The raw stream will not be useful, since it has been pre-compiled into a
binary format.

On Thu, May 20, 2010 at 8:07 PM, Nathan <[email protected]> wrote:

> This will work to get a parser:
>
> XmlResourceParser xpp = getResources().getXml(R.xml.file_name);
>
> To get an underlying stream, you might be able to use openRawResource
> or get the AssetManager to get to the input streams.
>
> getResources.getAssets().getInputStream(fileName)
>
> I haven't tried it but its worth a try.
>
> Nathan
>
>
> On May 16, 7:59 pm, tiany <[email protected]> wrote:
> > I know when the xml file is under /res/raw,I can do this by
> > context.getResources().openRawResource(rid); but when the xml file is
> > under /res/xml,how can I do it?
> >
> > thx in advance
> >
> > tiany
> >
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-- 
Dianne Hackborn
Android framework engineer
[email protected]

Note: please don't send private questions to me, as I don't have time to
provide private support, and so won't reply to such e-mails.  All such
questions should be posted on public forums, where I and others can see and
answer them.

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