Question

我正在使用项目反应器来使用rest从Web服务加载数据。这与多个线程并行完成。我开始在Web服务上达到速率限制，因此我希望每秒发送最多10个请求以避免出现这些错误。我怎么用反应堆呢？

使用zipWith（Mono.delayMillis（100））？或者有更好的方法吗？

谢谢

Answer 1

您可以使用delayElements代替整个zipwith。

Answer 2

以下代码将以每秒10个请求的速度在GET上执行https://www.google.com/。为了支持服务器无法在1秒内处理所有10个请求的情况，您必须进行其他更改。当服务器仍处理前一秒的请求时，您可以跳过发送请求。

@Test
void parallelHttpRequests() {
    // this is just for limiting the test running period otherwise you don't need it
    int COUNT = 2;

    // use whatever (blocking) http client you desire;
    // when using e.g. WebClient (Spring, non blocking client)
    // the example will slightly change for no longer use
    // subscribeOn(Schedulers.elastic())
    RestTemplate client = new RestTemplate();

    var exit = new AtomicBoolean(false);
    var lock = new ReentrantLock();
    var condition = lock.newCondition();

    MessageFormat message = new MessageFormat("#batch: {0}, #req: {1}, resultLength: {2}");
    Flux.interval(Duration.ofSeconds(1L))
            .take(COUNT) // this is just for limiting the test running period otherwise you don't need it
            .doOnNext(batch -> debug("#batch", batch)) // just for debugging
            .flatMap(batch -> Flux.range(1, 10) // 10 requests per 1 second
                            .flatMap(i -> Mono.fromSupplier(() ->
                                    client.getForEntity("https://www.google.com/", String.class).getBody()) // your request goes here (1 of 10)
                                    .map(s -> message.format(new Object[]{batch, i, s.length()})) // here the request's result will be the output of message.format(...)
                                    .doOnSubscribe(s -> debug("doOnSubscribe: #batch = " + batch + ", i = " + i)) // just for debugging
                                    .subscribeOn(Schedulers.elastic()) // one I/O thread per request
                            )
            )
            .subscribe(
                    s -> debug("received", s) // do something with the above request's result
                    e -> {
                        debug("error", e.getMessage());
                        signalAll(exit, condition, lock);
                    },
                    () -> {
                        debug("done");
                        signalAll(exit, condition, lock);
                    }
            );

    await(exit, condition, lock);
}

private void await(AtomicBoolean exit, Condition condition, Lock lock) {
    lock.lock();
    while (!exit.get()) {
        try {
            condition.await();
        } catch (InterruptedException e) {
            // maybe spurious wakeup
            e.printStackTrace();
        }
    }
    lock.unlock();
    debug("exit");
}

private void signalAll(AtomicBoolean exit, Condition condition, Lock lock) {
    exit.set(true);
    try {
        lock.lock();
        condition.signalAll();
    } finally {
        lock.unlock();
    }
}

使用Reactor限制请求率

2 个答案: