我在使用时遇到了我不明白的行为Sinks.Many<String>
向多个订阅者通知某些事件:
fun main() {
val sink : Sinks.Many<String> = Sinks.many().multicast().onBackpressureBuffer()
val flux = sink.asFlux().log()
val d = flux.subscribe {
println("--> $it")
}
sink.emitNext("1", Sinks.EmitFailureHandler.FAIL_FAST)
val d2 = flux.subscribe {
println("--2> $it")
}
sink.emitNext("2", Sinks.EmitFailureHandler.FAIL_FAST)
}
此代码显示第一个订阅者获取值 1 和 2,第二个订阅者获取值 2。到目前为止一切顺利:
11:49:06.936 [main] INFO reactor.Flux.EmitterProcessor.1 - onSubscribe(EmitterProcessor.EmitterInner)
11:49:06.938 [main] INFO reactor.Flux.EmitterProcessor.1 - request(unbounded)
11:49:06.942 [main] INFO reactor.Flux.EmitterProcessor.1 - onNext(1)
--> 1
11:49:06.942 [main] INFO reactor.Flux.EmitterProcessor.1 - onSubscribe(EmitterProcessor.EmitterInner)
11:49:06.942 [main] INFO reactor.Flux.EmitterProcessor.1 - request(unbounded)
11:49:06.943 [main] INFO reactor.Flux.EmitterProcessor.1 - onNext(2)
--> 2
11:49:06.943 [main] INFO reactor.Flux.EmitterProcessor.1 - onNext(2)
--2> 2
现在,假设第一个订阅者在第一次发射后处置(取消)其订阅,我期望第一个订阅者获得 1,第二个订阅者获得 2:
val sink : Sinks.Many<String> = Sinks.many().multicast().onBackpressureBuffer()
val flux = sink.asFlux().log()
val d = flux.subscribe {
println("--> $it")
}
sink.emitNext("1", Sinks.EmitFailureHandler.FAIL_FAST)
d.dispose()
val d2 = flux.subscribe {
println("--2> $it")
}
sink.emitNext("2", Sinks.EmitFailureHandler.FAIL_FAST)
}
11:51:48.684 [main] INFO reactor.Flux.EmitterProcessor.1 - onSubscribe(EmitterProcessor.EmitterInner)
11:51:48.685 [main] INFO reactor.Flux.EmitterProcessor.1 - request(unbounded)
11:51:48.689 [main] INFO reactor.Flux.EmitterProcessor.1 - onNext(1)
--> 1
11:51:48.689 [main] INFO reactor.Flux.EmitterProcessor.1 - cancel()
11:51:48.689 [main] INFO reactor.Flux.EmitterProcessor.1 - onSubscribe(EmitterProcessor.EmitterInner)
11:51:48.689 [main] INFO reactor.Flux.EmitterProcessor.1 - request(unbounded)
11:51:48.690 [main] INFO reactor.Flux.EmitterProcessor.1 - onComplete()
然而,当第二个订阅者尝试订阅时,通量被视为已完成。为什么会发生这种情况?我需要 Sinks.Many 可以随时订阅和取消订阅,而无需取消。