从 Room 到 UiState 的流程<List>

Question

class HomeViewModel(
    savedStateHandle: SavedStateHandle,
    private val categoryRepository: CategoryRepository,
    private val expenseRepository: ExpenseRepository
): ViewModel() {
    private val month: Int = savedStateHandle[HomeDestination.monthArgs] ?: Calendar.MONTH
    private val year: Int = savedStateHandle[HomeDestination.yearArgs] ?: Calendar.YEAR

    private val _categories: Flow<List<Category>> = categoryRepository.getAllCategoriesStream()
    private val _expenses: Flow<List<Expense>> = expenseRepository.getAllExpensesBetweenStream(
        LocalDate.of(year, month, 1).toEpochDay(),
        LocalDate.of(year, if (month + 1 == 13) 1 else month + 1, 1).toEpochDay()
    )

    private var _uiState = MutableStateFlow(UiState())
    val uiState: StateFlow<UiState> = _uiState.asStateFlow()

    data class UiState(
        val categories: List<CategoryUiState> = listOf(),
        val expenses: List<ExpenseUiState> = listOf()
    )
}

我想将

_expenses

 的值链接到 

expenses

 中的 

UiState

 并执行此操作，以便在 

_expenses

 内容更新时更新费用。

Answer 1

假设问题不是如何将

Expense

 对象存储为 

ExpenseUiState

，我希望有一些辅助函数可以从 

UiState

 和 

Category

 创建正确的 

Expense

。我们称之为 

UiState.of(...)

。

然后您可以将

_uiState

 和 

uiState

 替换为：

val uiState: StateFlow<UiState> = combine(_categories, _expenses) { category, expenses ->
    UiState.of(category, expenses)
}.stateIn(
    scope = viewModelScope,
    started = SharingStarted.WhileSubscribed(5_000),
    initialValue = UiState.Default,
)

这使用

UiState.of

 将多个流合并为单个流以创建结果内容。然后将其转换为

StateFlow

。这需要一个初始值，所以我发明了一个

UiState.Default

。将其替换为您希望流程返回的任何内容，直到从 

_categories

 和 

_expenses

 生成第一个值。如果你愿意的话，可以是

null

。