我有开始日期和结束日期。两个日期之间的持续时间应为年，月和日。我是java的新手。 当我运行以下方法时，我获得的是0年，12个月1天。 请建议一种替代方案，以获得年，月和日的准确差异。
import java.sql.Date;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.*;
public class Duration {
private String getAssignmentDuration(java.util.Date oldDate, java.util.Date newDate) {
Calendar c1 = Calendar.getInstance();
Calendar c2 = Calendar.getInstance();
if (oldDate.compareTo(newDate) > 0) {
c1.setTime(newDate);
c2.setTime(oldDate);
} else {
System.out.println("invalid");
return "Invalid selection";
}
int year = 0;
int month = 0;
int days = 0;
boolean doneMonth = false;
boolean doneYears = false;
while (c1.before(c2)) {
//log.debug("Still in Loop");
if (!doneYears) {
c1.add(Calendar.YEAR, 1);
year++;
}
if (c1.after(c2) || doneYears) {
if (!doneYears) {
doneYears = true;
year--;
c1.add(Calendar.YEAR, -1);
}
if (!doneMonth) {
c1.add(Calendar.MONTH, 1);
month++;
}
if (c1.after(c2) || doneMonth) {
if (!doneMonth) {
doneMonth = true;
month--;
c1.add(Calendar.MONTH, -1);
}
c1.add(Calendar.DATE, 1);
days++;
if (c1.after(c2)) {
days--;
}
// this will not be executed
if (days == 31 || month==12) {
break;
}
}
}
}
System.out.println(year + " years, " + month + " months, " + days + " days");
return year + " years, " + month + " months, " + days + " days";
}
public static void main(String[] args) {
Duration d1= new Duration();
SimpleDateFormat sdf = new SimpleDateFormat("yyyy/MM/dd");
java.util.Date oldDate = null;
try {
oldDate = sdf.parse("2012/08/29");
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
java.util.Date newDate = null;
try {
newDate = sdf.parse("2013/08/31");
} catch (ParseException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
d1.getAssignmentDuration(oldDate, newDate);
}
}
答案 0 :(得分：10)
public static String getDateDifferenceInDDMMYYYY(Date from, Date to) {
Calendar fromDate=Calendar.getInstance();
Calendar toDate=Calendar.getInstance();
fromDate.setTime(from);
toDate.setTime(to);
int increment = 0;
int year,month,day;
System.out.println(fromDate.getActualMaximum(Calendar.DAY_OF_MONTH));
if (fromDate.get(Calendar.DAY_OF_MONTH) > toDate.get(Calendar.DAY_OF_MONTH)) {
increment =fromDate.getActualMaximum(Calendar.DAY_OF_MONTH);
}
System.out.println("increment"+increment);
// DAY CALCULATION
if (increment != 0) {
day = (toDate.get(Calendar.DAY_OF_MONTH) + increment) - fromDate.get(Calendar.DAY_OF_MONTH);
increment = 1;
} else {
day = toDate.get(Calendar.DAY_OF_MONTH) - fromDate.get(Calendar.DAY_OF_MONTH);
}
// MONTH CALCULATION
if ((fromDate.get(Calendar.MONTH) + increment) > toDate.get(Calendar.MONTH)) {
month = (toDate.get(Calendar.MONTH) + 12) - (fromDate.get(Calendar.MONTH) + increment);
increment = 1;
} else {
month = (toDate.get(Calendar.MONTH)) - (fromDate.get(Calendar.MONTH) + increment);
increment = 0;
}
// YEAR CALCULATION
year = toDate.get(Calendar.YEAR) - (fromDate.get(Calendar.YEAR) + increment);
return year+"\tYears\t\t"+month+"\tMonths\t\t"+day+"\tDays";
}
public static void main(String[] args) {
Calendar calendar = Calendar.getInstance();
calendar.set(1999,01,8);
/* Calendar calendar1 = Calendar.getInstance();
calendar1.set(2012,01,23);*/
System.out.println(getDateDifferenceInDDMMYYYY(calendar.getTime(),new Date()));
}
答案 1 :(得分：4)
Joda Time有一个时间Interval的概念，你可以使用，如：
Interval interval = new Interval(oldDate.getTime(), newDate.getTime());
然后使用Period对象，例如：
Period period = interval.toPeriod().normalizedStandard(PeriodType.yearMonthDay());
PeriodFormatter formatter = new PeriodFormatterBuilder()
.appendYears()
.appendSuffix(" year ", " years ")
.appendSeparator(" and ")
.appendMonths()
.appendSuffix(" month ", " months ")
.appendSeparator(" and ")
.appendDays()
.appendSuffix(" day ", " days ")
.toFormatter();
System.out.println(formatter.print(period));
您可以轻松地在数年和数月内打印您的差异。
可能你在发布问题时改变了一些东西，因为要修复你的代码（注意我没有测试你的代码是否适用于所有类型的范围），你只需要正确初始化Calendar对象并反过来无效的选择检查：
Calendar c1 = Calendar.getInstance();
Calendar c2 = Calendar.getInstance();
if (oldDate.compareTo(newDate) < 0) {
c2.setTime(newDate);
c1.setTime(oldDate);
} else {
System.out.println("invalid");
return "Invalid selection";
}
答案 2 :(得分：4)
假设您已Date date1, date2，并且已将其初始化为date1>date2。
long diff = date1.getTime() - date2.getTime(); //this is going to give you the difference in milliseconds
Date result = new Date(diff);
Format frmt = new SimpleDateFormat("yy MM dd HH:mm:ss");
return frmt.format(result).toString();//or if you want system.out.println(...);
答案 3 :(得分：2)
Period.between(
LocalDate.of( 2017 , Month.JANUARY , 23 ) ,
LocalDate.of( 2017 , Month.MARCH , 27 )
)
呼叫：
.getYears()
.getMonths()
.getDays()
你使用麻烦的旧日期时间类，现在是旧的，取而代之的是java.time类。
LocalDate类表示没有时间且没有时区的仅限日期的值。
时区对于确定日期至关重要。对于任何给定的时刻，日期在全球范围内因地区而异。例如，在Paris France午夜后的几分钟是新的一天，而Montréal Québec中仍然是“昨天”。
以continent/region的格式指定proper time zone name，例如America/Montreal，Africa/Casablanca或Pacific/Auckland。切勿使用诸如EST或IST之类的3-4字母缩写，因为它们不是真正的时区，不是标准化的，甚至不是唯一的（！）。
ZoneId z = ZoneId.of( "America/Montreal" );
LocalDate today = LocalDate.now( z );
today.toString（）：2017-05-05
对于我们的示例，我们创建了另一个LocalDate。
LocalDate earlier = today.minusMonths( 2 ).minusWeeks( 3 ).minusDays( 2 ) ;
early.toString（）：2017-02-10
要以年 - 月 - 日的粒度表示未附加到时间轴的时间跨度，请使用Period类。
Period p = Period.between( earlier , today ) ;
int years = p.getYears();
int months = p.getMonths();
int days = p.getDays();
请参阅this code run live at IdeOne.com。
ISO 8601标准定义了日期时间值的文本表示的格式。对于年 - 月 - 天的持续时间，模式为PnYnMnDTnHnMnS，其中P标记开头，T将年 - 月 - 日部分与小时 - 分 - 秒部分分开。< / p>
在解析/生成字符串时，java.time类默认使用标准格式。 Period类在其toString方法中生成此特定模式。
String output = p.toString() ;
p.toString（）：P2M25D
答案 4 :(得分：1)
long diff = today.getTimeInMillis() - birth.getTimeInMillis();
// Calculate difference in seconds
long Seconds = diff / 1000;
// Calculate difference in minutes
long Minutes = diff / (60 * 1000);
// Calculate difference in hours
long Hours = diff / (60 * 60 * 1000);
// Calculate difference in days
long Days = diff / (24 * 60 * 60 * 1000);
long Months = diff / (24 * 60 * 60 * 12 * 1000);
//lblTsec, lblTmint, lblthours,lblTdays;
System.out.println("Seconds : " + Seconds + "");
System.out.println("Minutes : " + Minutes + "");
System.out.println("Hours : " + Hours + "");
System.out.println("Days : " + Days + "");
答案 5 :(得分：0)
public static long[] differenceBetweenDates(Date fromDate, Date toDate) {
Calendar startDate = Calendar.getInstance();
startDate.setTime(fromDate);
long years = 0;
long months = 0;
long days = 0;
Calendar endDate = Calendar.getInstance();
endDate.setTime(toDate);
Calendar tmpdate = Calendar.getInstance();
tmpdate.setTime(startDate.getTime());
tmpdate.add(Calendar.YEAR, 1);
while (tmpdate.compareTo(endDate) <= 0) {
startDate.add(Calendar.YEAR, 1);
tmpdate.add(Calendar.YEAR, 1);
years++;
}
tmpdate.setTime(startDate.getTime());
tmpdate.add(Calendar.MONTH, 1);
while (tmpdate.compareTo(endDate) <= 0) {
startDate.add(Calendar.MONTH, 1);
tmpdate.add(Calendar.MONTH, 1);
months++;
}
tmpdate.setTime(startDate.getTime());
tmpdate.add(Calendar.DATE, 1);
while (tmpdate.compareTo(endDate) <= 0) {
startDate.add(Calendar.DATE, 1);
tmpdate.add(Calendar.DATE, 1);
days++;
}
return new long[]{days, months, years};
}
答案 6 :(得分：0)
您只需计算两个日期毫秒之间的差，然后除以秒，分钟，小时，天和月即可。
假设您希望在几年之间有所不同，
public int findDiff(Date fromDate, Date toDate) {
if(fromDate == null || toDate == null) {
return -1;
}
long diff = toDate.getTime() - fromDate.getTime();
int diffInYears = (int) (diff / (60 * 60 * 1000 * 24 * 30.41666666 * 12));
return diffInYears;
}
假设您希望月份之间的差额从分频器中删除12（意味着月份）。同样，您可以获得几天，几小时，几分钟。.