Java 指定时间内如果请求没有返回值就重试怎么实现（异步）

也有使用 Mono.defer 似乎好像也不太行？

completablefuture timedGet()可以设置超时时间,超时后会抛出 TimeoutException 异常你捕获该异常进行重试就可以了

resilience4j ？？？

我记得 guava 的 retry 是可以写条件判断是否重试的

可以看看 project reactor 文档，webflux 也是在这上边实现的，里边有个 timeout()还有其它的可以实现。

用 reactor 、rxjava 这些框架比较好实现

for 循环 + countDownLatch.await 指定超时时间

You can retry requests using the Java 11 HTTP client by composing the returned future with re-schedules in a loop. Here’s an example:

public static <T> CompletableFuture<T> retry(Supplier<CompletableFuture<T>> action, int maxRetries) {
return action.get().handle((result, throwable) -> {
if (throwable != null && maxRetries > 0) {
return retry(action, maxRetries - 1);
} else if (throwable != null) {
throw new RuntimeException(throwable);
} else {
return result;
}
}).thenCompose(Function.identity());
}
This method takes a Supplier that returns a CompletableFuture, and an integer maxRetries that specifies the maximum number of retries. It returns a new CompletableFuture that retries the action if it fails with an exception.

Here’s how you can use this method to retry an HTTP request:

HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
.uri(URI.create("http://example.com"))
.build();

CompletableFuture<HttpResponse<String>> respOnse= retry(() -> client.sendAsync(request, HttpResponse.BodyHandlers.ofString()), 3);
This code retries the HTTP request up to three times if it fails with an exception.

I hope this helps! Let me know if you have any other questions.

Mono.defer(() -> {
System.out.println(LocalDateTime.now());
val integer = Random.Default.nextInt(100);
System.out.println(integer);
if (integer > 80) {
return Mono.just(100);
} else {
return Mono.never();
}
})
.timeout(Duration.ofSeconds(1))
.retry(10)

defer 就行

楼上的回答都有同样的问题，就是应用挂掉重试就不会进行了