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
如果你想修复跨越夏时制边界的日期范围的问题(例如,一个日期在夏季,另一个日期在冬季),你可以使用这个来获得天数的差异:
public static long calculateDifferenceInDays(Date start, Date end, Locale locale) {
Calendar cal = Calendar.getInstance(locale);
cal.setTime(start);
cal.set(Calendar.HOUR_OF_DAY, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
cal.set(Calendar.MILLISECOND, 0);
long startTime = cal.getTimeInMillis();
cal.setTime(end);
cal.set(Calendar.HOUR_OF_DAY, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
cal.set(Calendar.MILLISECOND, 0);
long endTime = cal.getTimeInMillis();
// calculate the offset if one of the dates is in summer time and the other one in winter time
TimeZone timezone = cal.getTimeZone();
int offsetStart = timezone.getOffset(startTime);
int offsetEnd = timezone.getOffset(endTime);
int offset = offsetEnd - offsetStart;
return TimeUnit.MILLISECONDS.toDays(endTime - startTime + offset);
}
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
另一个纯Java变体:
public boolean isWithin30Days(Calendar queryCalendar) {
// 1. Take the date you are checking, and roll it back N days
Calendar queryCalMinus30Days = Calendar.getInstance();
queryCalMinus30Days.setTime(queryCalendar.getTime());
queryCalMinus30Days.add(Calendar.DATE, -30); // subtract 30 days from the calendar
// 2. Get respective milliseconds for the two Calendars: now & queryCal minus N days
long nowL = Calendar.getInstance().getTimeInMillis();
long queryCalMinus30DaysL = queryCalMinus30Days.getTimeInMillis();
// 3. if nowL is still less than the queryCalMinus30DaysL, it means queryCalendar is more than 30 days into future
boolean isWithin30Days = nowL >= queryCalMinus30DaysL;
return isWithin30Days;
}
感谢这里的入门代码:https://stackoverflow.com/a/30207726/2162226
下面的代码可以给你想要的输出:
String startDate = "Jan 01 2015";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("MMM dd yyyy");
LocalDate date = LocalDate.parse(startDate, formatter);
String currentDate = "Feb 11 2015";
LocalDate date1 = LocalDate.parse(currentDate, formatter);
System.out.println(date1.toEpochDay() - date.toEpochDay());
只需要对每个函数调用getTime,取其差值,然后除以一天中的毫秒数。
以下是一种解决方案,因为我们有许多方法可以实现这一点:
import java.util.*;
int syear = 2000;
int eyear = 2000;
int smonth = 2;//Feb
int emonth = 3;//Mar
int sday = 27;
int eday = 1;
Date startDate = new Date(syear-1900,smonth-1,sday);
Date endDate = new Date(eyear-1900,emonth-1,eday);
int difInDays = (int) ((endDate.getTime() - startDate.getTime())/(1000*60*60*24));