如何从 kotlin 的 Serializable 类中获取值？

Question

我有一个fragment，它正在处理objects。所以我需要将值 Instance 发送到 OnCreateview() 所以我想转换我的可序列化表单，所以我用 getter/setter 创建这些类

BeanDemo.kt

class BeanDemo : Serializable {

    var MyAppDatabase: AppDatabase ? = null

    constructor() {

    }

    //secoutry constructor

    constructor(appDatabase: AppDatabase){
        this. MyAppDatabase =  appDatabase
        Log.d("appDatabase : Bean", "appDatabase$appDatabase")
    }

    //getter/setter methods

    fun getName(): AppDatabase? {
        Log.d("appDatabase : getName", "appDatabase$MyAppDatabase")
        return MyAppDatabase
    }

    fun setName(NEWAPPDB: AppDatabase) {
        Log.d("appDatabase : NEWAPPDB", "appDatabase$NEWAPPDB")
        MyAppDatabase = NEWAPPDB
    }
}

SendingFragment.kt

class SendingFragment: Fragment(),Serializable
{

    private var linearLayoutManager: LinearLayoutManager? = null

    companion object {
        /**
         * new instance pattern for fragment
         */
        @JvmStatic
        fun newInstance(myObject: List<TransactionEntity>?, cc: Context, appDatabase: AppDatabase, networkDefinitionProvider: NetworkDefinitionProvider, incoming: TransactionAdapterDirection): SendingFragment {

            val gson = Gson()
            val gson1 = GsonBuilder().create()

            val sampleVar = BeanDemo(appDatabase)
            sampleVar.setName(appDatabase)

            val bundle = Bundle()
            bundle.putSerializable("serializedObject",sampleVar)
            val sendFragament = SendingFragment()
            sendFragament.arguments = bundle
            return sendFragament
        }
    }


    override fun onCreateView(inflater: LayoutInflater, container: ViewGroup?, savedInstanceState: Bundle?): View? {

        val beanDemo: BeanDemo
        val bundle = arguments
        beanDemo = bundle!!.getSerializable("serializedObject") as BeanDemo
        val name = beanDemo.getName()

        // Inflate the layout for this fragment
        val rootView = inflater.inflate(R.layout.send_fragment, container, false)
        val recyclerView = rootView.findViewById<RecyclerView>(R.id.transaction_recycler_out) as RecyclerView
        linearLayoutManager = LinearLayoutManager(activity, LinearLayout.VERTICAL, false)
        recyclerView.layoutManager = linearLayoutManager
        //recyclerView.adapter = DuplicateTransactionRecyclerAdapter(NewIT,NewAPPDB,NewINCOMTYPE,NewNETWORKPROVSTR)
        recyclerView.setHasFixedSize(true)
        return rootView
    }
}

问题我可以在片段中的类的实例中设置值，但我无法在 OnCretaview() 中获取数据。返回空值。请指导我犯了什么错误。提前考虑

更新

@Database(entities = {AddressBookEntry.class, Token.class, Balance.class, TransactionEntity.class}, version = 1)
@TypeConverters({RoomTypeConverters.class})
public abstract class AppDatabase extends RoomDatabase implements Serializable {

    public abstract AddressBookDAO getAddressBook();

    public abstract TokenDAO getTokens();

    public abstract TransactionDAO getTransactions();

    public abstract BalanceDAO getBalances();
}

Answer 1

如果类AppDatabase 本身不是可序列化的，则不能对BeanDemo 进行序列化。

还要考虑到在 kotlin 中使用类似 Bean 的对象时，您可以使用数据类 (https://kotlinlang.org/docs/reference/data-classes.html) 来避免编写如此多的样板文件。通过这样做：

data class BeanDemo(var MyAppDatabase: AppDatabase? = null) : Serializable()

除其他外，您还可以获得：setter、getter、equals()、hashCode() 和 toString() 明智的实现。

此外，如果您使用的是 Android，请考虑使用 Parcelable 和“Kotlin Android 扩展”（此处的文档：https://kotlinlang.org/docs/tutorials/android-plugin.html）来处理 Bundle。