在viewmodel中使用mutableState和jetpack compose时的最佳实践 &g

Question

我有以下ViewModel

@HiltViewModel
class ShareViewModel @Inject constructor(
private val taskRepository: TaskRepository
): ViewModel() {
private val searchAppBarStateMutableState: MutableState<SearchAppBarState> = mutableStateOf(SearchAppBarState.CLOSED)
val searchAppBarState: State<SearchAppBarState> = searchAppBarStateMutableState
private val listOfTaskMutableStateFlow = MutableStateFlow<List<TodoTaskEntity>>(emptyList())
val listOfTaskStateFlow = listOfTaskMutableStateFlow.asStateFlow()
}

我从未像上面的例子那样公开mutableStateFlow。当这样做时，SonarLint会显示一个警告。

MutableStateFlow" and "MutableSharedFlow" should not be exposed

我将同样的技术应用于mutableState

但是，如果我像下面这样做，我不会得到任何警告。

val searchAppBarStateMutableState: MutableState<SearchAppBarState> = mutableStateOf(SearchAppBarState.CLOSED)

只是想知道在jetpack compose中使用MutableState的最佳实践是什么。

Answer 1

要在viewmodel中使用mutableState，请在viewmodel中使用private setter定义mutableState，例如-

var isVisible by mutableState(false)
private set

通过以上操作，我们可以从视图模型外部读取可变状态，但不更新它。要在视图模型中更新创建一个公共函数，请执行-

fun setVisibility(value: Boolean) {
isVisible = value
}

通过创建setter函数，我们遵循了关注点分离，并且在编辑mutableState时拥有单一的事实来源。

Answer 2

我认为错误在于您设置了

val searchAppBarState: State<SearchAppBarState> = searchAppBarStateMutableState

如果你想共享私有值，你不应该把它设置为相等，你可以使用get修饰符

val searchAppBarState: State<SearchAppBarState> get() = searchAppBarStateMutableState

最好用下划线命名，因为许多开发人员都习惯了:

private val _searchAppBarState: MutableState<SearchAppBarState> = mutableStateOf(SearchAppBarState.CLOSED)
val searchAppBarState: State<SearchAppBarState> get() = _searchAppBarState