您可以分两步解决这个问题:
- 计算最新的最早开始时间之间的区别。计算0这种差异之间的随机值,并将它添加到的最早开始时间,以获得一个随机的开始时间
- 从一个随机数天,在这个数字是0和90之间,你的情况创建
Duration
(当然,添加此持续时间加到开始时间将产生结束时刻)
下面是一个例子:
import java.time.Duration;
import java.time.OffsetDateTime;
import java.time.temporal.ChronoUnit;
import java.util.Random;
public class RandomOffsetDateTimeRange
{
public static void main(String[] args)
{
OffsetDateTime earliestStartTime =
OffsetDateTime.parse("2017-04-01T00:00:00+00:00");
OffsetDateTime latestStartTime =
OffsetDateTime.parse("2017-05-31T23:59:59+00:00");
int maxDays = 90;
Random random = new Random(0);
OffsetDateTime startTime =
computeRandomOffsetDateTime(
earliestStartTime, latestStartTime, random);
Duration duration = randomDaysDuration(maxDays, random);
OffsetDateTime endTime = startTime.plus(duration);
System.out.println("startTime " + startTime);
System.out.println("endTime " + endTime);
System.out.println("duration " + duration.toDays() + " days");
}
private static Duration randomDaysDuration(int maxDays, Random random)
{
int days = random.nextInt(maxDays);
return Duration.ofDays(days);
}
private static OffsetDateTime computeRandomOffsetDateTime(
OffsetDateTime min, OffsetDateTime max, Random random)
{
long differenceInSeconds = min.until(max, ChronoUnit.SECONDS);
long seconds = nextLong(random, differenceInSeconds);
return min.plusSeconds(seconds);
}
// From https://stackoverflow.com/a/2546186/3182664
private static long nextLong(Random rng, long n) {
// error checking and 2^x checking removed for simplicity.
long bits, val;
do {
bits = (rng.nextLong() << 1) >>> 1;
val = bits % n;
} while (bits-val+(n-1) < 0L);
return val;
}
}
输出是
startTime 2017-04-13T08:03:53Z
endTime 2017-06-01T08:03:53Z
duration 49 days
(注:在该实例中,我计算以秒为单位的差异。根据应用程序的情况下,您可能需要使用不同的单元,例如,整个天)
是持续时间较毫秒一个int?秒?分钟?什么? – Catchwa
持续时间代表一天。谢谢。 – leo