Question

由于Observable转换了很多功能，我创建了BehaviourSubject。现在我想分享Observable的值，因此不会为每个新订户重新执行函数链。此外，我希望共享副本的行为与原始副本相同，即新到达的订阅者应该在订阅后获得最后一次发布的值。

在0.20.x中，可以将multicast(subjectFactory).refCount()与BehaviourSubject的工厂一起使用，或者只使用share(initialValue)，BehaviourSubject依次使用PublishSubject而不是1.0.x。

如何在{{1}}中实现相同的行为？

Answer 1

我认为您可以multicast(behaviorSubjectFactory).refCount()替换replay(1).refCount()。

为了使讨论更具体，这里有一个完整的例子（在Scala中）：

@volatile var startTime: Long = 0
def printTimestamped(s: String) {
  println(s"[t=${System.currentTimeMillis-startTime}] $s")
}

// Suppose for simplicity that the UI Events are just ticks of a
// hot timer observable.
val uiEvents = Observable.timer(1000 millis, 1000 millis)
        .doOnEach(i => printTimestamped("producing " + i))
        .publish

// Now apply all the transformations
val transformed = uiEvents.map(i => i + 101)
        .doOnEach(i => printTimestamped("transformed to " + i))

// And set a default start value
val o1 = transformed.startWith(100)

// Share and make sure new subscribers get the last element replayed
// immediately after they subscribe:
val o2 = o1.replay(1).refCount

// startTime is just before we start the uiEvents observable
startTime = System.currentTimeMillis
val subscriptionUiEvents = uiEvents.connect

Thread.sleep(500)

printTimestamped("subscribing A")
val subscriptionA = o2.subscribe(i => printTimestamped("A got " + i))

Thread.sleep(2000)

printTimestamped("subscribing B")
val subscriptionB = o2.subscribe(i => printTimestamped("B got " + i))

Thread.sleep(2000)

printTimestamped("unsubscribing B")
subscriptionB.unsubscribe()

Thread.sleep(2000)

printTimestamped("unsubscribing A")
subscriptionA.unsubscribe()

// Now the transformations will stop being executed, but the UI
// events will still be produced

Thread.sleep(2000)

// Finally, also stop the UI events:
subscriptionUiEvents.unsubscribe()

输出：

[t=505] subscribing A
[t=519] A got 100
[t=1002] producing 0
[t=1003] transformed to 101
[t=1003] A got 101
[t=2002] producing 1
[t=2002] transformed to 102
[t=2002] A got 102
[t=2520] subscribing B
[t=2521] B got 102
[t=3003] producing 2
[t=3003] transformed to 103
[t=3003] A got 103
[t=3003] B got 103
[t=4002] producing 3
[t=4002] transformed to 104
[t=4002] A got 104
[t=4002] B got 104
[t=4521] unsubscribing B
[t=5003] producing 4
[t=5003] transformed to 105
[t=5003] A got 105
[t=6002] producing 5
[t=6002] transformed to 106
[t=6002] A got 106
[t=6522] unsubscribing A
[t=7003] producing 6
[t=8002] producing 7

原始回答：

引用release notes for 1.0.0：

删除了任何带有初始值的方法重载，因为startWith运算符已经允许这样做。

因此，只需使用share(initialValue)而不是share().startWith(initialValue)。

RxJava与新订阅者的最后一个值共享

1 个答案: