我在Scala中使用Java的Java .util.Date类,并希望比较Date对象和当前时间。我知道我可以通过使用getTime()来计算delta:
(new java.util.Date()).getTime() - oldDate.getTime()
然而,这只给我留下一个长表示毫秒。有没有更简单,更好的方法来得到时间?
我在Scala中使用Java的Java .util.Date类,并希望比较Date对象和当前时间。我知道我可以通过使用getTime()来计算delta:
(new java.util.Date()).getTime() - oldDate.getTime()
然而,这只给我留下一个长表示毫秒。有没有更简单,更好的方法来得到时间?
当前回答
先回答最初的问题:
将以下代码放入Long getAge(){}这样的函数中
Date dahora = new Date();
long MillisToYearsByDiv = 1000l *60l * 60l * 24l * 365l;
long javaOffsetInMillis = 1990l * MillisToYearsByDiv;
long realNowInMillis = dahora.getTime() + javaOffsetInMillis;
long realBirthDayInMillis = this.getFechaNac().getTime() + javaOffsetInMillis;
long ageInMillis = realNowInMillis - realBirthDayInMillis;
return ageInMillis / MillisToYearsByDiv;
这里最重要的是在乘法和除法时处理长数字。当然,还有Java在日期演算中应用的偏移量。
:)
其他回答
让我来看看Joda Interval和Days之间的差异:
DateTime start = new DateTime(2012, 2, 6, 10, 44, 51, 0);
DateTime end = new DateTime(2012, 2, 6, 11, 39, 47, 1);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
System.out.println(period.getYears() + " years, " + period.getMonths() + " months, " + period.getWeeks() + " weeks, " + period.getDays() + " days");
System.out.println(period.getHours() + " hours, " + period.getMinutes() + " minutes, " + period.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *1 weeks, 1 days*
//0 hours, 54 minutes, 56 seconds
//Period can set PeriodType,such as PeriodType.yearMonthDay(),PeriodType.yearDayTime()...
Period p = new Period(start, end, PeriodType.yearMonthDayTime());
System.out.println(p.getYears() + " years, " + p.getMonths() + " months, " + p.getWeeks() + " weeks, " + p.getDays() + "days");
System.out.println(p.getHours() + " hours, " + p.getMinutes() + " minutes, " + p.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *0 weeks, 8 days*
//0 hours, 54 minutes, 56 seconds
因为这个问题用Scala做了标记,
import scala.concurrent.duration._
val diff = (System.currentTimeMillis() - oldDate.getTime).milliseconds
val diffSeconds = diff.toSeconds
val diffMinutes = diff.toMinutes
val diffHours = diff.toHours
val diffDays = diff.toDays
Since dates can contain hours and minutes, final result will be rounded down, which will result in incorrect value. For example, you calculate difference between today at 22:00 p.m and day after tomorrow 00:00 a.m, so the final result will be 1, because in reality it was 1.08 or smth difference, then it gets rounded down when calling TimeUnit.MILLISECONDS.toDays(..). That's why you need to take that in account, so in my solution I subtract the remainder of milliseconds from milliseconds in a day. Additionally, if you want to count the end date, you need to +1 it.
import java.util.Date;
import java.util.concurrent.TimeUnit;
public static long getDaysBetween(Date date1, Date date2, boolean includeEndDate) {
long millisInDay = 60 * 60 * 24 * 1000;
long difference = Math.abs(date1.getTime() - date2.getTime());
long add = millisInDay - (difference % millisInDay);//is used to calculate true number of days, because by default hours, minutes are also counted
return TimeUnit.MILLISECONDS.toDays(difference + add) + (includeEndDate ? 1 : 0);
}
测试:
Date date1 = new Date(121, Calendar.NOVEMBER, 27); //2021 Nov 27
Date date2 = new Date(121, Calendar.DECEMBER, 29); //2021 Dec 29
System.out.println( getDaysBetween(date1, date2, false) ); //32 days difference
System.out.println( getDaysBetween(date1, date2, true) ); //33 days difference
这是另一个样本。基本上适用于用户定义的模式。
public static LinkedHashMap<String, Object> checkDateDiff(DateTimeFormatter dtfObj, String startDate, String endDate)
{
Map<String, Object> dateDiffMap = new HashMap<String, Object>();
DateTime start = DateTime.parse(startDate,dtfObj);
DateTime end = DateTime.parse(endDate,dtfObj);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
dateDiffMap.put("ISO-8601_PERIOD_FORMAT", period);
dateDiffMap.put("YEAR", period.getYears());
dateDiffMap.put("MONTH", period.getMonths());
dateDiffMap.put("WEEK", period.getWeeks());
dateDiffMap.put("DAY", period.getWeeks());
dateDiffMap.put("HOUR", period.getHours());
dateDiffMap.put("MINUTE", period.getMinutes());
dateDiffMap.put("SECOND", period.getSeconds());
return dateDiffMap;
}
@Michael Borgwardt的回答实际上在Android上不能正常运行。存在舍入错误。例如5月19日到21日是1天,因为它是1.99:1。在转换为int型之前使用round。
Fix
int diffInDays = (int)Math.round(( (newerDate.getTime() - olderDate.getTime())
/ (1000 * 60 * 60 * 24) ))
请注意,这适用于UTC日期,因此如果查看本地日期,差异可能会减小一天。由于夏时制的原因,要让它在本地日期上正确工作,需要一种完全不同的方法。