Reactor StepVerifier.withVirtualTime 无限期阻塞
Reactor StepVerifier.withVirtualTime blocks indefinitely
我正在尝试使用 Reactor 的虚拟时间功能,但测试无限期阻塞(没有超时)或抛出 AssertionError
(有超时):
@Test
public void test() {
StepVerifier.withVirtualTime(() ->
Flux.just(1, 2, 3, 4).delayElements(Duration.ofSeconds(1)))
.expectSubscription()
.expectNextCount(4)
.expectComplete()
.verify(Duration.ofSeconds(10));
}
例外情况是:
java.lang.AssertionError: VerifySubscribertimed out on reactor.core.publisher.FluxConcatMap$ConcatMapImmediate@66d1af89
相同的实时示例按预期运行:
@Test
public void test2() {
StepVerifier.create(Flux.just(1, 2, 3, 4).delayElements(Duration.ofSeconds(1)))
.expectSubscription()
.expectNextCount(4)
.expectComplete()
.verify(Duration.ofSeconds(10));
}
我在参考 Manipulating Time 之后的第一个示例中看不到错误。
怎么了?
你需要使用.thenAwait(Duration)
,否则(虚拟)时钟不会移动,延迟永远不会发生。您也可以在 expectSubscription()
.
之后使用 .expectNoEvent(Duration)
after
例如:
@Test
public void test() {
StepVerifier.withVirtualTime(() ->
Flux.just(1, 2, 3, 4).delayElements(Duration.ofSeconds(1)))
.expectSubscription() //t == 0
//move the clock forward by 1s, and check nothing is emitted in the meantime
.expectNoEvent(Duration.ofSeconds(1))
//so this effectively verifies the first value is delayed by 1s:
.expectNext(1)
//and so on...
.expectNoEvent(Duration.ofSeconds(1))
.expectNext(2)
//or move the clock forward by 2s, allowing events to take place,
//and check last 2 values where delayed
.thenAwait(Duration.ofSeconds(2))
.expectNext(3, 4)
.expectComplete()
//trigger the verification and check that in realtime it ran in under 200ms
.verify(Duration.ofMillis(200));
}
我正在尝试使用 Reactor 的虚拟时间功能,但测试无限期阻塞(没有超时)或抛出 AssertionError
(有超时):
@Test
public void test() {
StepVerifier.withVirtualTime(() ->
Flux.just(1, 2, 3, 4).delayElements(Duration.ofSeconds(1)))
.expectSubscription()
.expectNextCount(4)
.expectComplete()
.verify(Duration.ofSeconds(10));
}
例外情况是:
java.lang.AssertionError: VerifySubscribertimed out on reactor.core.publisher.FluxConcatMap$ConcatMapImmediate@66d1af89
相同的实时示例按预期运行:
@Test
public void test2() {
StepVerifier.create(Flux.just(1, 2, 3, 4).delayElements(Duration.ofSeconds(1)))
.expectSubscription()
.expectNextCount(4)
.expectComplete()
.verify(Duration.ofSeconds(10));
}
我在参考 Manipulating Time 之后的第一个示例中看不到错误。
怎么了?
你需要使用.thenAwait(Duration)
,否则(虚拟)时钟不会移动,延迟永远不会发生。您也可以在 expectSubscription()
.
.expectNoEvent(Duration)
after
例如:
@Test
public void test() {
StepVerifier.withVirtualTime(() ->
Flux.just(1, 2, 3, 4).delayElements(Duration.ofSeconds(1)))
.expectSubscription() //t == 0
//move the clock forward by 1s, and check nothing is emitted in the meantime
.expectNoEvent(Duration.ofSeconds(1))
//so this effectively verifies the first value is delayed by 1s:
.expectNext(1)
//and so on...
.expectNoEvent(Duration.ofSeconds(1))
.expectNext(2)
//or move the clock forward by 2s, allowing events to take place,
//and check last 2 values where delayed
.thenAwait(Duration.ofSeconds(2))
.expectNext(3, 4)
.expectComplete()
//trigger the verification and check that in realtime it ran in under 200ms
.verify(Duration.ofMillis(200));
}