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
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?
其他回答
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;
}
另一个纯Java变体:
public boolean isWithin30Days(Calendar queryCalendar) {
// 1. Take the date you are checking, and roll it back N days
Calendar queryCalMinus30Days = Calendar.getInstance();
queryCalMinus30Days.setTime(queryCalendar.getTime());
queryCalMinus30Days.add(Calendar.DATE, -30); // subtract 30 days from the calendar
// 2. Get respective milliseconds for the two Calendars: now & queryCal minus N days
long nowL = Calendar.getInstance().getTimeInMillis();
long queryCalMinus30DaysL = queryCalMinus30Days.getTimeInMillis();
// 3. if nowL is still less than the queryCalMinus30DaysL, it means queryCalendar is more than 30 days into future
boolean isWithin30Days = nowL >= queryCalMinus30DaysL;
return isWithin30Days;
}
感谢这里的入门代码:https://stackoverflow.com/a/30207726/2162226
public static void main(String[] args) {
String dateStart = "01/14/2012 09:29:58";
String dateStop = "01/14/2012 10:31:48";
SimpleDateFormat format = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = format.parse(dateStart);
d2 = format.parse(dateStop);
DateTime date11 = new DateTime(d1);
DateTime date22 = new DateTime(d2);
int days = Days.daysBetween(date11.withTimeAtStartOfDay(), date22.withTimeAtStartOfDay()).getDays();
int hours = Hours.hoursBetween(date11, date22).getHours() % 24;
int minutes = Minutes.minutesBetween(date11, date22).getMinutes() % 60;
int seconds = Seconds.secondsBetween(date11, date22).getSeconds() % 60;
if (hours > 0 || minutes > 0 || seconds > 0) {
days = days + 1;
}
System.out.println(days);
} catch (Exception e) {
e.printStackTrace();
}
}
这将为同一天提供日期差异
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。
Since dates can contain hours and minutes, final result will be rounded down, which will result in incorrect value. For example, you calculate difference between today at 22:00 p.m and day after tomorrow 00:00 a.m, so the final result will be 1, because in reality it was 1.08 or smth difference, then it gets rounded down when calling TimeUnit.MILLISECONDS.toDays(..). That's why you need to take that in account, so in my solution I subtract the remainder of milliseconds from milliseconds in a day. Additionally, if you want to count the end date, you need to +1 it.
import java.util.Date;
import java.util.concurrent.TimeUnit;
public static long getDaysBetween(Date date1, Date date2, boolean includeEndDate) {
long millisInDay = 60 * 60 * 24 * 1000;
long difference = Math.abs(date1.getTime() - date2.getTime());
long add = millisInDay - (difference % millisInDay);//is used to calculate true number of days, because by default hours, minutes are also counted
return TimeUnit.MILLISECONDS.toDays(difference + add) + (includeEndDate ? 1 : 0);
}
测试:
Date date1 = new Date(121, Calendar.NOVEMBER, 27); //2021 Nov 27
Date date2 = new Date(121, Calendar.DECEMBER, 29); //2021 Dec 29
System.out.println( getDaysBetween(date1, date2, false) ); //32 days difference
System.out.println( getDaysBetween(date1, date2, true) ); //33 days difference