Question

I am using this intent to open a chooser to pick up a file:

Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
intent.setDataAndType(Uri.parse(getExternalFilesDir(null).toString()), "*/*");
intent.addCategory(Intent.CATEGORY_OPENABLE);
Intent chooser = Intent.createChooser(intent, "Choose File");
startActivityForResult(chooser, FILE_SELECT);

I don’t fully understand it, but it works. I copied it from some example in the web
The result is an uri, but I find the processing of the uri, also copied from the example, too cumbersome

if(uri != null) {
    if("content".equalsIgnoreCase(uri.getScheme())) {
        String[] projection = {"_data"};
        Cursor cursor = null;
        try {
            cursor = context.getContentResolver().query(uri, projection, null, null, null);
            int column_index = cursor.getColumnIndexOrThrow("_data");
            if(cursor.moveToFirst()) {
                return cursor.getString(column_index);
            }
        }
        catch (Exception e) {}
    } else if("file".equalsIgnoreCase(uri.getScheme())) {
        return uri.getPath();
    }
}

Is this the simplest way to let the user pick a file and get its path?
Why do I need to check if the uri is “content”? I just want to open a text or zipped file the user chose in the intent.
Can I ignore the “content” check, just assume “file” and simply use uri.getPath() ?

Edit I: Is there a way to force choosing with uri scheme “file”?

Answer 1

Yes. This is correct. You are basically starting an intent to get access to the native app. which in this case is the file system. The next would be a part of the onActivityResultData() which would then use the uri on the intent and get the contents of the file that has been selected.