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"));

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"));

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。

一种简单的方法，将整数值赋给enum

示例如下:

public enum MyEnum {

    TYPE_ONE(1), TYPE_TWO(2), TYPE_THREE(3);

    private int value;

    MyEnum(int value) {
        this.value = value;
    }

    public int getValue() {
        return value;
    }

}

发件人页面：

Intent nextIntent = new Intent(CurrentActivity.this, NextActivity.class);
nextIntent.putExtra("key_type", MyEnum.TYPE_ONE.getValue());
startActivity(nextIntent);

接收端:

Bundle mExtras = getIntent().getExtras();
int mType = 0;
if (mExtras != null) {
    mType = mExtras.getInt("key_type", 0);
}

/* OR
    Intent mIntent = getIntent();
    int mType = mIntent.getIntExtra("key_type", 0);
*/

if(mType == MyEnum.TYPE_ONE.getValue())
    Toast.makeText(NextActivity.this, "TypeOne", Toast.LENGTH_SHORT).show();
else if(mType == MyEnum.TYPE_TWO.getValue())
    Toast.makeText(NextActivity.this, "TypeTwo", Toast.LENGTH_SHORT).show();
else if(mType == MyEnum.TYPE_THREE.getValue())
    Toast.makeText(NextActivity.this, "TypeThree", Toast.LENGTH_SHORT).show();
else
    Toast.makeText(NextActivity.this, "Wrong Key", Toast.LENGTH_SHORT).show();

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);