[android] How to get the android Path string to a file on Assets folder?

I need to know the string path to a file on assets folder, because I'm using a map API that needs to receive a string path, and my maps must be stored on assets folder

This is the code i'm trying:

    MapView mapView = new MapView(this);
    mapView.setClickable(true);
    mapView.setBuiltInZoomControls(true);
    mapView.setMapFile("file:///android_asset/m1.map");
    setContentView(mapView);

Something is going wrong with "file:///android_asset/m1.map" because the map is not being loaded.

Which is the correct string path file to the file m1.map stored on my assets folder?

Thanks

EDIT for Dimitru: This code doesn't works, it fails on is.read(buffer); with IOException

        try {
            InputStream is = getAssets().open("m1.map");
            int size = is.available();
            byte[] buffer = new byte[size];
            is.read(buffer);
            is.close();
            text = new String(buffer);
        } catch (IOException e) {throw new RuntimeException(e);}

Just to add on Jacek's perfect solution. If you're trying to do this in Kotlin, it wont work immediately. Instead, you'll want to use this:

@Throws(IOException::class)
fun getSplashVideo(context: Context): File {
    val cacheFile = File(context.cacheDir, "splash_video")
    try {
        val inputStream = context.assets.open("splash_video")
        val outputStream = FileOutputStream(cacheFile)
        try {
            inputStream.copyTo(outputStream)
        } finally {
            inputStream.close()
            outputStream.close()
        }
    } catch (e: IOException) {
        throw IOException("Could not open splash_video", e)
    }
    return cacheFile
}

You can use this method.

    public static File getRobotCacheFile(Context context) throws IOException {
        File cacheFile = new File(context.getCacheDir(), "robot.png");
        try {
            InputStream inputStream = context.getAssets().open("robot.png");
            try {
                FileOutputStream outputStream = new FileOutputStream(cacheFile);
                try {
                    byte[] buf = new byte[1024];
                    int len;
                    while ((len = inputStream.read(buf)) > 0) {
                        outputStream.write(buf, 0, len);
                    }
                } finally {
                    outputStream.close();
                }
            } finally {
                inputStream.close();
            }
        } catch (IOException e) {
            throw new IOException("Could not open robot png", e);
        }
        return cacheFile;
    }

You should never use InputStream.available() in such cases. It returns only bytes that are buffered. Method with .available() will never work with bigger files and will not work on some devices at all.

In Kotlin (;D):

@Throws(IOException::class)
fun getRobotCacheFile(context: Context): File = File(context.cacheDir, "robot.png")
    .also {
        it.outputStream().use { cache -> context.assets.open("robot.png").use { it.copyTo(cache) } }
    }

Have a look at the ReadAsset.java from API samples that come with the SDK.

       try {
        InputStream is = getAssets().open("read_asset.txt");

        // We guarantee that the available method returns the total
        // size of the asset...  of course, this does mean that a single
        // asset can't be more than 2 gigs.
        int size = is.available();

        // Read the entire asset into a local byte buffer.
        byte[] buffer = new byte[size];
        is.read(buffer);
        is.close();

        // Convert the buffer into a string.
        String text = new String(buffer);

        // Finally stick the string into the text view.
        TextView tv = (TextView)findViewById(R.id.text);
        tv.setText(text);
    } catch (IOException e) {
        // Should never happen!
        throw new RuntimeException(e);
    }