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
查看示例http://www.roseindia.net/java/beginners/DateDifferent.shtml 这个例子给出了天、小时、分钟、秒和毫秒的差异:)。
import java.util.Calendar;
import java.util.Date;
public class DateDifferent {
public static void main(String[] args) {
Date date1 = new Date(2009, 01, 10);
Date date2 = new Date(2009, 07, 01);
Calendar calendar1 = Calendar.getInstance();
Calendar calendar2 = Calendar.getInstance();
calendar1.setTime(date1);
calendar2.setTime(date2);
long milliseconds1 = calendar1.getTimeInMillis();
long milliseconds2 = calendar2.getTimeInMillis();
long diff = milliseconds2 - milliseconds1;
long diffSeconds = diff / 1000;
long diffMinutes = diff / (60 * 1000);
long diffHours = diff / (60 * 60 * 1000);
long diffDays = diff / (24 * 60 * 60 * 1000);
System.out.println("\nThe Date Different Example");
System.out.println("Time in milliseconds: " + diff + " milliseconds.");
System.out.println("Time in seconds: " + diffSeconds + " seconds.");
System.out.println("Time in minutes: " + diffMinutes + " minutes.");
System.out.println("Time in hours: " + diffHours + " hours.");
System.out.println("Time in days: " + diffDays + " days.");
}
}
其他回答
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()
这可能是最直接的方法了——也许是因为我已经用Java编写了一段时间了(它的日期和时间库确实很笨拙),但对我来说,代码看起来“简单而漂亮”!
您是否对以毫秒为单位返回的结果感到满意,或者您的问题的一部分是希望以某种替代格式返回?
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;
}
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。
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";
}
}