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ScheduledThreadPoolExecutor

timer（不建议用）

timer也可以进行延迟运行，但是会有很多问题。

比如task1运行时间超过task2延迟时间。

@Slf4j(topic = "c.Main")
public class Main {public static void main(String[] args) {Timer timer = new Timer();TimerTask task1 = new TimerTask() {@Overridepublic void run() {log.debug("task 1");try {sleep(2000);} catch (InterruptedException e) {e.printStackTrace();}}};TimerTask task2 = new TimerTask() {@Overridepublic void run() {log.debug("task 2");}};log.debug("start...");timer.schedule(task1, 1000);timer.schedule(task2, 1000);}
}

当task1出异常，task2不会运行。

ScheduledThreadPoolExecutor

Executors就可以创建。

延迟执行：

@Slf4j(topic = "c.Main")
public class Main {public static void main(String[] args) {ScheduledExecutorService scheduledExecutorService = Executors.newScheduledThreadPool(1);scheduledExecutorService.schedule(() -> {log.debug("task1");int i = 1 / 0; // 模拟异常}, 1, TimeUnit.SECONDS);scheduledExecutorService.schedule(() -> {log.debug("task2");}, 1, TimeUnit.SECONDS);}
}

可以看出通过控制线程数可以解决timer的缺点。
同时出现异常也不会影响其他任务运行。

定时运行：

@Slf4j(topic = "c.Main")
public class Main {public static void main(String[] args) {ScheduledExecutorService scheduledExecutorService = Executors.newScheduledThreadPool(1);log.debug("start...");scheduledExecutorService.scheduleAtFixedRate(() -> {log.debug("running...");}, 1, 1, TimeUnit.SECONDS); // 每隔1s反复执行}
}

处理异常

try catch

ExecutorService pool = Executors.newFixedThreadPool(1);
pool.submit(() -> {try {log.debug("task1");int i = 1 / 0;} catch (Exception e) {log.error("error:", e);}
});

通过返回值判断

ExecutorService pool = Executors.newFixedThreadPool(1);Future<Boolean> f = pool.submit(() -> {log.debug("task1");int i = 1 / 0; // 模拟异常return true;
});
log.debug("result:{}", f.get());

若正常执行返回true，
若出现异常get会返回异常信息。

应用

每周四定时运行

public class Main {public static void main(String[] args) throws ExecutionException, InterruptedException {// 获得当前时间LocalDateTime now = LocalDateTime.now();// 获取本周四 18点时间LocalDateTime startTime =now.with(DayOfWeek.THURSDAY).withHour(18).withMinute(0).withSecond(0).withNano(0);// 如果当前时间已经超过 本周四 那么找下周四if(now.compareTo(startTime) >= 0) {startTime = startTime.plusWeeks(1);}// 计算时间差，延时执行时间long initialDelay = Duration.between(now, startTime).toMillis();// 执行间隔 1周long period = 1000 * 60 * 60 * 24 * 7;ScheduledExecutorService executor = Executors.newScheduledThreadPool(2);executor.scheduleAtFixedRate(() -> {System.out.println("running..");}, initialDelay, period, TimeUnit.MILLISECONDS);}
}