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
int daysDiff = (date1.getTime() - date2.getTime()) / MILLIS_PER_DAY;
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?
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.");
}
}
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 );
因为这个问题用Scala做了标记,
import scala.concurrent.duration._
val diff = (System.currentTimeMillis() - oldDate.getTime).milliseconds
val diffSeconds = diff.toSeconds
val diffMinutes = diff.toMinutes
val diffHours = diff.toHours
val diffDays = diff.toDays