如何将Enum对象添加到Android Bundle中?


当前回答

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?)
}

其他回答

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];
    }
}

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。

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

我创建了一个Koltin扩展:

fun Bundle.putEnum(key: String, enum: Enum<*>) {
    this.putString( key , enum.name )
}

inline fun <reified T: Enum<T>> Intent.getEnumExtra(key:String) : T {
    return enumValueOf( getStringExtra(key) )
}

创建一个bundle并添加:

Bundle().also {
   it.putEnum( "KEY" , ENUM_CLAS.ITEM )
}

并获得:

intent?.getEnumExtra< ENUM_CLAS >( "KEY" )?.let{}

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!