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
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";
}
}
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
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。
:)
试试这个:
int epoch = (int) (new java.text.SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse("01/01/1970 00:00:00").getTime() / 1000);
你可以在parse()方法参数中编辑字符串。
简单的diff(不含lib)
/**
* Get a diff between two dates
* @param date1 the oldest date
* @param date2 the newest date
* @param timeUnit the unit in which you want the diff
* @return the diff value, in the provided unit
*/
public static long getDateDiff(Date date1, Date date2, TimeUnit timeUnit) {
long diffInMillies = date2.getTime() - date1.getTime();
return timeUnit.convert(diffInMillies,TimeUnit.MILLISECONDS);
}
然后你可以调用:
getDateDiff(date1,date2,TimeUnit.MINUTES);
以分钟为单位获取两个日期的差值。
TimeUnit是java.util.concurrent。TimeUnit,一个从纳米到天的标准Java枚举。
人类可读的差异(不含lib)
public static Map<TimeUnit,Long> computeDiff(Date date1, Date date2) {
long diffInMillies = date2.getTime() - date1.getTime();
//create the list
List<TimeUnit> units = new ArrayList<TimeUnit>(EnumSet.allOf(TimeUnit.class));
Collections.reverse(units);
//create the result map of TimeUnit and difference
Map<TimeUnit,Long> result = new LinkedHashMap<TimeUnit,Long>();
long milliesRest = diffInMillies;
for ( TimeUnit unit : units ) {
//calculate difference in millisecond
long diff = unit.convert(milliesRest,TimeUnit.MILLISECONDS);
long diffInMilliesForUnit = unit.toMillis(diff);
milliesRest = milliesRest - diffInMilliesForUnit;
//put the result in the map
result.put(unit,diff);
}
return result;
}
http://ideone.com/5dXeu6
输出类似Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0},单位是有序的。
您只需将该映射转换为用户友好的字符串。
警告
上面的代码段计算两个瞬间之间的简单差。它会在夏令时切换期间导致问题,就像这篇文章中解释的那样。这意味着如果你计算没有时间的日期之间的差异,你可能会少了一天/小时。
在我看来,日期的差异是主观的,尤其是在日子上。你可以:
计算经过24小时的时间:day+1 - day = 1 day = 24h 计算经过的时间,考虑到夏令时:天+1 -天= 1 = 24小时(但使用午夜时间和夏令时,它可以是0天和23小时) 计算日开关的数量,这意味着一天+1 1pm -一天11am = 1天,即使经过的时间只有2h(如果有夏令时则为1h:p)
我的答案是有效的,如果你的日期差异的定义天匹配第一种情况
与JodaTime
如果你正在使用JodaTime,你可以得到2个瞬间的差异(millies支持ReadableInstant)日期:
Interval interval = new Interval(oldInstant, new Instant());
但是你也可以得到本地日期/时间的差异:
// returns 4 because of the leap year of 366 days
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5), PeriodType.years()).getYears()
// this time it returns 5
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5+1), PeriodType.years()).getYears()
// And you can also use these static methods
Years.yearsBetween(LocalDate.now(), LocalDate.now().plusDays(365*5)).getYears()
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。