将两个状态流合并为新的状态流

Question

我有两个状态流。是否可以将它们组合起来并获得新的状态流？从逻辑上讲，这应该是可能的，因为两个状态流都有初始值，但正如我所见，组合函数 returns 只是 Flow 而不是 StateFlow。

Answer 1

使用combine运算符，需要两个流和一个转换函数来合并两个流的结果。

 val int = MutableStateFlow(2)
 val double = MutableStateFlow(1.8)
 int.combine(double){ i, d ->
            i + d             
 }.collect(::println)

Answer 2

到目前为止我创建了函数：

fun <T1, T2, R> combineState(
        flow1: StateFlow<T1>,
        flow2: StateFlow<T2>,
        scope: CoroutineScope = GlobalScope,
        sharingStarted: SharingStarted = SharingStarted.Eagerly,
        transform: (T1, T2) -> R
): StateFlow<R> = combine(flow1, flow2) {
    o1, o2 -> transform.invoke(o1, o2)
}.stateIn(scope, sharingStarted, transform.invoke(flow1.value, flow2.value))

Answer 3

您可以使用 combine 运算符，然后对由此产生的 Flow 使用 stateIn 函数。

来自 kotlinx 协程存储库中的 stateIn documentation：

stateIn 函数“将 cold Flow 转换为 hot StateFlow在给定的协程范围内启动，与多个下游订阅者共享上游流的单个运行实例最近发出的值。"

在撰写本文时，它的签名是：

fun <T> Flow<T>.stateIn(
  scope: CoroutineScope,
  started: SharingStarted,
  initialValue: T
): StateFlow<T> (source)

因此，您应该能够对 Flow 进行任何需要的转换，包括组合它们，然后最终使用 stateIn 将它们转换回 StateFlow。

它可能看起来像这样（也许创建一个拼字游戏游戏点数计算器）：

val wordFlow = MutableStateFlow("Hi")
val pointFlow = MutableStateFlow(5)

val stateString = wordFlow.combine(pointFlow) { word, points ->
    "$word is worth $points points"
}.stateIn(viewModelScope, SharingStarted.Eagerly, "Default is worth 0 points")

stateString 将是 StateFlow<String> 类型，您已成功将另外两个 StateFlows 合并为一个 StateFlow.

Answer 4

合并 n 个状态流

@Suppress("CHANGING_ARGUMENTS_EXECUTION_ORDER_FOR_NAMED_VARARGS")
inline fun <reified T, R> combineStateFlow(
    vararg flows: StateFlow<T>,
    scope: CoroutineScope = GlobalScope,
    sharingStarted: SharingStarted = SharingStarted.Eagerly,
    crossinline transform: (Array<T>) -> R
): StateFlow<R> = combine(flows = flows) {
    transform.invoke(it)
}.stateIn(
    scope = scope,
    started = sharingStarted,
    initialValue = transform.invoke(flows.map {
        it.value
    }.toTypedArray())
)

使用：

data class A(val a: String)
data class B(val b: Int)

private val test1 = MutableStateFlow(A("a"))
private val test2 = MutableStateFlow(B(2))
@Suppress("CHANGING_ARGUMENTS_EXECUTION_ORDER_FOR_NAMED_VARARGS")
private val _isValidForm = combineStateFlow(
    flows = arrayOf(test1, test2),
    scope = viewModelScope
) { combinedFlows: Array<Any> ->
    combinedFlows.map {
        val doSomething = when (it) {
            is A -> true
            is B -> false
            else -> false
        }
    }
}

Gist

Answer 5

与@Nikola Despotoski 类似的解决方案，但采用扩展函数的形式

/**
 * Combines two [StateFlow]s into a single [StateFlow]
 */
fun <T1, T2, R> StateFlow<T1>.combineState(
  flow2: StateFlow<T2>,
  scope: CoroutineScope = GlobalScope,
  sharingStarted: SharingStarted = SharingStarted.Eagerly,
  transform: (T1, T2) -> R
): StateFlow<R> = combine(this, flow2) { o1, o2 -> transform.invoke(o1, o2) }
  .stateIn(scope, sharingStarted, transform.invoke(this.value, flow2.value))

将两个状态流合并为新的状态流

Combine two state flows into new state flow

collections

kotlin

kotlin-flow

合并 n 个状态流

使用：