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。

对于Intent，你可以这样使用:

意图:kotlin

FirstActivity:

val intent = Intent(context, SecondActivity::class.java)
intent.putExtra("type", typeEnum.A)
startActivity(intent)

SecondActivity:

override fun onCreate(savedInstanceState: Bundle?) {
     super.onCreate(savedInstanceState) 
     //...
     val type = (intent.extras?.get("type") as? typeEnum.Type?)
}

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

enum YourEnum { 类型1, 类型2 ｝

传递bundle as bundleOf("TYPE" to youenum . type1)

收到, 参数?。let {it.getSerializable(B"TYPE") as YourEnum}

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。

另一个选择:

public enum DataType implements Parcleable {
    SIMPLE, COMPLEX;

    public static final Parcelable.Creator<DataType> CREATOR = new Creator<DataType>() {

        @Override
        public DataType[] newArray(int size) {
            return new DataType[size];
        }

        @Override
        public DataType createFromParcel(Parcel source) {
            return DataType.values()[source.readInt()];
        }
    };

    @Override
    public int describeContents() {
        return 0;
    }

    @Override
    public void writeToParcel(Parcel dest, int flags) {
        dest.writeInt(this.ordinal());
    }
}