Question

我想实现一个容纳2次，开始和结束的小时间容器，，我想使用它在多天的事件上迭代它，以检查是否在该特定时间范围内发生了事件，我将只需要在一分钟/范围内工作，所以该对象看起来像是这样的事情。 timeContainer（小时 - 分钟开始，小时 - 分钟结束）;

时间范围通常为几分钟，如下午13:00至13:10 我遇到的问题是毫秒，因为它们代表一个静态的最后时刻，它们不能用于多天的迭代，我会用它作为伪代码：

select events that has happened in the timeframe specified in the constructor  regardless of the day and month,

事件包含一个Calendar实例，我想针对小时和分钟执行匹配，任何建议？我正试图用joda-time来解决这个问题，但到目前为止还没有找到方法

谢谢

Answer 1

如果你想用毫秒来做，只需使用mod（％）和一天中的毫秒数。如果你想和Joda一起去，你可能会得到一些更具可读性的东西。见下面的例子。

public class Test {

    static class TimeContainer {
        private static final long second = 1000;
        private static final long minute = 60 * second;
        private static final long hour = 60 * minute;
        private static final long day = 24 * hour;

        private final long starttime;
        private final long endtime;

        public TimeContainer(long startHour, long startMinutes, long endHour, long endMinutes) {
            starttime = startHour * hour + startMinutes * minute;
            endtime = endHour * hour + endMinutes * minute + minute; 
        }

        public boolean test(long timeToTest) {
            long hoursInDay = timeToTest % day;
            return hoursInDay >= starttime && hoursInDay <= endtime;
        }
    }

    static class JodaContainer {
        private final LocalTime starttime;
        private final LocalTime endtime;

        public JodaContainer (LocalTime start, LocalTime end) {
            starttime = start;
            endtime = end;
        }

        public boolean test(long timeToTest) {
            LocalTime lt = new LocalTime(timeToTest);
            return lt.equals(starttime) || lt.equals(endtime) || (lt.isAfter(starttime) && lt.isBefore(endtime));
        }
    }

    public static void main(String[] args) {

        long[] testTimes1 = new long[5];
        long[] testTimes2 = new long[5];

        Calendar test1 = Calendar.getInstance(TimeZone.getTimeZone("Etc/Zulu"));
        Calendar test2 = Calendar.getInstance();

        TimeContainer timeContainer = new TimeContainer(13, 0, 13, 10);
        JodaContainer jodaContainer = new JodaContainer(new LocalTime(13,0), new LocalTime(13,10));

        test1.set(2010, 10, 5, 13, 6, 20);
        test2.set(2010, 10, 5, 13, 6, 20);
        testTimes1[0] = test1.getTimeInMillis();
        testTimes2[0] = test2.getTimeInMillis();

        test1.set(2012, 9, 6, 13, 1, 24);
        testTimes1[1] = test1.getTimeInMillis();
        test2.set(2012, 9, 6, 13, 1, 24);
        testTimes2[1] = test2.getTimeInMillis();

        test1.set(2010, 11, 22, 13, 9, 1);
        testTimes1[2] = test1.getTimeInMillis();
        test2.set(2010, 11, 22, 13, 9, 1);
        testTimes2[2] = test2.getTimeInMillis();

        test1.set(2012, 10, 5, 13, 26, 20);
        testTimes1[3] = test1.getTimeInMillis();
        test2.set(2012, 10, 5, 13, 26, 20);
        testTimes2[3] = test2.getTimeInMillis();

        test1.set(2010, 10, 5, 14, 6, 20);
        testTimes1[4] = test1.getTimeInMillis();
        test2.set(2010, 10, 5, 14, 6, 20);
        testTimes2[4] = test2.getTimeInMillis();

        for (long t : testTimes1) {
            System.out.println(t + "=" + timeContainer.test(t));
        }
        System.out.println();
        for (long t : testTimes2) {
            System.out.println(t + "=" + jodaContainer.test(t));
        }

    }

 }

Answer 2

听起来你真的想要使用Joda Time的LocalTime课程：

timeContainer(LocalTime startTime, LocalTime endTime)

然后在您的代码中，您只需要为每个事件LocalTime提取DateTime，并检查它是否等于startTime之前和endTime之前

Answer 3

您可以调用Date.before（）或Date.after（）来检查时间是否在范围内。

boolean isTimeInRange(Calendar startTime, Calendar endTime, Date date ){

            if (date.before(endTime.getTime())
                    && date.after(startTime.getTime())) {

                // time in range
                return true;

            } else {

                return false;

            }

}

用于检查事件发生的时间容器

3 个答案: