在芬兰湾的科特林:

enum class MyEnum {
  NAME, SURNAME, GENDER
}

把这个枚举放在一个Bundle中:

Bundle().apply {
  putInt(MY_ENUM_KEY, MyEnum.ordinal)
}

从Bundle中获取enum:

val ordinal = getInt(MY_ENUM_KEY, 0)
MyEnum.values()[ordinal]

完整的例子:

class MyFragment : Fragment() {

    enum class MyEnum {
        NAME, SURNAME, GENDER
    }

    companion object {
        private const val MY_ENUM_KEY = "my_enum_key"

        fun newInstance(myEnum: MyEnum) = MyFragment().apply {
            arguments = Bundle().apply {
                putInt(MY_ENUM_KEY, myEnum.ordinal)
            }
        }
    }

    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        with(requireArguments()) {
            val ordinal = getInt(MY_ENUM_KEY, 0)
            val myEnum = MyEnum.values()[ordinal]
        }
    }
}

在Java中:

public final class MyFragment extends Fragment {
    private static final String MY_ENUM_KEY = "my_enum";

    public enum MyEnum {
        NAME,
        SURNAME,
        GENDER
    }

    public final MyFragment newInstance(MyEnum myEnum) {
        Bundle bundle = new Bundle();
        bundle.putInt(MY_ENUM_KEY, myEnum.ordinal());
        MyFragment fragment = new MyFragment();
        fragment.setArguments(bundle);
        return fragment;
    }

    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        Bundle arguments = this.requireArguments();
        int ordinal = arguments.getInt(MY_ENUM_KEY, 0);
        MyEnum myEnum = MyEnum.values()[ordinal];
    }
}

最好从myEnumValue.name()中将其作为字符串传递，并从yourenum . valueof (s)中恢复，否则必须保留枚举的顺序!

更详细的解释:从枚举序数转换为枚举类型

枚举是可序列化的，所以没有问题。

给定以下enum:

enum YourEnum {
  TYPE1,
  TYPE2
}

包:

// put
bundle.putSerializable("key", YourEnum.TYPE1);

// get 
YourEnum yourenum = (YourEnum) bundle.get("key");

目的:

// put
intent.putExtra("key", yourEnum);

// get
yourEnum = (YourEnum) intent.getSerializableExtra("key");

使用包。putSerializable(String key, Serializable s)和bundle。getSerializable (String键):

enum Mode = {
  BASIC, ADVANCED
}

Mode m = Mode.BASIC;

bundle.putSerializable("mode", m);

...

Mode m;
m = bundle.getSerializable("mode");

文档:http://developer.android.com/reference/android/os/Bundle.html

这对我来说很容易:

enum class MyEnum {
    FOO,
    BAR
}


val bundle = Bundle()
bundle.putAll(bundleOf("myKey", MyEnum.FOO))

// to read
val myEnum = bundle.get("myKey") as MyEnumClass

注意，如果你从onCreate得到这个，你会想使用as?防止任何空异常。

在芬兰湾的科特林:

enum class MyEnum {
  NAME, SURNAME, GENDER
}

把这个枚举放在一个Bundle中:

Bundle().apply {
  putInt(MY_ENUM_KEY, MyEnum.ordinal)
}

从Bundle中获取enum:

val ordinal = getInt(MY_ENUM_KEY, 0)
MyEnum.values()[ordinal]

完整的例子:

class MyFragment : Fragment() {

    enum class MyEnum {
        NAME, SURNAME, GENDER
    }

    companion object {
        private const val MY_ENUM_KEY = "my_enum_key"

        fun newInstance(myEnum: MyEnum) = MyFragment().apply {
            arguments = Bundle().apply {
                putInt(MY_ENUM_KEY, myEnum.ordinal)
            }
        }
    }

    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        with(requireArguments()) {
            val ordinal = getInt(MY_ENUM_KEY, 0)
            val myEnum = MyEnum.values()[ordinal]
        }
    }
}

在Java中:

public final class MyFragment extends Fragment {
    private static final String MY_ENUM_KEY = "my_enum";

    public enum MyEnum {
        NAME,
        SURNAME,
        GENDER
    }

    public final MyFragment newInstance(MyEnum myEnum) {
        Bundle bundle = new Bundle();
        bundle.putInt(MY_ENUM_KEY, myEnum.ordinal());
        MyFragment fragment = new MyFragment();
        fragment.setArguments(bundle);
        return fragment;
    }

    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        Bundle arguments = this.requireArguments();
        int ordinal = arguments.getInt(MY_ENUM_KEY, 0);
        MyEnum myEnum = MyEnum.values()[ordinal];
    }
}