java - 如何制作这个特殊的 View - com.android.colorpicker.ColorPickerPalette？

Question

我正在尝试在这里使用这个库，但我发现了一些困难，请帮帮我 - https://materialdoc.com/components/pickers/#color-picker .它给了我这个错误，我似乎无法找到解决方法: - 必需:com.android.colorpicker.ColorPickerPalett - 到目前为止我有这个:当我尝试膨胀 View 时出现错误。

    int[] color_array = this.getResources().
            getIntArray(R.array.default_color_choice_values);

    ColorPickerDialog colorPickerDialog = new ColorPickerDialog();
    colorPickerDialog.initialize(
            R.string.color_picker_default_title, color_array, 0, 2, color_array.length);

    colorPickerDialog.show(getFragmentManager(), "cal");

    LayoutInflater layoutInflater = LayoutInflater.from(this);
    ColorPickerPalette colorPickerPalette = layoutInflater.inflate(R.layout.custom_picker, null);

    ColorPickerSwatch.OnColorSelectedListener mOnColorSelectedListener = new ColorPickerSwatch.OnColorSelectedListener() {
        @Override
        public void onColorSelected(int color) {
            Log.v(LOG_TAG, "Will fill the code later on");
        }
    };
    colorPickerPalette.init(color_array.length, 3, mOnColorSelectedListener);

    colorPickerPalette.drawPalette(color_array, 0);


}

Answer 1

layoutInflater.inflate(R.layout.custom_picker, null); 的结果仅当您在 xml (custom_picker) 的根标记恰好是 <com.android.colorpicker.ColorPickerPalette .... 时，才会是 ColorPallet 类型

否则，如果您有 LinearLayout 或任何其他 ViewGroup，则必须更改代码以找到内部 View (假设您有选择器的标签 @+id)

 ViewGroup resultLayout = layoutInflater.inflate(R.layout.custom_picker, null);
 ColorPickerPalette colorPickerPalette = resultLayout.findViewById(R.id.color_picker_id_at_xml);