I created a Java application which opens an xml file that looks something like this:
Try this,
this.getResources().getString(R.xml.test); // returns you the path , in string,invoked on activity object
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();
}
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.