Java FilePath Android

Question

I have an android app that I am trying to get a file path for a SAX parser. I have the following structure:

assets:(Where my xml file is)
src(same level as assets)
   com
       msi
          androidrss(The calling java file is in here)

I tried several variations of this:

InputSource is = new InputSource("file://../../../../assets/Rss.xml");

But I always get a FNF Exception

Any suggestions?

typo.pl · Accepted Answer · 2011-03-11 00:08:39Z

1

You can't use File APIs to get at data in the assets directory when you're running from your APK. You have to use the Android - AssetManager class. You get an instance of the AssetManager class from the Context object. See Context object

answered Mar 11, 2011 at 0:08

typo.pl

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Ted Hopp · Accepted Answer · 2011-03-11 03:31:05Z

1

To elaborate on the answer by @typo.pl:

InputStream in = getContext().getAssets().open("Rss.xml");
// test that in is not null
try {
    // set up your SAX parser and handler
    parser.parse(in, handler);
} finally {
    try { in.close(); }
    catch (IOException ignored) {}
}

answered Mar 11, 2011 at 3:31

Ted Hopp

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Comments

Macarse · Accepted Answer · 2011-03-11 10:30:19Z

1

Place the xml at /res/raw and open it with:

getResources().openRawResource(R.raw.xmlName);

edited Mar 11, 2011 at 10:30

answered Mar 11, 2011 at 0:12

Macarse

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1 Comment

Ted Hopp Over a year ago

I think that should be xmlResourceId instead of xmlName.

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