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

Question

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);}

I found a way to do it refer to my answer here

Shailendra Madda
– Shailendra Madda

2019-06-05 07:30:57 +00:00
Commented Jun 5, 2019 at 7:30 — Shailendra Madda
– Shailendra Madda, Commented Jun 5, 2019 at 7:30

Phantômaxx · Accepted Answer · 2015-11-07 20:02:29Z

102

AFAIK the files in the assets directory don't get unpacked. Instead, they are read directly from the APK (ZIP) file.

So, you really can't make stuff that expects a file accept an asset 'file'.

Instead, you'll have to extract the asset and write it to a seperate file, like Dumitru suggests:

  File f = new File(getCacheDir()+"/m1.map");
  if (!f.exists()) try {

    InputStream is = getAssets().open("m1.map");
    int size = is.available();
    byte[] buffer = new byte[size];
    is.read(buffer);
    is.close();


    FileOutputStream fos = new FileOutputStream(f);
    fos.write(buffer);
    fos.close();
  } catch (Exception e) { throw new RuntimeException(e); }

  mapView.setMapFile(f.getPath());

edited Nov 7, 2015 at 20:02

Phantômaxx

38.1k21 gold badges88 silver badges122 bronze badges

answered Dec 12, 2011 at 13:33

Jacob Nordfalk

3,5431 gold badge24 silver badges21 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

NullPointerException Over a year ago

i get exception: java.io.FileNotFoundException: m1.map

NullPointerException Over a year ago

the exception is on the line InputStream is = getAssets().open("m1.map");

NullPointerException Over a year ago

ok, solved, but now i got the same exception that with Dumitru answer, 12-12 15:06:41.452: DEBUG/asset(3760): Data exceeds UNCOMPRESS_DATA_MAX (3491923 vs 1048576)

Danuofr Over a year ago

Nice that is exactly correct, the FILES in the assets directory are compressed so you have to extract to a byte array, write to a temp file and load that file, THANKS!

Danuofr Over a year ago

Oh yeah you can use "file:///android_asset/FILENAME" for a url

|

Jacek Marchwicki · Accepted Answer · 2018-08-08 16:59:01Z

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) } }
    }

Dumitru Hristov · Accepted Answer · 2011-12-12 13:18:14Z

10

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);
    }

answered Dec 12, 2011 at 13:18

Dumitru Hristov

1,5392 gold badges13 silver badges31 bronze badges

3 Comments

NullPointerException Over a year ago

It doesn't works for me, the line is.read(buffer); gives me IOException

Dumitru Hristov Over a year ago

What exception exactly does it throw?

NullPointerException Over a year ago

12-12 15:06:41.452: DEBUG/asset(3760): Data exceeds UNCOMPRESS_DATA_MAX (3491923 vs 1048576)

CheesusCrust · Accepted Answer · 2017-06-29 15:55:10Z

5

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
}

answered Jun 29, 2017 at 15:55

CheesusCrust

871 silver badge2 bronze badges

Collectives™ on Stack Overflow

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

4 Answers 4

6 Comments

Comments

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

6 Comments

Comments

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related