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
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
其他回答
这是另一个样本。基本上适用于用户定义的模式。
public static LinkedHashMap<String, Object> checkDateDiff(DateTimeFormatter dtfObj, String startDate, String endDate)
{
Map<String, Object> dateDiffMap = new HashMap<String, Object>();
DateTime start = DateTime.parse(startDate,dtfObj);
DateTime end = DateTime.parse(endDate,dtfObj);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
dateDiffMap.put("ISO-8601_PERIOD_FORMAT", period);
dateDiffMap.put("YEAR", period.getYears());
dateDiffMap.put("MONTH", period.getMonths());
dateDiffMap.put("WEEK", period.getWeeks());
dateDiffMap.put("DAY", period.getWeeks());
dateDiffMap.put("HOUR", period.getHours());
dateDiffMap.put("MINUTE", period.getMinutes());
dateDiffMap.put("SECOND", period.getSeconds());
return dateDiffMap;
}
使用GMT时区获取一个Calendar实例,使用Calendar类的set方法设置时间。GMT时区偏移量为0(并不重要),夏令时标志设置为false。
final Calendar cal = Calendar.getInstance(TimeZone.getTimeZone("GMT"));
cal.set(Calendar.YEAR, 2011);
cal.set(Calendar.MONTH, 9);
cal.set(Calendar.DAY_OF_MONTH, 29);
cal.set(Calendar.HOUR, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
final Date startDate = cal.getTime();
cal.set(Calendar.YEAR, 2011);
cal.set(Calendar.MONTH, 12);
cal.set(Calendar.DAY_OF_MONTH, 21);
cal.set(Calendar.HOUR, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
final Date endDate = cal.getTime();
System.out.println((endDate.getTime() - startDate.getTime()) % (1000l * 60l * 60l * 24l));
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 );
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
一个稍微简单一点的选择:
System.currentTimeMillis() - oldDate.getTime()
至于“更好”,你到底需要什么?将时间持续时间表示为小时数和天数等的问题是,由于日期的复杂性,它可能导致不准确和错误的期望(例如,由于夏令时,一天可能有23或25小时)。