Java實現系統(tǒng)限流的示例代碼

更新時間：2023年09月04日 10:23:42 作者：不焦躁的程序員

限流是保障系統(tǒng)高可用的方式之一,也是大廠高頻面試題,它在微服務系統(tǒng)中,緩存、限流、熔斷是保證系統(tǒng)高可用的三板斧,所以本文我們就來聊聊如何實現系統(tǒng)限流吧

限流是保障系統(tǒng)高可用的方式之一，也是大廠高頻面試題，如果面試官問一句，“如何實現每秒鐘1000個請求的限流？”，你要是分分鐘給他寫上幾種限流方案，那豈不香哉，哈哈！話不多說，我來列幾種常用限流實現方式。

1、Guava RateLimiter

Guava是Java領域很優(yōu)秀的開源項目，包含了日常開發(fā)常用的集合、String、緩存等，其中RateLimiter是常用限流工具。

RateLimiter是基于令牌桶算法實現的，如果每秒10個令牌，內部實現，會每100ms生產1個令牌。

使用Guava RateLimiter，如下：

1.引入pom依賴：

<dependency>
??<groupId>com.google.guava</groupId>
??<artifactId>guava</artifactId>
??<version>23.0</version>
</dependency>

2.代碼：

public?class?GuavaRateLimiterTest?{
????//比如每秒生產10個令牌，相當于每100ms生產1個令牌
????private?RateLimiter?rateLimiter?=?RateLimiter.create(10);
????/**
?????*?模擬執(zhí)行業(yè)務方法
?????*/
????public?void?exeBiz()?{
????????if?(rateLimiter.tryAcquire(1))?{
????????????try?{
????????????????Thread.sleep(500);
????????????}?catch?(InterruptedException?e)?{
????????????????e.printStackTrace();
????????????}
????????????System.out.println("線程"?+?Thread.currentThread().getName()?+?"：執(zhí)行業(yè)務邏輯");
????????}?else?{
????????????System.out.println("線程"?+?Thread.currentThread().getName()?+?"：被限流");
????????}
????}
????public?static?void?main(String[]?args)?throws?InterruptedException?{
????????GuavaRateLimiterTest?limiterTest?=?new?GuavaRateLimiterTest();
????????Thread.sleep(500);//等待500ms，讓limiter生產一些令牌
????????//模擬瞬間生產100個線程請求
????????for?(int?i?=?0;?i?<?100;?i++)?{
????????????new?Thread(limiterTest::exeBiz).start();
????????}
????}
}

2、滑窗計數

打個比方，某接口每秒允許100個請求，設置一個滑窗，窗口中有10個格子，每個格子占100ms，每100ms移動一次?；瑒哟翱诘母褡觿澐值脑蕉啵瑒哟翱诘臐L動就越平滑，限流的統(tǒng)計就會越精確。

代碼如下：

/**
?*?滑窗計數器
?*/
public?class?SliderWindowRateLimiter?implements?Runnable?{
????//每秒允許的最大訪問數
????private?final?long?maxVisitPerSecond;
????//將每秒時間劃分N個塊
????private?final?int?block;
????//每個塊存儲的數量
????private?final?AtomicLong[]?countPerBlock;
????//滑動窗口劃到了哪個塊兒，可以理解為滑動窗口的起始下標位置
????private?volatile?int?index;
????//目前總的數量
????private?AtomicLong?allCount;
????/**
?????*?構造函數
?????*
?????*?@param?block，每秒鐘劃分N個窗口
?????*?@param?maxVisitPerSecond?每秒最大訪問數量
?????*/
????public?SliderWindowRateLimiter(int?block,?long?maxVisitPerSecond)?{
????????this.block?=?block;
????????this.maxVisitPerSecond?=?maxVisitPerSecond;
????????countPerBlock?=?new?AtomicLong[block];
????????for?(int?i?=?0;?i?<?block;?i++)?{
????????????countPerBlock[i]?=?new?AtomicLong();
????????}
????????allCount?=?new?AtomicLong(0);
????}
????/**
?????*?判斷是否超過最大允許數量
?????*
?????*?@return
?????*/
????public?boolean?isOverLimit()?{
????????return?currentQPS()?>?maxVisitPerSecond;
????}
????/**
?????*?獲取目前總的訪問數
?????*
?????*?@return
?????*/
????public?long?currentQPS()?{
????????return?allCount.get();
????}
????/**
?????*?請求訪問進來，判斷是否可以執(zhí)行業(yè)務邏輯
?????*/
????public?void?visit()?{
????????countPerBlock[index].incrementAndGet();
????????allCount.incrementAndGet();
????????if?(isOverLimit())?{
????????????System.out.println(Thread.currentThread().getName()?+?"被限流"?+?"，currentQPS："?+?currentQPS()?+?"，index："?+?index);
????????}?else?{
????????????System.out.println(Thread.currentThread().getName()?+?"執(zhí)行業(yè)務邏輯"?+?"，currentQPS："?+?currentQPS()?+?"，index："?+?index);
????????}
????}
????/**
?????*?定時執(zhí)行器，
?????*?每N毫秒滑塊移動一次，然后再設置下新滑塊的初始化數字0，然后新的請求會落到新的滑塊上
?????*?同時總數減掉新滑塊上的數字，并且重置新的滑塊上的數量
?????*/
????@Override
????public?void?run()?{
????????index?=?(index?+?1)?%?block;
????????long?val?=?countPerBlock[index].getAndSet(0);
????????allCount.addAndGet(-val);
????}
????public?static?void?main(String[]?args)?{
????????SliderWindowRateLimiter?sliderWindowRateLimiter?=?new?SliderWindowRateLimiter(10,?100);
????????//固定的速率移動滑塊
????????ScheduledExecutorService?scheduledExecutorService?=?Executors.newSingleThreadScheduledExecutor();
????????scheduledExecutorService.scheduleAtFixedRate(sliderWindowRateLimiter,?100,?100,?TimeUnit.MILLISECONDS);
????????//模擬不同速度的請求
????????new?Thread(()?->?{
????????????while?(true)?{
????????????????sliderWindowRateLimiter.visit();
????????????????try?{
????????????????????Thread.sleep(10);
????????????????}?catch?(InterruptedException?e)?{
????????????????????e.printStackTrace();
????????????????}
????????????}
????????}).start();
????????//模擬不同速度的請求
????????new?Thread(()?->?{
????????????while?(true)?{
????????????????sliderWindowRateLimiter.visit();
????????????????try?{
????????????????????Thread.sleep(50);
????????????????}?catch?(InterruptedException?e)?{
????????????????????e.printStackTrace();
????????????????}
????????????}
????????}).start();
????}
}

3、信號量

利用Semaphore，每隔固定速率，釋放Semaphore的資源。線程獲取到資源，則執(zhí)行業(yè)務代碼。

代碼如下：

public?class?SemaphoreOne?{
????private?static?Semaphore?semaphore?=?new?Semaphore(10);
????public?static?void?bizMethod()?throws?InterruptedException?{
????????if?(!semaphore.tryAcquire())?{
????????????System.out.println(Thread.currentThread().getName()?+?"被拒絕");
????????????return;
????????}
????????System.out.println(Thread.currentThread().getName()?+?"執(zhí)行業(yè)務邏輯");
????????Thread.sleep(500);//模擬處理業(yè)務邏輯需要1秒
????????semaphore.release();
????}
????public?static?void?main(String[]?args)?{
????????Timer?timer?=?new?Timer();
????????timer.scheduleAtFixedRate(new?TimerTask()?{
????????????@Override
????????????public?void?run()?{
????????????????semaphore.release(10);
????????????????System.out.println("釋放所有鎖");
????????????}
????????},?1000,?1000);
????????for?(int?i?=?0;?i?<?10000;?i++)?{
????????????try?{
????????????????Thread.sleep(10);//模擬每隔10ms就有1個請求進來
????????????}?catch?(InterruptedException?e)?{
????????????????e.printStackTrace();
????????????}
????????????new?Thread(()?->?{
????????????????try?{
????????????????????SemaphoreOne.bizMethod();
????????????????}?catch?(InterruptedException?e)?{
????????????????????e.printStackTrace();
????????????????}
????????????}).start();
????????}
????}
}

4、令牌桶

令牌桶算法：一個存放固定容量令牌的桶，按照固定速率往桶里添加令牌，如有剩余容量則添加，沒有則放棄。如果有請求進來，則需要先從桶里獲取令牌，當桶里沒有令牌可取時，則拒絕任務。

令牌桶的優(yōu)點是：可以改變添加令牌的速率，一旦提高速率，則可以處理突發(fā)流量。

代碼如下：

public?class?TokenBucket?{
????/**
?????*?定義的桶
?????*/
????public?class?Bucket?{
????????//容量
????????int?capacity;
????????//速率，每秒放多少
????????int?rateCount;
????????//目前token個數
????????AtomicInteger?curCount?=?new?AtomicInteger(0);
????????public?Bucket(int?capacity,?int?rateCount)?{
????????????this.capacity?=?capacity;
????????????this.rateCount?=?rateCount;
????????}
????????public?void?put()?{
????????????if?(curCount.get()?<?capacity)?{
????????????????System.out.println("目前數量=="?+?curCount.get()?+?",?我還可以繼續(xù)放");
????????????????curCount.addAndGet(rateCount);
????????????}
????????}
????????public?boolean?get()?{
????????????if?(curCount.get()?>=?1)?{
????????????????curCount.decrementAndGet();
????????????????return?true;
????????????}
????????????return?false;
????????}
????}
????@Test
????public?void?testTokenBucket()?throws?InterruptedException?{
????????Bucket?bucket?=?new?Bucket(5,?2);
????????//固定線程，固定的速率往桶里放數據，比如每秒N個
????????ScheduledThreadPoolExecutor?scheduledCheck?=?new?ScheduledThreadPoolExecutor(1);
????????scheduledCheck.scheduleAtFixedRate(()?->?{
????????????bucket.put();
????????},?0,?1,?TimeUnit.SECONDS);
????????//先等待一會兒，讓桶里放點token
????????Thread.sleep(6000);
????????//模擬瞬間10個線程進來拿token
????????for?(int?i?=?0;?i?<?10;?i++)?{
????????????new?Thread(()?->?{
????????????????if?(bucket.get())?{
????????????????????System.out.println(Thread.currentThread()?+?"獲取到了資源");
????????????????}?else?{
????????????????????System.out.println(Thread.currentThread()?+?"被拒絕");
????????????????}
????????????}).start();
????????}
????????//等待，往桶里放token
????????Thread.sleep(3000);
????????//繼續(xù)瞬間10個線程進來拿token
????????for?(int?i?=?0;?i?<?10;?i++)?{
????????????new?Thread(()?->?{
????????????????if?(bucket.get())?{
????????????????????System.out.println(Thread.currentThread()?+?"獲取到了資源");
????????????????}?else?{
????????????????????System.out.println(Thread.currentThread()?+?"被拒絕");
????????????????}
????????????}).start();
????????}
????}
}