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
如果你需要一个格式化的返回字符串 “2天03h 42m 07s”,试试这个:
public String fill2(int value)
{
String ret = String.valueOf(value);
if (ret.length() < 2)
ret = "0" + ret;
return ret;
}
public String get_duration(Date date1, Date date2)
{
TimeUnit timeUnit = TimeUnit.SECONDS;
long diffInMilli = date2.getTime() - date1.getTime();
long s = timeUnit.convert(diffInMilli, TimeUnit.MILLISECONDS);
long days = s / (24 * 60 * 60);
long rest = s - (days * 24 * 60 * 60);
long hrs = rest / (60 * 60);
long rest1 = rest - (hrs * 60 * 60);
long min = rest1 / 60;
long sec = s % 60;
String dates = "";
if (days > 0) dates = days + " Days ";
dates += fill2((int) hrs) + "h ";
dates += fill2((int) min) + "m ";
dates += fill2((int) sec) + "s ";
return dates;
}
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
只需在两个Date对象上使用下面的方法。如果你想传递当前日期,只需传递new date()作为第二个参数,因为它是用当前时间初始化的。
public String getDateDiffString(Date dateOne, Date dateTwo)
{
long timeOne = dateOne.getTime();
long timeTwo = dateTwo.getTime();
long oneDay = 1000 * 60 * 60 * 24;
long delta = (timeTwo - timeOne) / oneDay;
if (delta > 0) {
return "dateTwo is " + delta + " days after dateOne";
}
else {
delta *= -1;
return "dateTwo is " + delta + " days before dateOne";
}
}
此外,除了从天数,如果,你也想要其他参数的差异,使用下面的片段,
int year = delta / 365;
int rest = delta % 365;
int month = rest / 30;
rest = rest % 30;
int weeks = rest / 7;
int days = rest % 7;
p.s. Code完全取自SO答案。
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));
你需要更清楚地定义你的问题。您可以只取两个Date对象之间的毫秒数,然后除以24小时内的毫秒数,例如……但是:
这将不考虑时区-日期总是UTC 这并没有考虑到日光节约时间(例如,有些日子可能只有23小时) 即使在UTC时间内,8月16日晚上11点到8月18日凌晨2点有多少天?只有27个小时,那意味着一天吗?还是应该是三天,因为它涵盖了三个日期?
public static String getDifferenceBtwTime(Date dateTime) {
long timeDifferenceMilliseconds = new Date().getTime() - dateTime.getTime();
long diffSeconds = timeDifferenceMilliseconds / 1000;
long diffMinutes = timeDifferenceMilliseconds / (60 * 1000);
long diffHours = timeDifferenceMilliseconds / (60 * 60 * 1000);
long diffDays = timeDifferenceMilliseconds / (60 * 60 * 1000 * 24);
long diffWeeks = timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 7);
long diffMonths = (long) (timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 30.41666666));
long diffYears = (long)(timeDifferenceMilliseconds / (1000 * 60 * 60 * 24 * 365));
if (diffSeconds < 1) {
return "one sec ago";
} else if (diffMinutes < 1) {
return diffSeconds + " seconds ago";
} else if (diffHours < 1) {
return diffMinutes + " minutes ago";
} else if (diffDays < 1) {
return diffHours + " hours ago";
} else if (diffWeeks < 1) {
return diffDays + " days ago";
} else if (diffMonths < 1) {
return diffWeeks + " weeks ago";
} else if (diffYears < 12) {
return diffMonths + " months ago";
} else {
return diffYears + " years ago";
}
}