java RxJava - 每秒发出一个 observable
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RxJava - emit an observable every second
提问by fergdev
I am making a timer in Android using RxJava. I need to make a timer in RxJava to emit an observable every second. I have tried the following but with no luck. Any thoughts on what I am doing wrong?
我正在使用 RxJava 在 Android 中制作计时器。我需要在 RxJava 中创建一个计时器，以每秒发出一个可观察对象。我尝试了以下但没有运气。关于我做错了什么的任何想法？
Observable.interval(1000L, TimeUnit.MILLISECONDS)
.timeInterval()
.observeOn(AndroidSchedulers.mainThread())
.subscribe({Log.d(LOG_TAG, "&&&& on timer") })
回答by hotkey
Your code seems not to be called. Check whether it is executed and when. As of working with Observable, it is completely correct.
您的代码似乎没有被调用。检查它是否被执行以及何时执行。在使用时Observable，它是完全正确的。
For example, I put your snippet inside onCreate(...)of my MainActivity:
例如，我将您的代码段放在onCreate(...)my 中MainActivity：
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
Observable.interval(1000L, TimeUnit.MILLISECONDS)
.timeInterval()
.observeOn(AndroidSchedulers.mainThread())
.subscribe { Log.d("tag", "&&&& on timer") }
// ...
}
And it works:
它有效：
Also, probably you don't need .timeInterval()because Observable.interval(...)itself emits sequential numbers within the specified rate, and .timeInterval()just transforms it to emit the time intervals elapsed between the emissions.
此外，您可能不需要，.timeInterval()因为Observable.interval(...)它本身会在指定的速率内发出序列号，而.timeInterval()只是将其转换为发出两次发射之间经过的时间间隔。
回答by Vesko
In your subscribe()you don't consume the longTimeIntervalobject that's returned by the timeInterval()operator.
在您中，subscribe()您不使用运算符longTimeInterval返回的对象timeInterval()。
Correct version:
正确版本：
.subscribe(longTimeInterval -> {
Log.d(LOG_TAG, "&&&& on timer");
}
Also I think you don't need the timeInterval()operator at all. Observable.interval()will emit an observable every second in your case, which I guess is what you want. timeInterval()transforms that to an observable that holds the exact time difference between two events occur, I doubt you'll need that.
另外我认为您根本不需要timeInterval()操作员。Observable.interval()在您的情况下，每秒都会发出一个 observable，我想这就是您想要的。timeInterval()将其转换为可观察到的两个事件发生之间的确切时间差，我怀疑您会需要它。
回答by kosiara - Bartosz Kosarzycki
In Kotlin & RxJava 2.0.2 simply define an Observable Interval with an initialDelay0 and period1 which will emit list item each second.
在 Kotlin & RxJava 2.0.2 中，只需定义一个带有initialDelay0 和period1的 Observable Interval ，它将每秒发出列表项。
val list = IntRange(0, 9).toList()
val disposable = Observable
.interval(0,1, TimeUnit.SECONDS)
.map { i -> list[i.toInt()] }
.take(list.size.toLong())
.subscribe {
println("item: $it")
}
Thread.sleep(11000)
