jetpack compose 传递参数给 viewModel

Question

如何在 jetpack compose 中将参数传递给 viewModel？

这是我的可组合项

@Composable
fun UsersList() {
  val myViewModel: MyViewModel = viewModel("db2name") // pass param like this
}

这是视图模型

class MyViewModel(private val dbname) : ViewModel() {
    private val users: MutableLiveData<List<User>> by lazy {
        MutableLiveData<List<User>>().also {
            loadUsers()
        }
    }

    fun getUsers(): LiveData<List<User>> {
        return users
    }

    private fun loadUsers() {
        // Do an asynchronous operation to fetch users.
    }
}

Answer 1

您需要像这样创建一个工厂来将动态参数传递给 ViewModel：

class MyViewModelFactory(private val dbname: String) :
    ViewModelProvider.NewInstanceFactory() {
    override fun <T : ViewModel?> create(modelClass: Class<T>): T = MyViewModel(dbname) as T
}

然后在可组合函数中像这样使用您的工厂：

@Composable
fun UsersList() {
    val myViewModel: MyViewModel =
        viewModel(factory = MyViewModelFactory("db2name")) // pass param like this
}

现在您可以访问 ViewModel 中的 dbname 参数：

class MyViewModel(private val dbname) : ViewModel() {
    // ...rest of the viewModel logics here
}

Answer 2

通常情况下，您不需要这样做。在 android 中，MVVM 视图模型通过依赖注入从存储库中获取数据。

这是推荐的 android 架构的官方文档：https://developer.android.com/jetpack/guide#recommended-app-arch

Answer 3

正如@Secret Keeper 提到的，您需要创建工厂。

If your ViewModel has dependencies, viewModel() takes an optional ViewModelProvider.Factory as a parameter.

class MyViewModelFactory(
    private val dbname: String
) : ViewModelProvider.Factory {
    @Suppress("UNCHECKED_CAST")
    override fun <T : ViewModel> create(modelClass: Class<T>): T {
        if (modelClass.isAssignableFrom(MyViewModel::class.java)) {
            return MyViewModel(dbname) as T
        }
        throw IllegalArgumentException("Unknown ViewModel class")
    }
}

要创建您的 viewModel，您需要传递可选参数。在您的可组合项中，您可以执行类似的操作。

val viewModel: MyViewModel = viewModel(
factory = MyViewModelFactory(
    dbname = "myDbName"
)

Answer 4

这里有一些 Jetpack Compose/Kotlin-specific 实现相同语法的语法：

ui/settings/SettingsViewModel.kt

class SettingsViewModel(
    private val settingsRepository: SettingsRepository
) : ViewModel() {
    /* Your implementation */
}

class SettingsViewModelFactory(
    private val settingsRepository: SettingsRepository
) : ViewModelProvider.Factory {

    override fun <T : ViewModel> create( modelClass: Class<T> ): T {
        if( modelClass.isAssignableFrom( SettingsViewModel::class.java ) ) {
            @Suppress( "UNCHECKED_CAST" )
            return SettingsViewModel( settingsRepository ) as T
        }
        throw IllegalArgumentException( "Unknown ViewModel Class" )
    }    

}

然后：

MainActivity.kt


/* dataStore by preferencesDataStore */

class MainActivity : ComponentActivity() {
    private lateinit var settingsRepository: SettingsRepository
    
    // Here we instantiate our ViewModel leveraging delegates and
    // a trailing lambda
    private val settingsViewModel by viewModels<SettingsViewModel> {
        SettingsViewModelFactory(
            settingsRepository
        )
    }

    /* onCreate -> setContent -> etc */
}

Answer 5

其他解决方案可行，但您必须为每个 ViewModel 创建一个工厂，这似乎有点过分了。

比较通用的方案是这样的：

inline fun <VM : ViewModel> viewModelFactory(crossinline f: () -> VM) =
    object : ViewModelProvider.Factory {
        override fun <T : ViewModel> create(aClass: Class<T>):T = f() as T
    }

并像这样使用它：

@Composable
fun MainScreen() {
    val viewModel: MyViewModel = viewModel(factory = viewModelFactory {
        MyViewModel("Test Name")
    })
}

对于这样的 ViewModel：

class MyViewModel(
  val name: String
):ViewModel() {}

jetpack compose 传递参数给 viewModel

jetpack compose pass parameter to viewModel

android

kotlin

android-viewmodel

android-jetpack-compose