parsing XML in Android

Question

In non-Android java, I used to use:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();

File file = new File("res/xml/xml_sample.xml");
Document document = builder.parse(file);

Now in Android the xml seems to get into R.xml.xml_parse, at least it's defined into the R.java file. However, I can't seem to access that path to use it as the File constructor argument.

I've seen multiple solutions around which involve using getResources() like this one:

http://blog.dimond.de/?p=35

However, since this code is in a independent class which is not an Activity, it doesn't work.

Any help is appreciated.

Can I reccomend using Simple XML to do your XML parsing instead? It is much easier to work with: massaioli.wordpress.com/2011/04/21/… — Robert Massaioli
– Robert Massaioli, Commented Oct 24, 2015 at 0:39

antlersoft · Accepted Answer · 2011-06-20 15:11:27Z

1

I think the other answers are ignoring that you want to use this XML (in a DOM) in non-Android specific code.

If you put the XML under res/xml, you can only use the XML pull parser on it, not the DOM parser. Usually that's what you want, and it's a lot more efficient, since most of the actual work of parsing is done at .apk build time; but if you need to use a DOM parser, put the XML file in res/raw. You can get an InputStream from a raw resource like this:

InputStream is = getResources().openRawResource(R.raw.hello)

and pass the input stream (instead of any kind of File or path) to the DOM parser.

Your other option is to use the pull parser events to build the DOM object, and then use that.

answered Jun 20, 2011 at 15:11

antlersoft

14.8k4 gold badges40 silver badges56 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

bluehallu Over a year ago

Now you're going in the direction I wanted :) The problem, however, is that I can't call getResources in this code. However, I've tried to pass the ApplicationContext as a parameter to the constructor of this class, to make it save this context and then be able to use it. Still, ctx.getResources().openRawResource(R.raw.sample), tells me there's no "sample" file inside R.raw. Even doing ctrl+space in Eclipse when typing "R.raw." doesn't recognize the sample.xml file...

antlersoft Over a year ago

You don't need to call getResources in the code that is doing the parsing; just pass in the InputStream (which is an ordinary java.io.InputStream) that you get from getResources. Can't help you with the missing R.raw.sample

bluehallu Over a year ago

Ok, done that, but still R.raw is empty, R.raw.sample tells me "sample cannot be resolved or is not a field". Do I have to force the R.java to compile or something?

antlersoft Over a year ago

Don't have my android dev system handy-- but files you add to res/raw or supposed to go to R.raw automagically--- You might also try using AssetManager (from docs): raw/ Arbitrary files to save in their raw form. To open these resources with a raw InputStream, call Resources.openRawResource() with the resource ID, which is R.raw.filename. However, if you need access to original file names and file hierarchy, you might consider saving some resources in the assets/ directory (instead of res/raw/). Files in assets/ are not given a resource ID, so you can read them only using AssetManager.

Heiko Rupp · Accepted Answer · 2011-06-20 14:58:34Z

1

Just pass in a Context object into your class and then do context.getResources() like:

public void myParser(Context ctx) {

    ctx.getResources()....
}

answered Jun 20, 2011 at 14:58

Heiko Rupp

31.1k13 gold badges86 silver badges120 bronze badges

Comments

JDx · Accepted Answer · 2011-06-20 14:57:48Z

0

If you put the xml file into your assets folder you can use:

 File file = new File("file:///android_asset/xml_sample.xml");

answered Jun 20, 2011 at 14:57

JDx

2,6553 gold badges24 silver badges33 bronze badges

Comments

DArkO · Accepted Answer · 2011-06-20 14:58:01Z

0

You need to pass the application or activity context and then you can: context.getResources()

answered Jun 20, 2011 at 14:58

DArkO

16.2k12 gold badges66 silver badges94 bronze badges

Comments

mah · Accepted Answer · 2011-06-20 14:58:04Z

0

Every android package has access to its resources, based off of its Context. If you need access to resources from an independent class, you must provide it a Context object so that it can use context.getResources().

There are ways to create a new context (so that you could, for example, get resources from it), however they require that you start with some valid context in the first place.

answered Jun 20, 2011 at 14:58

mah

40.1k9 gold badges79 silver badges94 bronze badges

Collectives™ on Stack Overflow

parsing XML in Android

5 Answers 5

4 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related