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
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?
看看Joda Time,它是一个改进的Java日期/时间API,应该可以很好地与Scala一起工作。
由于这里所有的答案都是正确的,但使用传统java或第三方库,如joda或类似的,我将放弃使用新java的另一种方式。Java 8及以后版本中的时间类。参见Oracle教程。
使用LocalDate和ChronoUnit:
LocalDate d1 = LocalDate.of(2017, 5, 1);
LocalDate d2 = LocalDate.of(2017, 5, 18);
long days = ChronoUnit.DAYS.between(d1, d2);
System.out.println( days );
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。
在阅读了许多关于这个问题的回答和评论后,我的印象是,要么使用Joda时间,要么考虑到日光节约时间的一些特点等等。由于这两种方法我都不想做,所以我最终编写了几行代码来计算两个日期之间的差异,而没有使用任何与日期或时间相关的Java类。
在下面的代码中,年、月和日的数字与现实生活中的数字相同。例如,2015年12月24日,年= 2015,月= 12,日= 24。
我想分享这些代码,以防其他人想要使用它。有3种方法:1)找出给定年份是否是闰年的方法2)计算给定年份1月1日的天数的方法3)计算任意两个日期之间天数的方法2(结束日期减去开始日期)。
方法如下:
1)
public static boolean isLeapYear (int year) {
//Every 4. year is a leap year, except if the year is divisible by 100 and not by 400
//For example 1900 is not a leap year but 2000 is
boolean result = false;
if (year % 4 == 0) {
result = true;
}
if (year % 100 == 0) {
result = false;
}
if (year % 400 == 0) {
result = true;
}
return result;
}
2)
public static int daysGoneSince (int yearZero, int year, int month, int day) {
//Calculates the day number of the given date; day 1 = January 1st in the yearZero
//Validate the input
if (year < yearZero || month < 1 || month > 12 || day < 1 || day > 31) {
//Throw an exception
throw new IllegalArgumentException("Too many or too few days in month or months in year or the year is smaller than year zero");
}
else if (month == 4 || month == 6 || month == 9 || month == 11) {//Months with 30 days
if (day == 31) {
//Throw an exception
throw new IllegalArgumentException("Too many days in month");
}
}
else if (month == 2) {//February 28 or 29
if (isLeapYear(year)) {
if (day > 29) {
//Throw an exception
throw new IllegalArgumentException("Too many days in month");
}
}
else if (day > 28) {
//Throw an exception
throw new IllegalArgumentException("Too many days in month");
}
}
//Start counting days
int days = 0;
//Days in the target month until the target day
days = days + day;
//Days in the earlier months in the target year
for (int i = 1; i < month; i++) {
switch (i) {
case 1: case 3: case 5:
case 7: case 8: case 10:
case 12:
days = days + 31;
break;
case 2:
days = days + 28;
if (isLeapYear(year)) {
days = days + 1;
}
break;
case 4: case 6: case 9: case 11:
days = days + 30;
break;
}
}
//Days in the earlier years
for (int i = yearZero; i < year; i++) {
days = days + 365;
if (isLeapYear(i)) {
days = days + 1;
}
}
return days;
}
3)
public static int dateDiff (int startYear, int startMonth, int startDay, int endYear, int endMonth, int endDay) {
int yearZero;
//daysGoneSince presupposes that the first argument be smaller or equal to the second argument
if (10000 * startYear + 100 * startMonth + startDay > 10000 * endYear + 100 * endMonth + endDay) {//If the end date is earlier than the start date
yearZero = endYear;
}
else {
yearZero = startYear;
}
return daysGoneSince(yearZero, endYear, endMonth, endDay) - daysGoneSince(yearZero, startYear, startMonth, startDay);
}