JAVA多线程限流解决并发问题
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
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Process finished with exit code 0
网站题目:JAVA多线程限流解决并发问题
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