How to read file from res/raw by name

With the help of the given links I was able to solve the problem myself. The correct way is to get the resource ID with

getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",                             "raw", getPackageName());

To get it as a InputStream

InputStream ins = getResources().openRawResource(            getResources().getIdentifier("FILENAME_WITHOUT_EXTENSION",            "raw", getPackageName()));

Here is example of taking XML file from raw folder:

 InputStream XmlFileInputStream = getResources().openRawResource(R.raw.taskslists5items); // getting XML

Then you can:

 String sxml = readTextFile(XmlFileInputStream);

when:

 public String readTextFile(InputStream inputStream) {        ByteArrayOutputStream outputStream = new ByteArrayOutputStream();        byte buf[] = new byte[1024];        int len;        try {            while ((len = inputStream.read(buf)) != -1) {                outputStream.write(buf, 0, len);            }            outputStream.close();            inputStream.close();        } catch (IOException e) {        }        return outputStream.toString();    }

You can read files in raw/res using getResources().openRawResource(R.raw.myfilename).

BUT there is an IDE limitation that the file name you use can only contain lower case alphanumeric characters and dot. So file names like XYZ.txt or my_data.bin will not be listed in R.