我有两个 DateTime 对象，需要找到差异的持续时间，
我有以下代码但不知道如何继续它以获得预期结果如下：
实施例
11/03/14 09:30:58
11/03/14 09:33:43
elapsed time is 02 minutes and 45 seconds
-----------------------------------------------------
11/03/14 09:30:58
11/03/15 09:30:58
elapsed time is a day
-----------------------------------------------------
11/03/14 09:30:58
11/03/16 09:30:58
elapsed time is two days
-----------------------------------------------------
11/03/14 09:30:58
11/03/16 09:35:58
elapsed time is two days and 05 mintues
代码
String dateStart = "11/03/14 09:29:58";
String dateStop = "11/03/14 09:33:43";
Custom date format
SimpleDateFormat format = new SimpleDateFormat("yy/MM/dd HH:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = format.parse(dateStart);
d2 = format.parse(dateStop);
} catch (ParseException e) {
e.printStackTrace();
}
// Get msec from each, and subtract.
long diff = d2.getTime() - d1.getTime();
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000);
System.out.println("Time in seconds: " + diffSeconds + " seconds.");
System.out.println("Time in minutes: " + diffMinutes + " minutes.");
System.out.println("Time in hours: " + diffHours + " hours.");
答案 0 :(得分：165)
使用Java内置类TimeUnit可以更好地处理日期差异转换。它提供了实用方法：
Date startDate = // Set start date
Date endDate = // Set end date
long duration = endDate.getTime() - startDate.getTime();
long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);
答案 1 :(得分：64)
尝试以下
{
Date dt2 = new DateAndTime().getCurrentDateTime();
long diff = dt2.getTime() - dt1.getTime();
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000);
int diffInDays = (int) ((dt2.getTime() - dt1.getTime()) / (1000 * 60 * 60 * 24));
if (diffInDays > 1) {
System.err.println("Difference in number of days (2) : " + diffInDays);
return false;
} else if (diffHours > 24) {
System.err.println(">24");
return false;
} else if ((diffHours == 24) && (diffMinutes >= 1)) {
System.err.println("minutes");
return false;
}
return true;
}
答案 2 :(得分：44)
使用Joda-Time库
DateTime startTime, endTime;
Period p = new Period(startTime, endTime);
long hours = p.getHours();
long minutes = p.getMinutes();
Joda Time有一个时间间隔的概念：
Interval interval = new Interval(oldTime, new Instant());
又一个例子 Date Difference
又一个Link
或Java-8（集成了Joda-Time概念）
Instant start, end;//
Duration dur = Duration.between(start, stop);
long hours = dur.toHours();
long minutes = dur.toMinutes();
答案 3 :(得分：12)
以下是Java 8中问题的解决方法，就像shamimz的回答一样。
来源：http://docs.oracle.com/javase/tutorial/datetime/iso/period.html
LocalDate today = LocalDate.now();
LocalDate birthday = LocalDate.of(1960, Month.JANUARY, 1);
Period p = Period.between(birthday, today);
long p2 = ChronoUnit.DAYS.between(birthday, today);
System.out.println("You are " + p.getYears() + " years, " + p.getMonths() + " months, and " + p.getDays() + " days old. (" + p2 + " days total)");
代码生成类似于以下内容的输出：
You are 53 years, 4 months, and 29 days old. (19508 days total)
我们必须使用LocalDateTime http://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html来获得小时，分钟和秒的差异。
答案 4 :(得分：6)
Date d2 = new Date();
Date d1 = new Date(1384831803875l);
long diff = d2.getTime() - d1.getTime();
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000);
int diffInDays = (int) diff / (1000 * 60 * 60 * 24);
System.out.println(diffInDays+" days");
System.out.println(diffHours+" Hour");
System.out.println(diffMinutes+" min");
System.out.println(diffSeconds+" sec");
答案 5 :(得分：6)
您可以创建类似
的方法public long getDaysBetweenDates(Date d1, Date d2){
return TimeUnit.MILLISECONDS.toDays(d1.getTime() - d2.getTime());
}
此方法将返回2天之间的天数。
答案 6 :(得分：5)
Michael Borgwardt在{{3}}写道：
int diffInDays = (int)( (newerDate.getTime() - olderDate.getTime()) / (1000 * 60 * 60 * 24) )请注意，这适用于UTC日期，因此差异可能是一天 如果你看看当地的日期。让它正常工作 由于日光，当地日期需要完全不同的方法 节省时间。
答案 7 :(得分：3)
在Java 8中，您可以制作DateTimeFormatter，Duration和LocalDateTime。这是一个例子：
final String dateStart = "11/03/14 09:29:58";
final String dateStop = "11/03/14 09:33:43";
final DateTimeFormatter formatter = new DateTimeFormatterBuilder()
.appendValue(ChronoField.MONTH_OF_YEAR, 2)
.appendLiteral('/')
.appendValue(ChronoField.DAY_OF_MONTH, 2)
.appendLiteral('/')
.appendValueReduced(ChronoField.YEAR, 2, 2, 2000)
.appendLiteral(' ')
.appendValue(ChronoField.HOUR_OF_DAY, 2)
.appendLiteral(':')
.appendValue(ChronoField.MINUTE_OF_HOUR, 2)
.appendLiteral(':')
.appendValue(ChronoField.SECOND_OF_MINUTE, 2)
.toFormatter();
final LocalDateTime start = LocalDateTime.parse(dateStart, formatter);
final LocalDateTime stop = LocalDateTime.parse(dateStop, formatter);
final Duration between = Duration.between(start, stop);
System.out.println(start);
System.out.println(stop);
System.out.println(formatter.format(start));
System.out.println(formatter.format(stop));
System.out.println(between);
System.out.println(between.get(ChronoUnit.SECONDS));
答案 8 :(得分：1)
这是代码：
String date1 = "07/15/2013";
String time1 = "11:00:01";
String date2 = "07/16/2013";
String time2 = "22:15:10";
String format = "MM/dd/yyyy HH:mm:ss";
SimpleDateFormat sdf = new SimpleDateFormat(format);
Date fromDate = sdf.parse(date1 + " " + time1);
Date toDate = sdf.parse(date2 + " " + time2);
long diff = toDate.getTime() - fromDate.getTime();
String dateFormat="duration: ";
int diffDays = (int) (diff / (24 * 60 * 60 * 1000));
if(diffDays>0){
dateFormat+=diffDays+" day ";
}
diff -= diffDays * (24 * 60 * 60 * 1000);
int diffhours = (int) (diff / (60 * 60 * 1000));
if(diffhours>0){
dateFormat+=diffhours+" hour ";
}
diff -= diffhours * (60 * 60 * 1000);
int diffmin = (int) (diff / (60 * 1000));
if(diffmin>0){
dateFormat+=diffmin+" min ";
}
diff -= diffmin * (60 * 1000);
int diffsec = (int) (diff / (1000));
if(diffsec>0){
dateFormat+=diffsec+" sec";
}
System.out.println(dateFormat);
出来了：
duration: 1 day 11 hour 15 min 9 sec
答案 9 :(得分：1)
它对我有用，可以尝试一下，希望对您有所帮助。让我知道是否有任何担忧。
Date startDate = java.util.Calendar.getInstance().getTime(); //set your start time
Date endDate = java.util.Calendar.getInstance().getTime(); // set your end time
long duration = endDate.getTime() - startDate.getTime();
long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);
Toast.makeText(MainActivity.this, "Diff"
+ duration + diffInDays + diffInHours + diffInMinutes + diffInSeconds, Toast.LENGTH_SHORT).show(); **// Toast message for android .**
System.out.println("Diff" + duration + diffInDays + diffInHours + diffInMinutes + diffInSeconds); **// Print console message for Java .**
答案 10 :(得分：1)
参考shamim的答案更新是一种无需使用任何第三方库即可执行任务的方法。只需复制方法并使用
public static String getDurationTimeStamp(String date) {
String timeDifference = "";
//date formatter as per the coder need
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
//parse the string date-ti
// me to Date object
Date startDate = null;
try {
startDate = sdf.parse(date);
} catch (ParseException e) {
e.printStackTrace();
}
//end date will be the current system time to calculate the lapse time difference
//if needed, coder can add end date to whatever date
Date endDate = new Date();
System.out.println(startDate);
System.out.println(endDate);
//get the time difference in milliseconds
long duration = endDate.getTime() - startDate.getTime();
//now we calculate the differences in different time units
//this long value will be the total time difference in each unit
//i.e; total difference in seconds, total difference in minutes etc...
long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);
//now we create the time stamps depending on the value of each unit that we get
//as we do not have the unit in years,
//we will see if the days difference is more that 365 days, as 365 days = 1 year
if (diffInDays > 365) {
//we get the year in integer not in float
//ex- 791/365 = 2.167 in float but it will be 2 years in int
int year = (int) (diffInDays / 365);
timeDifference = year + " years ago";
System.out.println(year + " years ago");
}
//if days are not enough to create year then get the days
else if (diffInDays > 1) {
timeDifference = diffInDays + " days ago";
System.out.println(diffInDays + " days ago");
}
//if days value<1 then get the hours
else if (diffInHours > 1) {
timeDifference = diffInHours + " hours ago";
System.out.println(diffInHours + " hours ago");
}
//if hours value<1 then get the minutes
else if (diffInMinutes > 1) {
timeDifference = diffInMinutes + " minutes ago";
System.out.println(diffInMinutes + " minutes ago");
}
//if minutes value<1 then get the seconds
else if (diffInSeconds > 1) {
timeDifference = diffInSeconds + " seconds ago";
System.out.println(diffInSeconds + " seconds ago");
}
return timeDifference;
// that's all. Happy Coding :)
}
答案 11 :(得分：0)
我最近使用一种简单的方法解决了类似的问题。
public static void main(String[] args) throws IOException, ParseException {
TimeZone utc = TimeZone.getTimeZone("UTC");
Calendar calendar = Calendar.getInstance(utc);
Date until = calendar.getTime();
calendar.add(Calendar.DAY_OF_MONTH, -7);
Date since = calendar.getTime();
long durationInSeconds = TimeUnit.MILLISECONDS.toSeconds(until.getTime() - since.getTime());
long SECONDS_IN_A_MINUTE = 60;
long MINUTES_IN_AN_HOUR = 60;
long HOURS_IN_A_DAY = 24;
long DAYS_IN_A_MONTH = 30;
long MONTHS_IN_A_YEAR = 12;
long sec = (durationInSeconds >= SECONDS_IN_A_MINUTE) ? durationInSeconds % SECONDS_IN_A_MINUTE : durationInSeconds;
long min = (durationInSeconds /= SECONDS_IN_A_MINUTE) >= MINUTES_IN_AN_HOUR ? durationInSeconds%MINUTES_IN_AN_HOUR : durationInSeconds;
long hrs = (durationInSeconds /= MINUTES_IN_AN_HOUR) >= HOURS_IN_A_DAY ? durationInSeconds % HOURS_IN_A_DAY : durationInSeconds;
long days = (durationInSeconds /= HOURS_IN_A_DAY) >= DAYS_IN_A_MONTH ? durationInSeconds % DAYS_IN_A_MONTH : durationInSeconds;
long months = (durationInSeconds /=DAYS_IN_A_MONTH) >= MONTHS_IN_A_YEAR ? durationInSeconds % MONTHS_IN_A_YEAR : durationInSeconds;
long years = (durationInSeconds /= MONTHS_IN_A_YEAR);
String duration = getDuration(sec,min,hrs,days,months,years);
System.out.println(duration);
}
private static String getDuration(long secs, long mins, long hrs, long days, long months, long years) {
StringBuffer sb = new StringBuffer();
String EMPTY_STRING = "";
sb.append(years > 0 ? years + (years > 1 ? " years " : " year "): EMPTY_STRING);
sb.append(months > 0 ? months + (months > 1 ? " months " : " month "): EMPTY_STRING);
sb.append(days > 0 ? days + (days > 1 ? " days " : " day "): EMPTY_STRING);
sb.append(hrs > 0 ? hrs + (hrs > 1 ? " hours " : " hour "): EMPTY_STRING);
sb.append(mins > 0 ? mins + (mins > 1 ? " mins " : " min "): EMPTY_STRING);
sb.append(secs > 0 ? secs + (secs > 1 ? " secs " : " secs "): EMPTY_STRING);
sb.append("ago");
return sb.toString();
}
按照预期打印：7 days ago。
答案 12 :(得分：0)
这是我写的一个程序，它获取两个日期之间的天数（此处没有时间）。
import java.util.Scanner;
public class HelloWorld {
public static void main(String args[]) {
Scanner s = new Scanner(System.in);
System.out.print("Enter starting date separated by dots: ");
String inp1 = s.nextLine();
System.out.print("Enter ending date separated by dots: ");
String inp2 = s.nextLine();
int[] nodim = {
0,
31,
28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31
};
String[] inpArr1 = split(inp1);
String[] inpArr2 = split(inp2);
int d1 = Integer.parseInt(inpArr1[0]);
int m1 = Integer.parseInt(inpArr1[1]);
int y1 = Integer.parseInt(inpArr1[2]);
int d2 = Integer.parseInt(inpArr2[0]);
int m2 = Integer.parseInt(inpArr2[1]);
int y2 = Integer.parseInt(inpArr2[2]);
if (y1 % 4 == 0) nodim[2] = 29;
int diff = m1 == m2 && y1 == y2 ? d2 - (d1 - 1) : (nodim[m1] - (d1 - 1));
int mm1 = m1 + 1, mm2 = m2 - 1, yy1 = y1, yy2 = y2;
for (; yy1 <= yy2; yy1++, mm1 = 1) {
mm2 = yy1 == yy2 ? (m2 - 1) : 12;
if (yy1 % 4 == 0) nodim[2] = 29;
else nodim[2] = 28;
if (mm2 == 0) {
mm2 = 12;
yy2 = yy2 - 1;
}
for (; mm1 <= mm2 && yy1 <= yy2; mm1++) diff = diff + nodim[mm1];
}
System.out.print("No. of days from " + inp1 + " to " + inp2 + " is " + diff);
}
public static String[] split(String s) {
String[] retval = {
"",
"",
""
};
s = s + ".";
s = s + " ";
for (int i = 0; i <= 2; i++) {
retval[i] = s.substring(0, s.indexOf("."));
s = s.substring((s.indexOf(".") + 1), s.length());
}
return retval;
}
}
答案 13 :(得分：0)
我仍然没有任何答案是最新的。因此，这是使用来自Java.time的Duration，现代Java日期和时间API的现代答案（MayurB和mkobit的答案提到了相同的类，但没有一个可以正确转换为天，小时，分钟和分钟。按要求）。
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yy/MM/dd HH:mm:ss");
String dateStart = "11/03/14 09:29:58";
String dateStop = "11/03/14 09:33:43";
ZoneId zone = ZoneId.systemDefault();
ZonedDateTime startDateTime = LocalDateTime.parse(dateStart, formatter).atZone(zone);
ZonedDateTime endDateTime = LocalDateTime.parse(dateStop, formatter).atZone(zone);
Duration diff = Duration.between(startDateTime, endDateTime);
if (diff.isZero()) {
System.out.println("0 minutes");
} else {
long days = diff.toDays();
if (days != 0) {
System.out.print("" + days + " days ");
diff = diff.minusDays(days);
}
long hours = diff.toHours();
if (hours != 0) {
System.out.print("" + hours + " hours ");
diff = diff.minusHours(hours);
}
long minutes = diff.toMinutes();
if (minutes != 0) {
System.out.print("" + minutes + " minutes ");
diff = diff.minusMinutes(minutes);
}
long seconds = diff.getSeconds();
if (seconds != 0) {
System.out.print("" + seconds + " seconds ");
}
System.out.println();
}
此示例代码段的输出为：
3分45秒
请注意，Duration始终将一天视为24小时。如果您想以不同的方式处理夏季异常等时间异常，则解决方案包括（1）使用ChronoUnit.DAYS（2）使用Period（3）Use LocalDateTime instead of ZonedDateTime`（可能会被视为黑客。）
上面的代码与Java 8和ThreeTen Backport一起使用，即从java.time向Java 6和7的反向移植。从Java 9可以使用方法toHoursPart更好地编写它， toMinutesPart和toSecondsPart添加在那里。
我会在有时间的时候再详细解释一下，也许要等到下周。
答案 14 :(得分：0)
您可以使用此功能获取两个DateTime之间的差异
DateTime startDate = DateTime.now();
DateTime endDate = DateTime.now();
Days daysBetween = Days.daysBetween(startDate, endDate);
System.out.println(daysBetween.toStandardSeconds());
答案 15 :(得分：0)
下面的代码将给出两个 DateTime 之间的差异（适用于 Java 8 及更高版本）
private long countDaysBetween(LocalDateTime startDate, LocalDateTime enddate)
{
if(startDate == null || enddate == null)
{
throw new IllegalArgumentException("No such a date");
}
long daysBetween = ChronoUnit.DAYS.between(startDate, enddate);
return daysBetween;
}
答案 16 :(得分：-2)
function canIWatch(age){
if (age <= 0) {
return "Invalid age.";
}
else if(age < 6) {
return "You are not allowed to watch Deadpool after 6.00pm.";
}
else if(age < 17) {
return "You must be accompanied by a guardian who is 21 or older.";
}
else if(age < 25) {
return "You are allowed to watch Deadpool, right after you show some ID.";
}
else {
return "Yay! You can watch Deadpool with no strings attached!";
}
}