational Structure" type="Page"/>
<node name="Experience" type="Page"/>
<node name="Expertise" type="Language">
<node name="Flex" type="Language"/>
<node name="Android" type="Language"/>
<node name="Java" type="Language"/>
</node>
<node name="Project" type="Project">
<node name="Sabre" type="Project"/>
<node name="Android Project" type="Project"/>
<node name="Media Framework" type="Project"/>
<node name="CCD" type="Project"/>
<node name="Redbus" type="Project"/>
<node name="India Today" type="Project"/>
<node name="Money Control" type="Project"/>
</node>
</node>
<node name="Reorganize" type="Action"/>
<node name="Add New Employee" type="Action">
<node name="Dir" type="Page"/>
<node name="P.M" type="Page"/>
<node name="S.E" type="Page"/>
<node name="JUN" type="Page"/>
<node name="EMP" type="Page"/>
</node>
<node name="Add Project Members" type="Action"/>
</node>

Following is my java code :

JAVA CODE :

import java.io.IOException;
        import java.io.StringWriter;
        import java.net.URL;
        import java.util.ArrayList;
        import java.util.Collections;
        import java.util.List;

        import javax.xml.parsers.DocumentBuilder;
        import javax.xml.parsers.DocumentBuilderFactory;
        import javax.xml.parsers.ParserConfigurationException;
        import javax.xml.transform.OutputKeys;
        import javax.xml.transform.Transformer;
        import javax.xml.transform.TransformerException;
        import javax.xml.transform.TransformerFactory;
        import javax.xml.transform.dom.DOMSource;
        import javax.xml.transform.stream.StreamResult;

        import org.w3c.dom.Document;
        import org.w3c.dom.Element;
        import org.w3c.dom.NamedNodeMap;
        import org.w3c.dom.Node;
        import org.w3c.dom.NodeList;
        import org.xml.sax.InputSource;
        import org.xml.sax.SAXException;

        import android.os.Bundle;
        import android.app.Activity;
        import android.view.Menu;

        public class MainActivity extends Activity {

            Document dom;

            public MainActivity() {
                MainActivity();

                ArrayList<Element> listGroup = getGroups();

            }




            private ArrayList getGroups() {
                try {
                    Element documentElement = (Element) dom.getDocumentElement();
                    System.out.println("root element is :"+documentElement.getNodeName());
                    NodeList childNodes = documentElement.getElementsByTagName("node");

                    for (int i = 0; i < childNodes.getLength(); i++) {

                        if (childNodes.item(i).getNodeType() == Node.ELEMENT_NODE) {
                            Element element = (Element) childNodes.item(i);
                            if(((Element) childNodes.item(i)).hasAttribute("name")) {
                            String name = element.getAttribute("name");
                            String type = element.getAttribute("type");
                            System.out.println("name : " + name);
                            System.out.println("type : " + type);
                        }}
                    }
                } catch (Exception e) {
                }
                return null;

            }

            private void MainActivity() {
                DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();


                try {
                    String sURL = "huhhadjdhjshdjks.com";
                    URL url = new URL(sURL);
                    DocumentBuilder db = dbf.newDocumentBuilder();

                    dom = db.parse(new InputSource(url.openStream()));

                } catch (ParserConfigurationException pce) {
                    pce.printStackTrace();
                } catch (SAXException se) {
                    se.printStackTrace();
                } catch (IOException ioe) {
                    ioe.printStackTrace();
                }
            }
        }

I am getting the following output :

logcat :

01-31 17:45:59.330: I/System.out(1871): root element is :node
01-31 17:45:59.340: I/System.out(1871): name : View
01-31 17:45:59.360: I/System.out(1871): type : Page
01-31 17:45:59.380: I/System.out(1871): name : Organisational Structure
01-31 17:45:59.380: I/System.out(1871): type : Page
01-31 17:45:59.400: I/System.out(1871): name : Experience
01-31 17:45:59.400: I/System.out(1871): type : Page
01-31 17:45:59.400: I/System.out(1871): name : Expertise
01-31 17:45:59.400: I/System.out(1871): type : Language
01-31 17:45:59.400: I/System.out(1871): name : Flex
01-31 17:45:59.410: I/System.out(1871): type : Language
01-31 17:45:59.410: I/System.out(1871): name : Android
01-31 17:45:59.420: I/System.out(1871): type : Language
01-31 17:45:59.420: I/System.out(1871): name : Java
01-31 17:45:59.430: I/System.out(1871): type : Language
01-31 17:45:59.430: I/System.out(1871): name : Project
01-31 17:45:59.430: I/System.out(1871): type : Project
01-31 17:45:59.440: I/System.out(1871): name : Sabre
01-31 17:45:59.440: I/System.out(1871): type : Project
01-31 17:45:59.440: I/System.out(1871): name : Android Project
01-31 17:45:59.450: I/System.out(1871): type : Project
01-31 17:45:59.450: I/System.out(1871): name : Media Framework
01-31 17:45:59.460: I/System.out(1871): type : Project
01-31 17:45:59.460: I/System.out(1871): name : CCD
01-31 17:45:59.460: I/System.out(1871): type : Project
01-31 17:45:59.460: I/System.out(1871): name : Redbus
01-31 17:45:59.460: I/System.out(1871): type : Project
01-31 17:45:59.460: I/System.out(1871): name : India Today
01-31 17:45:59.480: I/System.out(1871): type : Project
01-31 17:45:59.480: I/System.out(1871): name : Money Control
01-31 17:45:59.489: I/System.out(1871): type : Project
01-31 17:45:59.489: I/System.out(1871): name : Reorganize
01-31 17:45:59.500: I/System.out(1871): type : Action
01-31 17:45:59.500: I/System.out(1871): name : Add New Employee
01-31 17:45:59.500: I/System.out(1871): type : Action
01-31 17:45:59.500: I/System.out(1871): name : Dir
01-31 17:45:59.500: I/System.out(1871): type : Page
01-31 17:45:59.510: I/System.out(1871): name : P.M
01-31 17:45:59.510: I/System.out(1871): type : Page
01-31 17:45:59.510: I/System.out(1871): name : S.E
01-31 17:45:59.530: I/System.out(1871): type : Page
01-31 17:45:59.530: I/System.out(1871): name : JUN
01-31 17:45:59.530: I/System.out(1871): type : Page
01-31 17:45:59.530: I/System.out(1871): name : EMP
01-31 17:45:59.530: I/System.out(1871): type : Page
01-31 17:45:59.530: I/System.out(1871): name : Add Project Members
01-31 17:45:59.530: I/System.out(1871): type : Action

I need all the name comes in ascending order. Any help?

You can store all the names in an arraylist and then sort them using - Collections.sort(arraylist); — Anukool, Jan 31 '13 at 12:53
Generally you'd map the xml into an object type. Store them as an array or a list and use the built in sort function. — nikhil, Jan 31 '13 at 12:55

score 1 · Accepted Answer · answered Jan 31 '13 at 13:08

Create a variable items to store name and type:

Map<String, String> items = new HashMap<>();
items.put(name, type);

Sort a names:

List<String> keys = new ArrayList(items.keySet());
Collections.sort(keys);

Then output they:

for (String key : keys) {
    System.out.println("name : " + key);
    System.out.println("type : " + items.get(key));        
}

score 0 · Answer 2 · answered Jan 31 '13 at 12:53

0

you load it into an array and sort it. XML is read in two ways (generally).

1.) you build a tree (DOM), so its not in the accending order or anything. You can do tree traversals and get result sorted.

2.) SAX like parses where it pars tags, you need to do some processing , but takes very less amount of memory in reading the doc.

answered Jan 31 '13 at 12:53

dinesh707

12,106
22
84
134

Be it DOM or SAX, still he has to hold values in a collection (or something similar) and then sort it. – Richie Jan 31 '13 at 13:01
1

@Richie - not strictly true. You could in theory do N passes through the `childNode` list, picking out the "next smallest" each time. (It is not trivial to get right ... assuming you were silly enough to try.) – Stephen C Jan 31 '13 at 13:10

score 0 · Answer 3 · answered Jan 31 '13 at 13:10

This is a recursive element sorter. Unfortunately it will only work with XOM Elements.

private void sortElement(Element e) {
  // Remove them all.
  Nodes nodes = e.removeChildren();

  // Sort them.
  Map<String, List<Element>> sorted = new TreeMap<String, List<Element>>();
  for (int i = 0; i < nodes.size(); i++) {
    Node n = nodes.get(i);
    if (n instanceof Element) {
      Element it = (Element) n;
      // Get it sorted.
      sortElement(it);
      // Key it by name and then its xml.
      String key = it.getLocalName() + "!" + it.toXML();
      // Add it.
      List<Element> in = sorted.get(key);
      if (in == null) {
        in = new ArrayList<Element>();
      }
      in.add(it);
      sorted.put(key, in);
    }
  }
  // Add them all back in.
  for (Entry<String, List<Element>> entry : sorted.entrySet()) {
    for (Element a : entry.getValue()) {
      e.appendChild(a);
    }
  }
}

score 0 · Answer 4 · edited May 23 '17 at 11:44

0

Here you can find the static method sortNodes to sort NodeList by any attribute

And the simple XML example:

*** Sort by id ***
1 : 100.00
2 : 5.10
3 : 29.99
*** Sort by price ***
2 : 5.10
3 : 29.99
1 : 100.00

Hope it helps.

edited May 23 '17 at 11:44

Community

1
1

answered Sep 15 '15 at 22:55

Andrey Lebedenko

1,850
17
24

how to sort xml code in ascending order in java

4 Answers4