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JAVA多线程限流解决并发问题

时间:2019-08-01 12:27:47      阅读:169      评论:0      收藏:0      [点我收藏+]
package concurrent; import java.text.SimpleDateFormat; import java.util.Date; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.util.concurrent.Semaphore; import java.util.concurrent.TimeUnit; /** * Auth: zhouhongliang * Date:2019/8/1 * 并发限流功能 * Semaphore */ public class SemaphoreDemo { public static void main(String[] args) { ExecutorService executorService = Executors.newCachedThreadPool(); Semaphore semaphore = new Semaphore(3); for (int i = 0; i < 10; i++) { executorService.execute(() -> { try { //semaphore.acquire();//一直等待 if (semaphore.tryAcquire(3, TimeUnit.SECONDS)) {//等待3秒 play(); semaphore.release(); } else { System.out.println(Thread.currentThread().getName() + " 进入超时"); } } catch (Exception e) { e.printStackTrace(); } finally { } }); } executorService.shutdown(); } /** * 模拟任务 */ public static void play() { SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); System.out.println(simpleDateFormat.format(new Date()) + " " + Thread.currentThread().getName() + " 进入"); try { Thread.sleep(2000); } catch (InterruptedException e) { e.printStackTrace(); } System.out.println(simpleDateFormat.format(new Date()) + " " + Thread.currentThread().getName() + " 退出"); } }

输出结果:
2019-08-01 11:09:50 pool-1-thread-1 进入
2019-08-01 11:09:50 pool-1-thread-3 进入
2019-08-01 11:09:50 pool-1-thread-2 进入
2019-08-01 11:09:52 pool-1-thread-3 退出
2019-08-01 11:09:52 pool-1-thread-1 退出
2019-08-01 11:09:52 pool-1-thread-2 退出
2019-08-01 11:09:52 pool-1-thread-4 进入
2019-08-01 11:09:52 pool-1-thread-5 进入
2019-08-01 11:09:52 pool-1-thread-6 进入
pool-1-thread-7 进入超时
pool-1-thread-8 进入超时
pool-1-thread-9 进入超时
pool-1-thread-10 进入超时
2019-08-01 11:09:54 pool-1-thread-6 退出
2019-08-01 11:09:54 pool-1-thread-5 退出
2019-08-01 11:09:54 pool-1-thread-4 退出

Process finished with exit code 0

JAVA多线程限流解决并发问题

原文:https://blog.51cto.com/11147669/2425542

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