英文:
How to transform Flow<T> to StateFlow<List<T>> in Kotlin?
问题
Here are the translations of the code-related parts:
有一个这样的流
val flow = flow {
for (i in 1..100) {
emit(i)
}
}
我需要将此流中的所有元素显示在屏幕上,所以我需要将 Flow<Int>
转换为 StateFlow<List<Int>>
。
我应该怎么做?
注意:不要修改
flow
内部的代码,类似于emit(list)
第一个编辑
我需要使用 StateFlow<T>.collectAsStateWithLifecycle
来在屏幕上显示数据。
如果我使用 flowOf((1..100).toList())
,那么我需要两个 Flow
,这当然是一个解决方案。但我正在寻找的是直接将 Flow<Int>
转换为 StateFlow<List<Int>>
。
英文:
There is a Flow like this
val flow = flow {
for (i in 1..100) {
emit(i)
}
}
I need to display all the element in this flow to the screen, so I need to convert Flow<Int>
to StateFlow<List<Int>>
.
what should i do?
> Note: Do not modify the code inside flow
, similar to emit(list)
first edit
I need to use StateFlow<T>.collectAsStateWithLifecycle
to display data on Screen.
If i use flowOf((1..100).toList())
, then I need two Flow
, which is of course a solution. But what I am looking for is to directly convert Flow<Int>
to StateFlow<List<Int>>
.
答案1
得分: 2
I'm guessing your flow in your question is just an example and you want to know how to convert any arbitrary flow.
如果你的问题中的流程只是一个示例,你想知道如何将任何任意的流程转换成什么。
If you're using Compose, what you ultimately need is State, not StateFlow. You can convert a Flow directly to State using Flow.collectAsState()
or Flow.collectAsStateWithLifecycle()
. You have to provide some initial value. I'm using emptyList()
as a sample initial value.
如果您使用Compose,您最终需要的是State,而不是StateFlow。您可以使用 Flow.collectAsState()
或 Flow.collectAsStateWithLifecycle()
将流程直接转换为State。您必须提供一些初始值。我正在使用 emptyList()
作为示例初始值。
val state: State<List<Int> = someFlow.collectAsStateWithLifecycle(emptyList())
collectAsStateWithLifecycle
is preferred to collectAsState
when on Android to avoid the possibility of using resources to continue collecting while something is off-screen.
在Android上,当您要避免在屏幕外继续收集时,建议使用collectAsStateWithLifecycle
而不是collectAsState
,以避免使用资源的可能性。
If you already have a StateFlow, you can convert it to State using StateFlow.collectAsState()
or StateFlow.collectAsStateWithLifecycle()
, which does not require you to provide an initial value since StateFlow always has a value that it can use as the initial value.
如果您已经有了StateFlow,您可以使用 StateFlow.collectAsState()
或 StateFlow.collectAsStateWithLifecycle()
将其转换为State,因为StateFlow始终有一个可以用作初始值的值,所以无需提供初始值。
If you want to convert a Flow to StateFlow (which, again, is unnecessary if you're only using it for Compose), you can use stateIn
. This requires a CoroutineScope to run the coroutine that will collect the cold flow and make it hot. It also requires you to specify under what conditions it should start/keep it hot. And it requires an initial value.
如果您想要将Flow转换为StateFlow(如果您只在Compose中使用它,则不需要这样做),您可以使用 stateIn
。这需要一个CoroutineScope来运行将收集冷流并使其变为热流的协程。还需要您指定在什么条件下它应该启动/保持热流。并且需要提供一个初始值。
val stateFlow: StateFlow<List<Int>> = someFlow.stateIn(viewModelScope, SharingStarted.WhileSubscribed(5000L), emptyList())
英文:
I'm guessing your flow in your question is just an example and you want to know how to convert any arbitrary flow.
If you're using Compose, what you ultimately need is State, not StateFlow. You can convert a Flow directly to State using Flow.collectAsState()
or Flow.collectAsStateWithLifecycle()
. You have to provide some initial value. I'm using emptyList()
as a sample initial value.
val state: State<List<Int> = someFlow.collectAsStateWithLifecycle(emptyList())
collectAsStateWithLifecycle
is preferred to collectAsState
when on Android to avoid the possibility of using resources to continue collecting while something is off-screen.
If you already have a StateFlow, you can convert it to State using StateFlow.collectAsState()
or StateFlow.collectAsStateWithLifecycle()
, which does not require you to provide an initial value since StateFlow always has a value that it can use as the initial value.
If you want to convert a Flow to StateFlow (which, again, is unnecessary if you're only using it for Compose), you can use stateIn
. This requires a CoroutineScope to run the coroutine that will collect the cold flow and make it hot. It also requires you to specify under what conditions it should start/keep it hot. And it requires an initial value.
val stateFlow: StateFlow<List<Int>> = someFlow.stateIn(viewModelScope, SharingStarted.WhileSubscribed(5000L), emptyList())
答案2
得分: 0
没有比使用 toList 更好的方法。如果你真的想要的话,你可以使用 flow { emit(flow.toList()) }
。
英文:
There isn't a better way than using toList. If you really wanted, you could use flow { emit(flow.toList()) }
.
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