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

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

我知道这是一个老问题，但我也遇到了同样的问题，我想分享一下我是如何解决它的。关键是Miguel所说的:枚举是可序列化的。

给定以下enum:

enum YourEnumType {
    ENUM_KEY_1, 
    ENUM_KEY_2
}

Put:

Bundle args = new Bundle();
args.putSerializable("arg", YourEnumType.ENUM_KEY_1);

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];
    }
}

我认为将enum转换为int(对于普通enum)，然后设置在bundle上是最简单的方法。就像下面的代码:

myIntent.PutExtra("Side", (int)PageType.Fornt);

然后检查状态:

int type = Intent.GetIntExtra("Side",-1);
if(type == (int)PageType.Fornt)
{
    //To Do
}

但并不适用于所有枚举类型!

有一点需要注意，如果你使用bundle。将一个Bundle添加到通知中，你可能会遇到以下问题:

*** Uncaught remote exception!  (Exceptions are not yet supported across processes.)
    java.lang.RuntimeException: Parcelable encountered ClassNotFoundException reading a Serializable object.

...

要解决这个问题，你可以做以下事情:

public enum MyEnum {
    TYPE_0(0),
    TYPE_1(1),
    TYPE_2(2);

    private final int code;

    private MyEnum(int code) {
        this.code = navigationOptionLabelResId;
    }

    public int getCode() {
        return code;
    }

    public static MyEnum fromCode(int code) {
        switch(code) {
            case 0:
                return TYPE_0;
            case 1:
                return TYPE_1;
            case 2:
                return TYPE_2;
            default:
                throw new RuntimeException(
                    "Illegal TYPE_0: " + code);
        }
    }
}

然后可以这样使用:

// Put
Bundle bundle = new Bundle();
bundle.putInt("key", MyEnum.TYPE_0.getCode());

// Get 
MyEnum myEnum = MyEnum.fromCode(bundle.getInt("key"));