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
让我来看看Joda Interval和Days之间的差异:
DateTime start = new DateTime(2012, 2, 6, 10, 44, 51, 0);
DateTime end = new DateTime(2012, 2, 6, 11, 39, 47, 1);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
System.out.println(period.getYears() + " years, " + period.getMonths() + " months, " + period.getWeeks() + " weeks, " + period.getDays() + " days");
System.out.println(period.getHours() + " hours, " + period.getMinutes() + " minutes, " + period.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *1 weeks, 1 days*
//0 hours, 54 minutes, 56 seconds
//Period can set PeriodType,such as PeriodType.yearMonthDay(),PeriodType.yearDayTime()...
Period p = new Period(start, end, PeriodType.yearMonthDayTime());
System.out.println(p.getYears() + " years, " + p.getMonths() + " months, " + p.getWeeks() + " weeks, " + p.getDays() + "days");
System.out.println(p.getHours() + " hours, " + p.getMinutes() + " minutes, " + p.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *0 weeks, 8 days*
//0 hours, 54 minutes, 56 seconds
其他回答
我喜欢基于timeunit的方法,直到我发现它只覆盖了一个时间单元在下一个更高单位中有多少个单位是固定的这种微不足道的情况。当你想知道间隔了多少个月、多少年等时,这个问题就不成立了。
这里有一种计数方法,不像其他方法那么有效,但它似乎对我有用,而且还考虑到了夏令时。
public static String getOffsetAsString( Calendar cNow, Calendar cThen) {
Calendar cBefore;
Calendar cAfter;
if ( cNow.getTimeInMillis() < cThen.getTimeInMillis()) {
cBefore = ( Calendar) cNow.clone();
cAfter = cThen;
} else {
cBefore = ( Calendar) cThen.clone();
cAfter = cNow;
}
// compute diff
Map<Integer, Long> diffMap = new HashMap<Integer, Long>();
int[] calFields = { Calendar.YEAR, Calendar.MONTH, Calendar.DAY_OF_MONTH, Calendar.HOUR_OF_DAY, Calendar.MINUTE, Calendar.SECOND, Calendar.MILLISECOND};
for ( int i = 0; i < calFields.length; i++) {
int field = calFields[ i];
long d = computeDist( cAfter, cBefore, field);
diffMap.put( field, d);
}
final String result = String.format( "%dY %02dM %dT %02d:%02d:%02d.%03d",
diffMap.get( Calendar.YEAR), diffMap.get( Calendar.MONTH), diffMap.get( Calendar.DAY_OF_MONTH), diffMap.get( Calendar.HOUR_OF_DAY), diffMap.get( Calendar.MINUTE), diffMap.get( Calendar.SECOND), diffMap.get( Calendar.MILLISECOND));
return result;
}
private static int computeDist( Calendar cAfter, Calendar cBefore, int field) {
cBefore.setLenient( true);
System.out.print( "D " + new Date( cBefore.getTimeInMillis()) + " --- " + new Date( cAfter.getTimeInMillis()) + ": ");
int count = 0;
if ( cAfter.getTimeInMillis() > cBefore.getTimeInMillis()) {
int fVal = cBefore.get( field);
while ( cAfter.getTimeInMillis() >= cBefore.getTimeInMillis()) {
count++;
fVal = cBefore.get( field);
cBefore.set( field, fVal + 1);
System.out.print( count + "/" + ( fVal + 1) + ": " + new Date( cBefore.getTimeInMillis()) + " ] ");
}
int result = count - 1;
cBefore.set( field, fVal);
System.out.println( "" + result + " at: " + field + " cb = " + new Date( cBefore.getTimeInMillis()));
return result;
}
return 0;
}
如果你不想使用JodaTime或类似的,最好的解决方案可能是:
final static long MILLIS_PER_DAY = 24 * 3600 * 1000;
long msDiff= date1.getTime() - date2.getTime();
long daysDiff = Math.round(msDiff / ((double)MILLIS_PER_DAY));
每天的毫秒数并不总是相同的(因为日光节约时间和闰秒),但它非常接近,至少由于日光节约时间的偏差在较长时间内抵消了。因此,除法和舍入将给出正确的结果(至少只要所使用的本地日历不包含DST和闰秒以外的奇怪时间跳转)。
请注意,这仍然假设date1和date2被设置为一天中的同一时间。对于一天中的不同时间,你首先必须定义“日期差异”的含义,正如乔恩·斯基特指出的那样。
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);
}
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;
}
只需在两个Date对象上使用下面的方法。如果你想传递当前日期,只需传递new date()作为第二个参数,因为它是用当前时间初始化的。
public String getDateDiffString(Date dateOne, Date dateTwo)
{
long timeOne = dateOne.getTime();
long timeTwo = dateTwo.getTime();
long oneDay = 1000 * 60 * 60 * 24;
long delta = (timeTwo - timeOne) / oneDay;
if (delta > 0) {
return "dateTwo is " + delta + " days after dateOne";
}
else {
delta *= -1;
return "dateTwo is " + delta + " days before dateOne";
}
}
此外,除了从天数,如果,你也想要其他参数的差异,使用下面的片段,
int year = delta / 365;
int rest = delta % 365;
int month = rest / 30;
rest = rest % 30;
int weeks = rest / 7;
int days = rest % 7;
p.s. Code完全取自SO答案。