如何处理的行为类似于onActivityResult撰写导航?

Question

我有一个设置，我想导航到一个可组合的结果，然后根据该结果，我想从父启动另一个可组合的。下面是设置

ScreenA启动屏幕B以获取结果(用户在屏幕B中执行某事)->ScreenB设置一个结果并从堆栈中弹出自己->ScreenA查看结果，并根据该值启动ScreenC。

这里的问题是，一旦结果被ScreenB设置，我在结果上设置的观察者就会无限地发射。

下面是我的设置的简化版本:

fun AppNavHost(
navController: NavHostController
) {
NavHost(
navController = navController,
startDestination = "ScreenA",
) {
composable("ScreenA") {
ScreenA(
onClick = {
navController.navigate("ScreenB")
}
)
val result = navController.currentBackStackEntry
?.savedStateHandle
?.getLiveData<String>("key")
?.observeAsState()
result?.value?.let { str ->
// This seems to not remove the key or at least
// the live data value is not nulled out
navController.currentBackStackEntry
?.savedStateHandle
?.remove<String>("key")
if (str == "result") {
// This condition hits infinitely
navController.navigate("ScreenC")
// This logs infinitely
Log.d("TAG", "found result: $str")
}
}
}
composable("ScreenB") {
ScreenB(
onClick = {
navController.previousBackStackEntry
?.savedStateHandle
?.set("key", "result")
navController.popBackStack()
}
)
}
composable("ScreenC") { ScreenC() }
}
}

我已经看过这个问题，我在这里的设置是类似的。我如何修复这种行为，使屏幕b的结果只读取一次?

Answer 1

使用LaunchedEffect来避免在重组级别多次调用导航

NavHost(
navController = navController,
startDestination = "ScreenA",
) {
composable("ScreenA") {
Button(
onClick = {
navController.navigate("ScreenB")
},
modifier = Modifier.padding(16.dp)
) {
Text("Go to Screen B")
}
val result = navController.currentBackStackEntry
?.savedStateHandle
?.getLiveData<String>("key")
?.observeAsState()
result?.value?.let { str ->
navController.currentBackStackEntry
?.savedStateHandle
?.remove<String>("key")
LaunchedEffect(str) {
if (str == "result") {
navController.navigate("ScreenC")
Log.d("TAG", "found result: $str")
}
}
}
}
composable("ScreenB") {
Button(
onClick = {
navController.previousBackStackEntry
?.savedStateHandle
?.set("key", "result")
navController.popBackStack()
}
) {
Text("Screen B")
}
}
composable("ScreenC") { ScreenC() }
}