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
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));
其他回答
这是一个正确的Java 7解决方案,没有任何依赖。
public static int countDaysBetween(Date date1, Date date2) {
Calendar c1 = removeTime(from(date1));
Calendar c2 = removeTime(from(date2));
if (c1.get(YEAR) == c2.get(YEAR)) {
return Math.abs(c1.get(DAY_OF_YEAR) - c2.get(DAY_OF_YEAR)) + 1;
}
// ensure c1 <= c2
if (c1.get(YEAR) > c2.get(YEAR)) {
Calendar c = c1;
c1 = c2;
c2 = c;
}
int y1 = c1.get(YEAR);
int y2 = c2.get(YEAR);
int d1 = c1.get(DAY_OF_YEAR);
int d2 = c2.get(DAY_OF_YEAR);
return d2 + ((y2 - y1) * 365) - d1 + countLeapYearsBetween(y1, y2) + 1;
}
private static int countLeapYearsBetween(int y1, int y2) {
if (y1 < 1 || y2 < 1) {
throw new IllegalArgumentException("Year must be > 0.");
}
// ensure y1 <= y2
if (y1 > y2) {
int i = y1;
y1 = y2;
y2 = i;
}
int diff = 0;
int firstDivisibleBy4 = y1;
if (firstDivisibleBy4 % 4 != 0) {
firstDivisibleBy4 += 4 - (y1 % 4);
}
diff = y2 - firstDivisibleBy4 - 1;
int divisibleBy4 = diff < 0 ? 0 : diff / 4 + 1;
int firstDivisibleBy100 = y1;
if (firstDivisibleBy100 % 100 != 0) {
firstDivisibleBy100 += 100 - (firstDivisibleBy100 % 100);
}
diff = y2 - firstDivisibleBy100 - 1;
int divisibleBy100 = diff < 0 ? 0 : diff / 100 + 1;
int firstDivisibleBy400 = y1;
if (firstDivisibleBy400 % 400 != 0) {
firstDivisibleBy400 += 400 - (y1 % 400);
}
diff = y2 - firstDivisibleBy400 - 1;
int divisibleBy400 = diff < 0 ? 0 : diff / 400 + 1;
return divisibleBy4 - divisibleBy100 + divisibleBy400;
}
public static Calendar from(Date date) {
Calendar c = Calendar.getInstance();
c.setTime(date);
return c;
}
public static Calendar removeTime(Calendar c) {
c.set(HOUR_OF_DAY, 0);
c.set(MINUTE, 0);
c.set(SECOND, 0);
c.set(MILLISECOND, 0);
return c;
}
有很多方法可以找到日期和时间之间的区别。我所知道的最简单的方法之一是:
Calendar calendar1 = Calendar.getInstance();
Calendar calendar2 = Calendar.getInstance();
calendar1.set(2012, 04, 02);
calendar2.set(2012, 04, 04);
long milsecs1= calendar1.getTimeInMillis();
long milsecs2 = calendar2.getTimeInMillis();
long diff = milsecs2 - milsecs1;
long dsecs = diff / 1000;
long dminutes = diff / (60 * 1000);
long dhours = diff / (60 * 60 * 1000);
long ddays = diff / (24 * 60 * 60 * 1000);
System.out.println("Your Day Difference="+ddays);
打印语句只是一个示例—您可以按照自己喜欢的方式格式化它。
在某些地区使用毫秒方法可能会导致问题。
举个例子,03/24/2007和03/25/2007之间的差应该是1天;
然而,如果使用毫秒路径,你将得到0天,如果你在英国运行这个!
/** Manual Method - YIELDS INCORRECT RESULTS - DO NOT USE**/
/* This method is used to find the no of days between the given dates */
public long calculateDays(Date dateEarly, Date dateLater) {
return (dateLater.getTime() - dateEarly.getTime()) / (24 * 60 * 60 * 1000);
}
更好的实现方法是使用java.util.Calendar
/** Using Calendar - THE CORRECT WAY**/
public static long daysBetween(Calendar startDate, Calendar endDate) {
Calendar date = (Calendar) startDate.clone();
long daysBetween = 0;
while (date.before(endDate)) {
date.add(Calendar.DAY_OF_MONTH, 1);
daysBetween++;
}
return daysBetween;
}
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。
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