在科特林中使用房间作为单例

Question

我正在尝试将 Room 用作单例，因此我不必调用 Room.databaseBuilder() - 这很昂贵 - 不止一次。

@Database(entities = arrayOf(
        Price::class,
        StationOrder::class,
        TicketPrice::class,
        Train::class,
        TrainCategory::class
), version = 2)
@TypeConverters(Converters::class)
abstract class AppDatabase : RoomDatabase() {

    abstract fun dao(): TrainDao

companion object {
        fun createDatabase(context: Context): AppDatabase
                = Room.databaseBuilder(context, AppDatabase::class.java, "trains.db").build()
    }
}

注：

无法使用 Object，因为 Room 需要使用 abstract class。
单例必须是线程安全的，因为多个线程可能同时访问它。
必须能够将Context作为参数。

我查看了所有类似的 Whosebug 问题，其中 none 满足我的要求

不是线程安全的

使用对象

Answer 1

经过一些研究，我发现我有两个选择。

我考虑过实施其中之一，但这对 Kotlin 来说并不合适 - 样板代码太多。

经过更多的研究，我偶然发现了 this great article，它提供了一个很好的解决方案，它使用了双重检查锁定，但以一种优雅的方式。

companion object : SingletonHolder<AppDatabase, Context>({
       Room.databaseBuilder(it.applicationContext, AppDatabase::class.java, "train.db").build()
})

来自文章：

A reusable Kotlin implementation:

We can encapsulate the logic to lazily create and initialize a singleton with argument inside a SingletonHolder class. In order to make that logic thread-safe, we need to implement a synchronized algorithm and the most efficient one — which is also the hardest to get right — is the double-checked locking algorithm.

open class SingletonHolder<T, A>(creator: (A) -> T) {
    private var creator: ((A) -> T)? = creator
    @Volatile private var instance: T? = null

    fun getInstance(arg: A): T {
        val i = instance
        if (i != null) {
            return i
        }

        return synchronized(this) {
            val i2 = instance
            if (i2 != null) {
                i2
            } else {
                val created = creator!!(arg)
                instance = created
                creator = null
                created
            }
        }
    }
}

额外： 如果你想要带有两个参数的 Singleton

open class SingletonHolder2<out T, in A, in B>(creator: (A, B) -> T) {
    private var creator: ((A, B) -> T)? = creator
    @Volatile private var instance: T? = null

    fun getInstance(arg0: A, arg1: B): T {
        val i = instance
        if (i != null) return i

        return synchronized(this) {
            val i2 = instance
            if (i2 != null) {
                i2
            } else {
                val created = creator!!(arg0, arg1)
                instance = created
                creator = null
                created
            }
        }
    }
}

Answer 2

您可以使用 Kotlin 标准库的

fun <T> lazy(LazyThreadSafetyMode.SYNCHRONIZED, initializer: () -> T): Lazy<T>

companion object {
    private lateinit var context: Context
    private val database: AppDatabase by lazy(LazyThreadSafetyMode.SYNCHRONIZED) {
        Room.databaseBuilder(context, AppDatabase::class.java, "trains.db").build()
    }
    fun getDatabase(context: Context): AppDatabase {
        this.context = context.applicationContext
        return database
    }
}

但就我个人而言，我通常会在应用程序中添加依赖于 ApplicationContext 的单例，例如

<!-- AndroidManifest.xml -->
<manifest>
  <application android:name="MyApplication">
...

class MyApplication : Application() {
    val database: AppDatabase by lazy {
        Room.databaseBuilder(this, AppDatabase::class.java, "train.db").build()
    }
}

您甚至可以定义一个扩展方法以便于访问 context.database。

val Context.database
    get() =
        generateSequence(applicationContext) {
       (it as? ContextWrapper)?.baseContext
       }.filterIsInstance<MyApplication>().first().database

Answer 3

在这种特殊情况下，我会使用 Dagger 2, or some other dependency injection library like Koin or Toothpick。所有三个库都允许将依赖项作为单例提供。

这是 Dagger 2 模块的代码：

@Module
class AppModule constructor(private val context: Context) {
    @Provides
    @Singleton
    fun providesDatabase(): AppDatabase {
        return Room.databaseBuilder(
                context,
                AppDatabase::class.java,
                "train.db")
                .build()
    }
}

应用程序组件：

@Singleton
@Component(modules = arrayOf(
        AppModule::class
))
interface AppComponent {
    fun inject(viewModel: YourViewModel)
    fun inject(repository: YourRepository)
}

申请class提供注入：

class App : Application() {
    companion object {
        private lateinit var appComponent: AppComponent
        val component: AppComponent get() = appComponent
    }

    override fun onCreate() {
        super.onCreate()
        initializeDagger()
    }

    private fun initializeDagger() {
        component = DaggerAppComponent.builder()
                .appModule(AppModule(this))
                .build()
    }
}

然后将您的数据库作为单例注入到您需要的任何地方（例如在您的应用 repository 中）：

@Inject lateinit var appDatabase: AppDatabase

init {
    App.component.inject(this)
}

Answer 4

我是这样想的...

@Database(entities = [MyEntity::class], version = dbVersion, exportSchema = true)
abstract class AppDB : RoomDatabase() {

// First create a companion object with getInstance method
    companion object {
        fun getInstance(context: Context): AppDB = 
    Room.databaseBuilder(context.applicationContext, AppDB::class.java, dbName).build()
    }

    abstract fun getMyEntityDao(): MyEntityDao
}

// This is the Singleton class that holds the AppDB instance 
// which make the AppDB singleton indirectly
// Get the AppDB instance via AppDBProvider through out the app
object AppDBProvider {

private var AppDB: AppDB? = null

    fun getInstance(context: Context): AppDB {
        if (appDB == null) {
            appDB = AppDB.getInstance(context)
        }
       return appDB!!
    }

}

Answer 5

kotlin 中的单例真的很简单，只需这样做

companion object {
    @JvmStatic
    val DATABASE_NAME = "DataBase"

    @JvmField
    val database = Room.databaseBuilder(App.context(), DataBase::class.java, DataBase.DATABASE_NAME).build()

}

Answer 6

使用 @Volatile 实现线程安全。

public abstract class AppDatabase : RoomDatabase() {

   abstract fun trainDao(): trainDao

   companion object {
        @Volatile
        private var INSTANCE: AppDatabase? = null

        fun getDatabase(context: Context): Db = INSTANCE ?: synchronized(this){
            val instance = Room.databaseBuilder(
            context.applicationContext,
            AppDatabase ::class.java,
            "train-db"
          ).build()
          INSTANCE = instance
          instance
        }
   }
}

取自：https://developer.android.com/codelabs/android-room-with-a-view-kotlin#7

在科特林中使用房间作为单例

Using room as singleton in kotlin

singleton

android

kotlin

android-room