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根据服务器的频率和启动设置任务的启动

events java

GustyWind · 技术社区 · 15 年前

我有一个执行频率的频率和时间范围。例如,如果当前时间为凌晨2点,频率为360分钟,则必须假设

当前时间是从午夜开始的服务器启动时间。因此,如果服务器在早上5点启动,并且frequency是360,那么任务应该在早上6点运行

但是如果任务是凌晨5点,频率是240,因为从午夜开始已经过了240分钟下次跑步应该在早上8点

如果服务器在上午10点启动,频率为300,则从午夜开始,一次运行在上午5点结束,下一次运行在上午10点计算,因此将在上午10点立即启动

时间范围分为4个季度,上午12点至上午6点至下午12点,下午12点至下午6点,下午6点至次日12点下面提供了一些值:

 stNow----serverTime
sElapsed ---time elapsed from midnight
currFreq ----frequecy 
    protected int _getInitWorkerTime()
    {
         int vRun=0;
         STime stNow= new STime(PWMSystem.newDate());
         int currFreq = _findAlertFrequency()/60;
         int sElapsed =Constants.MINUTES_PER_DAY-stNow.elapsedMinutes(STime.midnight);
         _log.error("now time as 24------"+stNow.getHourNo24());
         _log.error("now currFreq------"+currFreq);
         _log.error("now sElapsed-----"+sElapsed);
         if(sElapsed == 1440)
             sElapsed=0;
         if(stNow.getHourNo24()>=0  && stNow.getHourNo24()<=6)
         {
             _log.error("now time as 24-inside cond-1-----"+stNow.getHourNo24());
                 if(currFreq>sElapsed)
                     vRun= currFreq-sElapsed;
                 else
                     vRun= -(currFreq-sElapsed);
         }
         if(stNow.getHourNo24()>6 && stNow.getHourNo24()<=12)
         {
             _log.error("now time as 24-inside cond-2-----"+stNow.getHourNo24());
             if(currFreq>sElapsed)
                  vRun=360-(currFreq-sElapsed);
              else
                  vRun=360-(-(currFreq-sElapsed));
         }
         if(stNow.getHourNo24()>12 && stNow.getHourNo24()<=18)
         {
             _log.error("now time as 24-inside cond-3-----"+stNow.getHourNo24());
             if(currFreq>sElapsed)
                  vRun=720-(currFreq-sElapsed);
              else
                  vRun=720-(-(currFreq-sElapsed));
         }
         if(stNow.getHourNo24()>18 && (stNow.getHourNo24()<=24 ||stNow.getHourNo24()<=0))
         {
             _log.error("now time as 24-inside cond-4-----"+stNow.getHourNo24());
             if(currFreq>sElapsed)
                  vRun=1080-(currFreq-sElapsed);
              else
                  vRun=1080-(-(currFreq-sElapsed));
         }
            // vRun=_MAX_FREQUENCY_DELAY_IN_SEC+ sElapsed*60+_findAlertFrequency()+_BOOT_DELAY_SECONDS_START;*/
        //vRun=stNow.elapsedMinutes(STime.midnight)*60+_findAlertFrequency()+_BOOT_DELAY_SECONDS_START;
        return (vRun*60 + _BOOT_DELAY_SECONDS_START);       
    }

2 回复 | 直到 15 年前

Jon Skeet 15 年前

这不就是说:

int nextTime = ((timeSinceMidnight / (period-1)) + 1) * period;

准时,例如下面的第三个测试用例。)

示例情况:

Period      Minutes since midnight        Result
360         120 (2am)                     360 (6am)
240         300 (5am)                     480 (8am)
300         600 (10am)                    600 (10am)

(请注意,我指的是周期而不是频率-通常较高的频率意味着发生了一些事情 较少的 .)

GustyWind 15 年前

我发现。。

protected int _getInitWorkerTime()
{   
        int vRun = 0;
        STime stNow = new STime(PWMSystem.newDate());
        int currFreq = _findAlertFrequency() / Constants.SECONDS_PER_MINUTE;
        int sElapsed = Constants.MINUTES_PER_DAY- stNow.elapsedMinutes(STime.midnight);
        int runsCompleted = sElapsed / currFreq;
        int remainLeft = (runsCompleted + 1) * currFreq;


                if (remainLeft > sElapsed)
                    vRun = remainLeft - sElapsed;
                else
                    vRun = sElapsed - remainLeft;


return (vRun * 60 + _BOOT_DELAY_SECONDS_START);