Android: How to detect a directory in the assets folder?

Question

I\'m retrieving files like this

String[] files = assetFiles.list(\"EngagiaDroid\"); 

How can we know whether it is a file or is a director

Answer 1

Another way relying on exceptions:

private void checkAssets(String path, AssetManager assetManager) {
    String TAG = "CheckAssets";
    String[] fileList;
    String text = "";
    if (assetManager != null) {
        try {
            fileList = assetManager.list(path);
        } catch (IOException e) {
            Log.e(TAG, "Invalid directory path " + path);
            return;
        }
    } else {
        fileList = new File(path).list();
    }

    if (fileList != null && fileList.length > 0) {
        for (String pathInFolder : fileList) {
            File absolutePath = new File(path, pathInFolder);

            boolean isDirectory = true;
            try {
                if (assetManager.open(absolutePath.getPath()) != null) {
                    isDirectory = false;
                }
            } catch (IOException ioe) {
                isDirectory = true;
            }

            text = absolutePath.getAbsolutePath() + (isDirectory ? " is Dir" : " is File");
            Log.d(TAG, text);
            if (isDirectory) {
                checkAssets(absolutePath.getPath(), assetManager);
            }
        }
    } else {
        Log.e(TAG, "Invalid directory path " + path);
    }
}

and then just call checkAssets("someFolder", getAssets()); or checkAssets("", getAssets()); if you want to check the root assets folder. But be aware that the root assets folder contains also other directories/files (Eg. webkit, images, etc.)

Answer 2

You can start from Android File

Answer 3

The appalling truth is that despite being asked nearly 10 years ago, no simple, elegant, roundly applauded method of determining whether an element in the array returned by AssetManager.list() is a file or a directory has been offered by any answer to date.

So, for example, if an asset directory contains a thousand elements, then seemingly a thousand I/O operations are necessary to isolate the directories.

Nor, for any element, does any native method exist for obtaining its parent directory - vital for something complex like an assets Browser / Picker - where you could end up looking at some seriously ugly code.

boolean isAssetDirectory = !elementName.contains(".");

The lateral approach that worked for me was to assume that any element without a dot (.) in its name was a directory. If the assumption is later proved wrong it can be easily rectified.

Asset files generally exist because you put them there. Deploy naming conventions that distinguish between directories and files.

Answer 4

I think a more general solution (in case you have subfolders etc.) would be something like this (based on the solution you linked to, I've added it there too):

...

copyFileOrDir("myrootdir");

...

private void copyFileOrDir(String path) {
    AssetManager assetManager = this.getAssets();
    String assets[] = null;
    try {
        assets = assetManager.list(path);
        if (assets.length == 0) {
            copyFile(path);
        } else {
            String fullPath = "/data/data/" + this.getPackageName() + "/" + path;
            File dir = new File(fullPath);
            if (!dir.exists())
                dir.mkdir();
            for (int i = 0; i < assets.length; ++i) {
                copyFileOrDir(path + "/" + assets[i]);
            }
        }
    } catch (IOException ex) {
        Log.e("tag", "I/O Exception", ex);
    }
}

private void copyFile(String filename) {
    AssetManager assetManager = this.getAssets();

    InputStream in = null;
    OutputStream out = null;
    try {
        in = assetManager.open(filename);
        String newFileName = "/data/data/" + this.getPackageName() + "/" + filename;
        out = new FileOutputStream(newFileName);

        byte[] buffer = new byte[1024];
        int read;
        while ((read = in.read(buffer)) != -1) {
            out.write(buffer, 0, read);
        }
        in.close();
        in = null;
        out.flush();
        out.close();
        out = null;
    } catch (Exception e) {
        Log.e("tag", e.getMessage());
    }

}

Answer 5

You can also try this, it works for me, since you cannot rely solely on .list()

public static boolean isDirectory(Context context, String path) throws IOException {
    //If list returns any entries, than the path is a directory
    String[] files = context.getAssets().list(path);
    if (files != null && files.length > 0) {
        return true;
    } else {
        try {
            //If we can open a stream then the path leads to a file
            context.getAssets().open(path);
            return false;
        } catch (Exception ex) {
            //.open() throws exception if it's a directory that you're passing as a parameter
            return true;
        }
    }
}

Answer 6

I've discovered this variant:

try {
    AssetFileDescriptor desc = getAssets().openFd(path);  // Always throws exception: for directories and for files
    desc.close();  // Never executes
} catch (Exception e) {
    exception_message = e.toString();
}

if (exception_message.endsWith(path)) {  // Exception for directory and for file has different message
    // Directory
} else {
    // File
}

It's a more faster as .list()