How to open a specific file or a folder in Android 10?

Question

I need to open a specific folder in Android 10, so that only the files inside that folder can be seen in my app.

I have tried this:

            try {
                Intent intent = new Intent(Intent.ACTION_OPEN_DOCUMENT);
                File f = new File(getExternalFilesDir(null) + "/MyFolder");
                Uri uri = FileProvider.getUriForFile(getApplicationContext(),getApplicationContext().getPackageName()+".provider",f);
                Toast.makeText(MainActivity.this,f.getPath(),Toast.LENGTH_LONG).show();
                intent.setDataAndType(uri,"text/plain");
                intent.addCategory(Intent.CATEGORY_OPENABLE);
                intent.addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION);
                startActivity(intent);
            }
            catch (Exception e) {
                Log.d("Exception",e.toString());
                Toast.makeText(MainActivity.this, e.toString(), Toast.LENGTH_LONG).show();
            }

But, it opens the entire tree.

What I actually want is to open that folder. The files in the folder should be openable by other apps. Any help will be appreciated.

Answer 1

I need to open a specific folder in Android 10, so that only the files inside that folder can be seen in my app

Sorry, but that is not an option.

The closest thing to what you want is EXTRA_INITIAL_URI. However:

• It takes a document or tree Uri that you obtained previously from the Storage Access Framework, not a FileProvider Uri; and

• It does not prevent users from navigating elsewhere — it just controls the starting point