android - 如何使用 kotlin 序列化我的类对象?
问题描述
我有片段,我需要将我的值构造函数传递给 OncretaeView() 方法,因此我在构造函数和 tr 中接收我的值以将我的对象转换为序列化对象,但它不会转换。这是我的代码。请指导我如何序列化和反序列化我的对象。提前致谢
import java.io.Serializable
class SendingFragment: Fragment(),Serializable {
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 model = myObject as List<TransactionEntity>
val IT = gson.toJson(model)
System.out.println("json representation :" + IT)
val bo = ByteArrayOutputStream()
val so = ObjectOutputStream(bo)
so.writeObject(appDatabase)
so.flush()
val serializedObject = String(Base64.encode(bo.toByteArray()))
val bundle = Bundle()
bundle.putString("bundleValue", IT)
bundle.putSerializable("serializedObject",serializedObject)
val sendFragament: SendingFragment = SendingFragment()
sendFragament.setArguments(bundle)
return sendFragament
}
}
override fun onCreateView(inflater: LayoutInflater, container: ViewGroup?, savedInstanceState: Bundle?): View? {
val mFragserializedObject = arguments!!.getSerializable("serializedObject")
System.out.println( "json serializedObject" + mFragserializedObject)
}
}
应用数据库.kt
@Database(entities = {AddressBookEntry.class, Token.class, Balance.class, TransactionEntity.class}, version = 1)
@TypeConverters({RoomTypeConverters.class})
public abstract class AppDatabase extends RoomDatabase {
public abstract AddressBookDAO getAddressBook();
public abstract TokenDAO getTokens();
public abstract TransactionDAO getTransactions();
public abstract BalanceDAO getBalances();
}
更新
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
}
}
**myfragent.kt**
fun newInstance(myObject: List<TransactionEntity>?, cc: Context, appDatabase: AppDatabase, networkDefinitionProvider: NetworkDefinitionProvider, incoming: TransactionAdapterDirection): SendingFragment {
val gson = Gson()
val gson1 = GsonBuilder().create()
val model = myObject as List<TransactionEntity>
val IT = gson.toJson(model)
// Here
val sampleVar = BeanDemo(appDatabase)
sampleVar.setName(appDatabase)
val bundle = Bundle()
bundle.putString("bundleValue", IT)
bundle.putSerializable("serializedObject",sampleVar)
val sendFragament: SendingFragment = SendingFragment()
sendFragament.setArguments(bundle)
return sendFragament
override fun onCreateView(inflater: LayoutInflater, container: ViewGroup?, savedInstanceState: Bundle?): View? {
val bundle = arguments
val obj = bundle!!.getSerializable("serializedObject") as BeanDemo
val name = obj.getName()
}
但是val obj =为 null,在 setter 方法中保存该值但不返回它。
解决方案
当你创建你的类Model/POJO
时,然后with 。extend
class
Serializable
例子-
class JsonData : Serializable {
@SerializedName("hasPreviousData")
var hasPreviousData: Boolean = false
@SerializedName("dataList")
var dataList: ArrayList<DataList>? = null
inner class DataList : Serializable {
@SerializedName("id")
var id: String? = null
@SerializedName("createdAt")
var createdAt: String? = null
@SerializedName("name")
var name: String? = null
}
}
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