使用WorkManager更新会议室中的数据库条目

问题描述 投票:4回答:1

我正在尝试在Android中使用WorkManager,因为到目前为止,我已经使用AsyncTask完成了插入更新。我正在尝试更新Room database中的用户。我已经在Repository中创建了一个类来在后台进行处理,但是我不知道这是否是正确的编码方式。我也有点困惑如何在userViewModel中调用该方法。

这是我的代码:

在存储库中:

inner class Test(ctx: Context, params: WorkerParameters) : Worker(ctx, params) {

    override fun doWork(): Result {
        fun updateUser(userId : Int , userName :String , userLastname: String){
            userDao.updateUser(userId, userName, userLastname)
        }
        return Result.success()
    }
}

在viewModel中:

private val workManager: WorkManager = WorkManager.getInstance()
fun updateUser(userID: Int, userName: String, userLastname: String)    {
    workManager.enqueue(OneTimeWorkRequest.from(Repository.Test::class.java))

}

也许这不是使用WorkManager的正确方法,所以如果有人可以帮助我,我将不胜感激。

android kotlin android-room android-jetpack android-workmanager
1个回答
6
投票

您可以对代码进行较小的更改。您可以使用WorkManager类将诸如userId,userName等的参数传递给Data。同样,也不需要使Worker类成为内部类。只需使其成为一个独立的类即可。因此,基本上,worker类将类似于:

TestWorker.kt

class TestWorker(ctx: Context, params: WorkerParameters) : Worker(ctx, params) {

    override fun doWork(): Result {
        val userId = inputData.getInt("USER_ID")
        val userName = inputData.getString("USER_NAME")
        val userLastname = inputData.getString("USER_LAST_NAME") 
        AppDatabase.getInstance().userDao.updateUser(userId, userName, userLastname)
        return Result.success()
    }
}

TestViewModel.kt

class TestViewModel : ViewModel() {
    fun updateUser(userId: Int, userName: String, userLastName: String) {
        val worker = OneTimeWorkRequest.Builder(TestWorker::class.java)
        val data = Data.Builder()
        //Add parameter in Data class. just like bundle. You can also add Boolean and Number in parameter.
        data.putInt("USER_ID", userId)
        data.putString("USER_NAME", userName)
        data.putString("USER_LAST_NAME", userLastName)
        //Set Input Data
        worker.setInputData(data.build())
        //enque worker
        WorkManager.getInstance().enqueue(worker.build())
    }
}

不要忘记使数据库像这样单调:

@Database(entities = [User::class],
    version = 1,
    exportSchema = false)
abstract class AppDatabase : RoomDatabase() {

    abstract fun userDao(): UserDao

    companion object {

        // For Singleton instantiation
        @Volatile
        private var instance: AppDatabase? = null

        fun getInstance(context: Context): AppDatabase {
            return instance ?: synchronized(this) {
                instance ?: buildDatabase(context).also { instance = it }
            }
        }

        private fun buildDatabase(context: Context) =
            Room.databaseBuilder(context, AppDatabase::class.java, "userdb")
                .build()
    }
}
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