Sunday, April 24, 2011

Re: [android-developers] OpenFileInput

On Sun, Apr 24, 2011 at 11:17 AM, Dalvin <singh.dalvin@gmail.com> wrote:
> I am trying to read a simple text file in my sample Android Application. I
> am using the below written code for reading the simple text file.
> 1. InputStream inputStream = openFileInput("test.txt");
> 2. InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
> 3. BufferedReader bufferedReader = new BufferedReader(inputStreamReader);
> My questions is :
> Where should I place this "test.txt" file in my project.

It does not go in your project.

> I have tried
> putting the file under "res/raw" and "asset" folder but I get the exception
> "FileNotFound" when first live of the code written above gets executed.

Of course.

If you want to read a raw resource, use getResources().openRawResource().

If you want to read an asset, use getResources().getAssets().open().

--
Mark Murphy (a Commons Guy)
http://commonsware.com | http://github.com/commonsguy
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Android 3.0 Programming Books: http://commonsware.com/books

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