How to find the duration of difference between two dates in java?
JavaDate ArithmeticJava Problem Overview
I have two objects of DateTime, which need to find the duration of their difference,
I have the following code but not sure how to continue it to get to the expected results as following:
Example:
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 minutes
Code:
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.");
Java Solutions
Solution 1 - Java
The date difference conversion could be handled in a better way using Java built-in class, TimeUnit. It provides utility methods to do that:
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);
Solution 2 - Java
try the following
{
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;
}
Solution 3 - Java
Use Joda-Time library
DateTime startTime, endTime;
Period p = new Period(startTime, endTime);
long hours = p.getHours();
long minutes = p.getMinutes();
Joda Time has a concept of time Interval:
Interval interval = new Interval(oldTime, new Instant());
One more example Date Difference
One more Link
or with Java-8 (which integrated Joda-Time concepts)
Instant start, end;//
Duration dur = Duration.between(start, stop);
long hours = dur.toHours();
long minutes = dur.toMinutes();
Solution 4 - Java
Here is how the problem can solved in Java 8 just like the answer by shamimz.
Source : 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)");
The code produces output similar to the following:
You are 53 years, 4 months, and 29 days old. (19508 days total)
We have to use LocalDateTime http://docs.oracle.com/javase/8/docs/api/java/time/LocalDateTime.html to get hour,minute,second differences.
Solution 5 - Java
You can create a method like
public long getDaysBetweenDates(Date d1, Date d2){
return TimeUnit.MILLISECONDS.toDays(d1.getTime() - d2.getTime());
}
This method will return the number of days between the 2 days.
Solution 6 - Java
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");
Solution 7 - Java
As Michael Borgwardt writes in his answer here:
> int diffInDays = (int)( (newerDate.getTime() - olderDate.getTime()) > / (1000 * 60 * 60 * 24) ) > > Note that this works with UTC dates, so the difference may be a day > off if you look at local dates. And getting it to work correctly with > local dates requires a completely different approach due to daylight > savings time.
Solution 8 - Java
It worked for me can try with this, hope it will be helpful . Let me know if any concern .
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 .**
Solution 9 - Java
In Java 8, you can make of DateTimeFormatter, Duration, and LocalDateTime. Here is an example:
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));
Solution 10 - Java
This is the code:
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);
and the out is:
duration: 1 day 11 hour 15 min 9 sec
Solution 11 - Java
I solved the similar problem using a simple method recently.
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();
}
And as expected it prints: 7 days ago.
Solution 12 - Java
with reference to shamim's answer update here is a method that does the task without using any third party library. Just copy the method and use
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 :)
}
Solution 13 - Java
java.time.Duration
I still didn’t feel any of the answers was quite up to date and to the point. So here is the modern answer using Duration from java.time, the modern Java date and time API (the answers by MayurB and mkobit mention the same class, but none of them correctly converts to days, hours, minutes and minutes as asked).
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();
}
Output from this example snippet is:
> 3 minutes 45 seconds
Note that Duration always counts a day as 24 hours. If you want to treat time anomalies like summer time transistions differently, solutions inlcude (1) use ChronoUnit.DAYS (2) Use Period (3) Use LocalDateTimeinstead ofZonedDateTime` (may be considered a hack).
The code above works with Java 8 and with ThreeTen Backport, that backport of java.time to Java 6 and 7. From Java 9 it may be possible to write it a bit more nicely using the methods toHoursPart, toMinutesPart and toSecondsPart added there.
I will elaborate the explanations further one of the days when I get time, maybe not until next week.
Solution 14 - Java
This is a program I wrote, which gets the number of days between 2 dates(no time here).
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;
}
}
Solution 15 - Java
You can get the difference between two DateTime using this
DateTime startDate = DateTime.now();
DateTime endDate = DateTime.now();
Days daysBetween = Days.daysBetween(startDate, endDate);
System.out.println(daysBetween.toStandardSeconds());
Solution 16 - Java
The below code will give the difference between two DateTime (Will work in Java 8 and above)
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;
}
Solution 17 - Java
If anyone wants a string with all of them together, this function can be used.
String getTimeDifference(long duration) {
StringBuilder timeRemaining = new StringBuilder();
long days = TimeUnit.MILLISECONDS.toDays(duration);
if (days >= 1) {
timeRemaining.append(days).append((days == 1) ? " day " : " days ");
}
duration -= TimeUnit.DAYS.toMillis(days);
long hours = TimeUnit.MILLISECONDS.toHours(duration);
if (hours >= 1) {
timeRemaining.append(hours).append((hours == 1) ? " hour " : " hours ");
}
duration -= TimeUnit.HOURS.toMillis(hours);
long minutes = TimeUnit.MILLISECONDS.toMinutes(duration);
if (minutes >= 1) {
timeRemaining.append(minutes).append((hours == 1) ? " minute " : " minutes ");
}
return timeRemaining.toString().trim();
}
Solution 18 - Java
// calculating the difference b/w startDate and endDate
String startDate = "01-01-2016";
String endDate = simpleDateFormat.format(currentDate);
date1 = simpleDateFormat.parse(startDate);
date2 = simpleDateFormat.parse(endDate);
long getDiff = date2.getTime() - date1.getTime();
// using TimeUnit class from java.util.concurrent package
long getDaysDiff = TimeUnit.MILLISECONDS.toDays(getDiff);