Open XML file from res/xml in Android

I created a Java application which opens an xml file that looks something like this:

<AnimalTree>
  <animal>
    <mammal>canine</mammal>
    <color>blue</color>
  </animal>
  <!-- ... -->
</AnimalTree>

And I can open it using:

File fXmlFile = getResources.getXml("res/xml/data.xml");
Do开发者_开发知识库cumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(fXmlFile);
doc.getDocumentElement().normalize();
NodeList animalNodes = doc.getElementsByTagName("animal");

Then I can simply create a node, push the object into a ListArray, then do what I want with the objects as I loop through the ListArray.

for (int temp = 0; temp < animalNodes.getLength(); temp++) {
Node nNode = animalNodes.item(temp);     
if (nNode.getNodeType() == Node.ELEMENT_NODE) {
Element eElement = (Element) nNode;
question thisAnimal = new animal();
thisAnimal.mammal = getTagValue("mammal",eElement);
// ...

Plain and simple! Now only, in Android I cannot simply read the file "res/xml/data.xml" because "File();" requires a String not an integer (id). This is where I am lost. Is there some way I can make "File();" open the file, or is this impossible without using SAXparser or XPP? (both of which I really cannot understand, no matter how hard I try.)

If I am forced to use those methods, can someone show me some simple code analogous to my example?

If it's in the resource tree, it'll get an ID assigned to it, so you can open a stream to it with the openRawResource function:

InputStream is = context.getResources().openRawResource(R.xml.data);

As for working with XML in Android, this link on ibm.com is incredibly thorough.

See Listing 9. DOM-based implementation of feed parser in that link.

Once you have the input stream (above) you can pass it to an instance of DocumentBuilder:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document dom = builder.parse(this.getInputStream());
Element root = dom.getDocumentElement();
NodeList items = root.getElementsByTagName("TheTagYouWant");

Keep in mind, I haven't done this personally -- I'm assuming the code provided by IBM works.

I tried the approach using openRawResource and got a SAXParseException. So, instead, I used getXml to get a XmlPullParser. Then I used next() to step through the parsing events. The actual file is res/xml/dinosaurs.xml.

XmlResourceParser parser = context.getResources().getXml(R.xml.dinosaurs);
int eventType = parser.getEventType();
while (eventType != XmlPullParser.END_DOCUMENT) {
    switch (eventType) {
        case XmlPullParser.START_DOCUMENT :
            Log.v(log_tag, "Start document");
            break;
        case XmlPullParser.START_TAG :
            Log.v(log_tag, "Start tag " + parser.getName() );
            break;
        case XmlPullParser.END_TAG :
            Log.v(log_tag, "End tag " + parser.getName() );
            break;
        case XmlPullParser.TEXT :
            Log.v(log_tag, "Text " + parser.getText() );
            break;
        default :
            Log.e(log_tag, "Unexpected eventType = " + eventType );
    }
    eventType = parser.next();
}

Try this,

this.getResources().getString(R.xml.test); // returns you the path , in string,invoked on activity object