我在Scala中使用Java的java.util.Date
类,想要比较一个Date
对象和当前时间。我知道可以通过使用getTime()
计算时间差:
(new java.util.Date()).getTime() - oldDate.getTime()
然而,这只给我留下了一个表示毫秒的long
。是否有更简单、更好的方法来获得时间差?
我在Scala中使用Java的java.util.Date
类,想要比较一个Date
对象和当前时间。我知道可以通过使用getTime()
计算时间差:
(new java.util.Date()).getTime() - oldDate.getTime()
然而,这只给我留下了一个表示毫秒的long
。是否有更简单、更好的方法来获得时间差?
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);
}
只需使用下面的方法,使用两个 Date
对象。如果您想传递当前日期,只需将 new Date()
作为第二个参数传递,因为它已初始化为当前时间。
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";
}
}
int year = delta / 365;
int rest = delta % 365;
int month = rest / 30;
rest = rest % 30;
int weeks = rest / 7;
int days = rest % 7;
注:代码完全摘自SO答案。
你可以尝试早期版本的Java。
public static String daysBetween(Date createdDate, Date expiryDate) {
Calendar createdDateCal = Calendar.getInstance();
createdDateCal.clear();
createdDateCal.setTime(createdDate);
Calendar expiryDateCal = Calendar.getInstance();
expiryDateCal.clear();
expiryDateCal.setTime(expiryDate);
long daysBetween = 0;
while (createdDateCal.before(expiryDateCal)) {
createdDateCal.add(Calendar.DAY_OF_MONTH, 1);
daysBetween++;
}
return daysBetween+"";
}
try this:
int epoch = (int) (new java.text.SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse("01/01/1970 00:00:00").getTime() / 1000);
由于您正在使用Scala,因此有一个非常好的Scala库Lamma。 使用Lamma,您可以直接使用-
运算符减去日期。
scala> Date(2015, 5, 5) - 2 // minus days by int
res1: io.lamma.Date = Date(2015,5,3)
scala> Date(2015, 5, 15) - Date(2015, 5, 8) // minus two days => difference between two days
res2: Int = 7
import java.util.*;
int syear = 2000;
int eyear = 2000;
int smonth = 2;//Feb
int emonth = 3;//Mar
int sday = 27;
int eday = 1;
Date startDate = new Date(syear-1900,smonth-1,sday);
Date endDate = new Date(eyear-1900,emonth-1,eday);
int difInDays = (int) ((endDate.getTime() - startDate.getTime())/(1000*60*60*24));
这是另一个示例。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;
}
dtfObj
等变量没有声明或赋值。最后,顺便提一下,我提供这个提示:您可以在第一行左侧使用更通用的“Map”接口,而不是“LinkedHashMap”。这是多态性的一个点/好处之一。 - Basil Bourquepublic 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();
}
}
这将同时给出同一天的日期差异
在Java中有一种简单的方法可以做到这一点。
//创建一个实用方法
public long getDaysBetweenDates(Date d1, Date d2){
return TimeUnit.MILLISECONDS.toDays(d1.getTime() - d2.getTime());
}
该方法将返回两个日期之间的天数。您可以使用默认的Java日期格式,也可以轻松地从任何日期格式进行转换。
@Michael Borgwardt的答案在Android上实际上并不正确。存在舍入误差。例如,从5月19日到21日显示为1天,因为它将1.99转换为1。在强制转换为int之前使用round函数。
修复方法
int diffInDays = (int)Math.round(( (newerDate.getTime() - olderDate.getTime())
/ (1000 * 60 * 60 * 24) ))
Date
、Calendar
和SimpleDateFormat
就太不应该了”。 - Basil Bourquejava.time.Period
和/或Duration
。请查看下面的Basil Bourque的答案。 - Ole V.V.