Question

我正在尝试计算两个日期之间的分钟数，同时排除任意定义并每周发生的时间段。我还需要能够计算反向，在给定时间的情况下，计算除了那些时间段之外的X分钟数。

例如，我可能有两个时期[星期五下午5点31分 - 星期六下午2点26分]和[星期二凌晨3点37分 - 星期四凌晨1点14分]，在计算两个日期之间的分钟时我不想算数并在计算前进时。

我目前只有一个代码可以解决这个问题，虽然它不是超级高效并且正在成为我系统的压力。我还需要适应我目前不做的多个定义的差距。

我在一个间隙中执行此操作的代码如下所示（hideStart和hideEnter是间隙的开始和结束日期时间，absoluteLowValue是我计算的开始时间之间的距离或时间）：

public int absoluteDistance(DateTime high){
    long totalMinutes = new Duration(absoluteLowValue,high).getStandardMinutes();

    if (!gapHider.isHidingGaps())
        return (int)totalMinutes;

    int minutesPerWeek = 10080;
    long minutesPerHide = new Duration(hideStart, hideEnd).getStandardMinutes();

    long numFullWeeks = totalMinutes/minutesPerWeek;
    long remainder = totalMinutes%minutesPerWeek;

    totalMinutes -= numFullWeeks*minutesPerHide;

    DateTime latestEnd = high;
    if (latestEnd.getDayOfWeek() == hideEnd.getDayOfWeek() && latestEnd.getSecondOfDay() < hideEnd.getSecondOfDay()){
        latestEnd = latestEnd.minusWeeks(1);
    }
    while (latestEnd.getDayOfWeek() != hideEnd.getDayOfWeek())
        latestEnd = latestEnd.minusDays(1);
    latestEnd = latestEnd.withTime(hideEnd.getHourOfDay(),
                                hideEnd.getMinuteOfHour(),
                                hideEnd.getSecondOfMinute(),
                                hideEnd.getMillisOfSecond());

    DateTime latestStart = high;
    if (latestStart.getDayOfWeek() == hideStart.getDayOfWeek() && latestStart.getSecondOfDay() < hideStart.getSecondOfDay()){
        latestStart = latestStart.minusWeeks(1);
    }
    while (latestStart.getDayOfWeek() != hideStart.getDayOfWeek())
        latestStart = latestStart.minusDays(1);
    latestStart = latestStart.withTime(hideStart.getHourOfDay(),
                                    hideStart.getMinuteOfHour(),
                                    hideStart.getSecondOfMinute(),
                                    hideStart.getMillisOfSecond());

    long timeToNearestEnd = new Duration(latestEnd, high).getStandardMinutes();
    long timeToNearestStart = new Duration(latestStart, high).getStandardMinutes();

    if (timeToNearestEnd < remainder){
        totalMinutes -= minutesPerHide;
    }else if (timeToNearestStart < remainder){
        totalMinutes -= new Duration(latestStart, high).getStandardMinutes();
    }

    return (int)totalMinutes;
}

public DateTime timeSinceAbsLow(int index){ 
    if (absoluteLowValue != null){

        if (!gapHider.isHidingGaps())
            return absoluteLowValue.plusMinutes(index);

        DateTime date = absoluteLowValue;
        long minutesPerWeek = 10080;
        long minutesPerHide = new Duration(hideStart, hideEnd).getStandardMinutes();
        int difference = (int)(minutesPerWeek - minutesPerHide);
        int count = 0;

        while (index - count >= difference){
            date = date.plusWeeks(1);
            count += difference;
        }

        int remaining = index - count;

        DateTime nextStart = date;

        while (nextStart.getDayOfWeek() != hideStart.getDayOfWeek())
            nextStart = nextStart.plusDays(1);
        nextStart = nextStart.withTime(hideStart.getHourOfDay(),
                                    hideStart.getMinuteOfHour(),
                                    hideStart.getSecondOfMinute(),
                                    hideStart.getMillisOfSecond());

        long timeDiff = new Duration(date, nextStart).getStandardMinutes();

        if (timeDiff < remaining){
            date = nextStart.plusMinutes((int)minutesPerHide);
            count+= timeDiff;
            remaining = index - count;
        }

        date = date.plusMinutes(remaining);
        return date;
    }
    return new DateTime(); 
}

是否有更好或更简单的方法来执行此过程？我想如果我添加一定数量的逻辑来循环遍历“间隙”列表，它只会减慢它的速度。我愿意不使用Jodatime，我恰好正在使用它。任何帮助表示赞赏！

Answer 1

如果我理解正确，您需要计算开始日期和结束日期之间的总时间，并减去一个或多个周期。因此，您可以使用Duration个对象，然后最终转换为您想要的任何内容（分钟，秒等）：

// compute all periods you don't want to count
List<Duration> hideList = new ArrayList<>();
// duration between friday and saturday (assuming they have the values of your example)
hideList.add(new Duration(friday, saturday));
// add as many periods you want

// total duration between start and end dates
Duration totalDuration = new Duration(start, end);

// subtract all periods from total
for (Duration duration : hideList) {
    totalDuration = totalDuration.minus(duration);
}

// convert to total number of minutes
long totalMinutes = totalDuration.getStandardMinutes();

我假设hideList中使用的所有日期都在开始日期和结束日期之间（但使用isAfter和isBefore方法很容易检查）

相反，您将totalMinutes添加到开始日期，然后将hideList中的所有持续时间加起来：

// sum total minutes to start
DateTime dt = start.plusMinutes((int) totalMinutes);

// sum all the periods
for (Duration duration : hideList) {
    dt = dt.plus(duration);
}
// dt is the end date

有一个细节：将Duration转换为分钟数时，您将失去秒和毫秒精度，因此在执行相反的算法时，这些字段将会丢失。如果这不是问题，请按上述方式进行。但是如果你想保留所有精度，只需先添加totalDuration对象而不是添加分钟数：

// sum totalDuratin to start (to preserve seconds and milliseconds precision)
DateTime dt = start.plus(totalDuration);

如何有效地计算排除时间块的时间之间的距离？

1 个答案: