问如何将URI转换为文件Android 10

EN

File f = getFile(getApplicationContext(), uri);

    public static File getFile(Context context, Uri uri) throws IOException {
    File destinationFilename = new File(context.getFilesDir().getPath() + File.separatorChar + queryName(context, uri));
    try (InputStream ins = context.getContentResolver().openInputStream(uri)) {
        createFileFromStream(ins, destinationFilename);
    } catch (Exception ex) {
        Log.e("Save File", ex.getMessage());
        ex.printStackTrace();
    }
    return destinationFilename;
}

public static void createFileFromStream(InputStream ins, File destination) {
    try (OutputStream os = new FileOutputStream(destination)) {
        byte[] buffer = new byte[4096];
        int length;
        while ((length = ins.read(buffer)) > 0) {
            os.write(buffer, 0, length);
        }
        os.flush();
    } catch (Exception ex) {
        Log.e("Save File", ex.getMessage());
        ex.printStackTrace();
    }
}

private static String queryName(Context context, Uri uri) {
    Cursor returnCursor =
            context.getContentResolver().query(uri, null, null, null, null);
    assert returnCursor != null;
    int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
    returnCursor.moveToFirst();
    String name = returnCursor.getString(nameIndex);
    returnCursor.close();
    return name;
}