This question already has answers here:
Timer - counting down and displaying result in specific format [duplicate]
(2 answers)
Closed 6 years ago.
So i am creating an application to countdown to a date in java.
So lets say i parse these 2 days:
Date Now = 2016/01/15 16:52:22
Date End = 2016/01/15 18:37:18
How would i calculate the differences so that i can get the output like so:
1 hour, 44 minutes and 56 seconds
instead of the total hours, total minutes and total seconds.
I guess the best way to explain it is that i need to get the seconds left of the minute, the minutes left of the hour, etc, etc.
You haven't specify your date format. Supposed you use the Date class, the following code gives you a difference in milliseconds.
long difference = dateEnd.getTime() - dateNow.getTime();
Then you approach to your result with a simple calculation:
int seconds = (int) (difference / 1000) % 60 ;
int minutes = (int) ((difference / (1000*60)) % 60);
int hours = (int) ((difference / (1000*60*60)) % 24);
Related
This question already has answers here:
Java 8: Difference between two LocalDateTime in multiple units
(11 answers)
Closed 2 years ago.
I decided to give myself a challenge on Java that implements this question's achievement.
The things I have to do is get LocalDateTime, convert the same code from the linked question's answers, then receiving a string from the function.
Here's what I've done so far:
public static String relTime(LocalDateTime now)
{
// accepted answer converted to Java
const int min = 60 * SECOND;
const int hour = 60 * MINUTE;
const int day = 24 * HOUR;
const int mon = 30 * DAY;
// still don't know how to convert this method
var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
if (delta < 1 * MINUTE)
return ts.Seconds == 1 ? "one second ago" : ts.Seconds + " seconds ago";
if (delta < 2 * MINUTE)
return "a minute ago";
if (delta < 45 * MINUTE)
return ts.Minutes + " minutes ago";
if (delta < 90 * MINUTE)
return "an hour ago";
if (delta < 24 * HOUR)
return ts.Hours + " hours ago";
if (delta < 48 * HOUR)
return "yesterday";
if (delta < 30 * DAY)
return ts.Days + " days ago";
if (delta < 12 * MONTH)
{
int months = Convert.ToInt32(Math.Floor((double)ts.Days / 30));
return months <= 1 ? "one month ago" : months + " months ago";
}
else
{
int years = Convert.ToInt32(Math.Floor((double)ts.Days / 365));
return years <= 1 ? "one year ago" : years + " years ago";
}
}
The only problem that I should encounter is from var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);.
Although I read 2 questions from Stack Overflow finding equivalents of TimeSpan and Ticks, I baely have any ideas how to properly convert the line of code. Also, I have to get a double which will need math.abs() to get TotalSeconds which I can't really find a proper way to deal with either, but I did find ZoneOffset.ofTotalSeconds and still don't know how to deal with it.
So how can I convert this properly?
var ts = new TimeSpan(DateTime.UtcNow.Ticks - yourDate.Ticks);
double delta = Math.Abs(ts.TotalSeconds);
You need to gain a deeper understanding of what this method actually does. Literally translating code from C# to Java won't give you a good solution and gets you stuck on language-specific details.
The two lines basically calculate the (absolute) difference in seconds of a timestamp to the current time. This can be written in Java as follows:
Duration duration = Duration.between(LocalDateTime.now(), timestamp);
long delta = duration.abs().getSeconds();
I'm just addressing your actual question here on how to transform these two lines. The provided snippet is not valid Java code and some parts are missing. delta is the difference in seconds which does not necessarily need to be a double. The argument you pass to your method should be named anything else than now because this is the timestamp you want to compare to the current time inside the method.
You could use SimpleDateFormat to create a nice display format (use something like "HH hours, mm minutes and ss seconds ago" for the format (not sure if this exact example works)). You could also use Instant to get the current time, and you can use Instant.now().minusSeconds(Instant.now().minusSeconds(seconds).getEpochSeconds()) for the time difference (or just use System.currentTimeMillis() and multiply by 1000).
Alternatively, you could use Duration and write a custom display format using getSeconds() and getHours() etc.
This question already has answers here:
TimeDelta java?
(3 answers)
Closed 3 years ago.
I have a duration stored in the database as a Double, but in my operations on LocalTime I need it to be of type Duration.
I need to be able to change from double to duration and vice versa.
I expect a duration of 3,5 hours to be 3 hours and 30 minutes.
// the Double value
Double someTime = 3.5;
// two ways of converting it to Duration
// (we need to use minutes because Duration.of expects a long, which cannot be fractional
Duration someDuration = Duration.ofMinutes((long) (someTime * 60));
Duration someDuration2 = Duration.of((long) (someTime * 60), ChronoUnit.MINUTES);
// convert the duration back to Double
Double someTimeAgain = (double)someDuration.toMinutes() / 60;
I assume your duration is hours as double?
Then you first need to calculate minutes from this.
This could either be
long durationInMinutes = durationFromDB * 60
or, if 0.5 is "half an hour" you need to calculate first the hours and then the minutes and add them... so 3.5h are 180 minutes + 30 minutes = 210 minutes
then you can use java.time.Duration#ofMinutes(long minutes) and get what you need.
I'm having Minutes in java.lang.Long and want to convert this value to java.math.BigDecimal, ie. as Hours.
BigDecimal hours = BigDecimal.valueOf(minutes)
.divide(BigDecimal.valueOf(DateTimeConstants.MINUTES_PER_HOUR))
.setScale(2,RoundingMode.HALF_DOWN);
Tried the above method. It return hours, but no the way actually i want it
How i need is :
240 Minutes : 4 Hours
230 Minutes : 3.50 hours
Any help?
I would convert your Minutes to a Period object:
Minutes minutes = ...;
Long millisec = minutes*60*1000;
Period period = new Period(millisec);
Then use the Period object you can ask the Hours. Anything you want...
Note: 230 minutes is not 3.50 hours, it's 3.83 hours, i'm assuming you mean "3 hours and 50 minutes".
So what you want is the hh:mm representation.
You don't need BigDecimals. Use this:
long minutes = 230;
long hours = minutes / 60;
long minnutesRemaining = minutes % 60;
System.out.println(hours + "." + minnutesRemaining);
I'm betting the OP actually wants to convert minutes into hours and minutes. This is as easy as:
int minutes = 230;
System.out.println(
String.format("%d Minutes: %d:%02d Hours", minutes, (minutes/60), (minutes%60)));
Just printing the minutes divided by 60 (using integer arithmetic) and the modulo of minutes divided by 60 (formatted as two digits with leading zeros by the "%02d" format.
You can do this using BigDecimal easy. You can use divideAndRemainder()
long minutes = 230L;
BigDecimal min = new BigDecimal(minutes);
BigDecimal constant = new BigDecimal(60);
BigDecimal[] val=min.divideAndRemainder(constant);
System.out.println(val[0]+"."+val[1]+" Hours");
Out put:
3.50 Hours
I don't know in what universe 230 minutes equals 3.5 hours, so I'm afraid that some string manipulation is your best bet:
BigDecimal hours = new BigDecimal(
String.format("%d.%d", minutes / 60, minutes % 60));
Printing out the value of hours yields 3.50, as per your requirement.
Use integer and modulo arithmetic:
long hours = minutes / 60; /*implicit round-down*/
long numberAfterDecimal = (minutes % 1.0 /*pull out the remainder*/) * 60;
Then format these two numbers as you wish.
This question already has an answer here:
How to calculate difference between two dates in years...etc with Joda-Time
(1 answer)
Closed 8 years ago.
I have a long-variable which represents an amount of delay in milliseconds. I want to transform this long to some kind of Date where it says how many hours, minutes, seconds, days, months, years have passed.
When using Date toString() from Java, as in new Date(5).toString, it says 5 milliseconds have passed from 1970. I need it to say 5 milliseconds have passed, and 0 minutes, hours, ..., years.
you cannot get direct values , without any reference date for your requirements, you need define first reference value like below:
String dateStart = "01/14/2012 09:29:58";
SimpleDateFormat format = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss")
Date d1 = format.parse(dateStart);
the above is your reference date , now you need to find the current date and time using following.
long currentDateTime = System.currentTimeMillis();
Date currentDate = new Date(currentDateTime);
Date d2.format(currentDate)
and the difference of these values like long diff=d2-d1 will gives values in milliseconds.
then
long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000) % 24;
long diffDays = diff / (24 * 60 * 60 * 1000);
and similarly for months and years.
you can also refer the example given on this link for more information http://javarevisited.blogspot.in/2012/12/how-to-convert-millisecond-to-date-in-java-example.html
From what I understand from your question you could achieve your goal by writing a method that will suit your needs i.e.:
static public String dateFromMili (long miliseconds) {
// constants that will hold the number of miliseconds
// in a given time unit (year, month etc.)
final int YEAR_IN_MILISECONDS = 12*30*24*60*60*1000;
final int MONTH_IN_MILISECONDS = 30*24*60*60*1000;
final int DAY_IN_MILISECONDS = 24*60*60*1000;
final int HOUR_IN_MILISECONDS = 60*60*1000;
final int MINUTE_IN_MILISECONDS = 60*1000;
final int SECONDS_IN_MILISECONDS = 1000;
// now use those constants to return an appropriate string.
return miliseconds +" miliseconds, "
+miliseconds/SECONDS_IN_MILISECONDS+" seconds, "
+miliseconds/MINUTE_IN_MILISECONDS+" minutes, "
+miliseconds/HOUR_IN_MILISECONDS+" hours, "
+miliseconds/DAY_IN_MILISECONDS+" days, "
+miliseconds/MONTH_IN_MILISECONDS+" months, "
+miliseconds/YEAR_IN_MILISECONDS+" years have passed";
}
Than you will have to pas the number of miliseconds as a parameter to your new function that will return the desired String (i.e for two seconds):
dateFromMili (2000);
You could also print your answer:
System.out.println(dateFromMili(2000));
The result would look like this:
2000 miliseconds, 2 seconds, 0 minutes, 0 hours, 0 days, 0 months, 0 years have passed
Note that this method will return Strings with integer value (you will not get for example "2.222333 years" but "2 years"). Furthermore, it could be perfected by changing the noun from plural to singular, when the context is appropriate ("months" to "month").
I hope my answer helped.
This is how I solved the problem:
I used a library called Joda-Time (http://www.joda.org/joda-time/) (credits to Keppil!)
Joda-Time has various data-structures for Date and Time. You can represent a date and time by a DateTime-object.
To represent the delay I was looking for, I had two options: a Period data-structure or a Duration data-structure. A good explanation of the difference between those two can be found here: Joda-Time: what's the difference between Period, Interval and Duration? .
I thus used a Duration-object, based on the current date of my DateTime-object. It has all the methods to convert the amount of milliseconds to years, months, weeks, days, hours, minutes and seconds.
i did some research, but still can't find how to get the days... Here is what I got:
int seconds = (int) (milliseconds / 1000) % 60 ;
int minutes = (int) ((milliseconds / (1000*60)) % 60);
int hours = (int) ((milliseconds / (1000*60*60)) % 24);
int days = ????? ;
Please help, I suck at math, thank's.
For simple cases like this, TimeUnit should be used. TimeUnit usage is a bit more explicit about what is being represented and is also much easier to read and write when compared to doing all of the arithmetic calculations explicitly. For example, to calculate the number days from milliseconds, the following statement would work:
long days = TimeUnit.MILLISECONDS.toDays(milliseconds);
For cases more advanced, where more finely grained durations need to be represented in the context of working with time, an all encompassing and modern date/time API should be used. For JDK8+, java.time is now included (here are the tutorials and javadocs). For earlier versions of Java joda-time is a solid alternative.
If you don't have another time interval bigger than days:
int days = (int) (milliseconds / (1000*60*60*24));
If you have weeks too:
int days = (int) ((milliseconds / (1000*60*60*24)) % 7);
int weeks = (int) (milliseconds / (1000*60*60*24*7));
It's probably best to avoid using months and years if possible, as they don't have a well-defined fixed length. Strictly speaking neither do days: daylight saving means that days can have a length that is not 24 hours.
Go for TImeUnit in java
In order to import use, java.util.concurrent.TimeUnit
long millisec=System.currentTimeMillis();
long seconds=TimeUnit.MILLISECONDS.toSeconds(millisec);
long minutes=TimeUnit.MILLISECONDS.toMinutes(millisec);
long hours=TimeUnit.MILLISECONDS.toMinutes(millisec);
long days=TimeUnit.MILLISECONDS.toDays(millisec);
java.time
You can use java.time.Duration which is modelled on ISO-8601 standards and was introduced with Java-8 as part of JSR-310 implementation. With Java-9 some more convenient methods were introduced.
Demo:
import java.time.Duration;
public class Main {
public static void main(String[] args) {
// Duration between the two instants
Duration duration = Duration.ofMillis(1234567890L);
// Print Duration#toString
System.out.println(duration);
// Custom format
// ####################################Java-8####################################
String formattedElapsedTime = String.format(
"%d Day %02d Hour %02d Minute %02d Second %d Millisecond (%d Nanosecond)", duration.toDays(),
duration.toHours() % 24, duration.toMinutes() % 60, duration.toSeconds() % 60,
duration.toMillis() % 1000, duration.toNanos() % 1000000000L);
System.out.println(formattedElapsedTime);
// ##############################################################################
// ####################################Java-9####################################
formattedElapsedTime = String.format("%d Day %02d Hour %02d Minute %02d Second %d Millisecond (%d Nanosecond)",
duration.toDaysPart(), duration.toHoursPart(), duration.toMinutesPart(), duration.toSecondsPart(),
duration.toMillisPart(), duration.toNanosPart());
System.out.println(formattedElapsedTime);
// ##############################################################################
}
}
A sample run:
PT342H56M7.89S
14 Day 06 Hour 56 Minute 07 Second 890 Millisecond (890000000 Nanosecond)
14 Day 06 Hour 56 Minute 07 Second 890 Millisecond (890000000 Nanosecond)
Learn more about the modern date-time API from Trail: Date Time.
int days = (int) (milliseconds / 86 400 000 )
public static final long SECOND_IN_MILLIS = 1000;
public static final long MINUTE_IN_MILLIS = SECOND_IN_MILLIS * 60;
public static final long HOUR_IN_MILLIS = MINUTE_IN_MILLIS * 60;
public static final long DAY_IN_MILLIS = HOUR_IN_MILLIS * 24;
public static final long WEEK_IN_MILLIS = DAY_IN_MILLIS * 7;
You could cast int but I would recommend using long.
You can’t. Sorry. Or more precisely: you can if you know a time zone and a start time (or end time). A day may have a length of 23, 24 or 25 hours or some other length. So there isn’t any sure-fire formula for converting from milliseconds to days. So while you can safely rely on 1000 milliseconds in a second, 60 seconds in a minute (reservation below) and 60 minutes in an hour, the conversion to days needs more context in order to be sure and accurate.
Reservation: In real life a minute is occasionally 61 seconds because of a leap second. Not in Java. Java always counts a minute as 60 seconds because common computer clocks don’t know leap seconds. Common operating systems and Java itself do know not only summer time (DST) but also many other timeline anomalies that cause a day to be shorter or longer than 24 hours.
To demonstrate. I am writing this on March 29, 2021, the day after my time zone, Europe/Copenhagen, and the rest of the EU switched to summer time.
ZoneId myTimeZone = ZoneId.of("Europe/Copenhagen");
ZonedDateTime now = ZonedDateTime.now(myTimeZone);
ZonedDateTime twoDaysAgo = now.minusDays(2);
ZonedDateTime inTwoDays = now.plusDays(2);
System.out.println(ChronoUnit.MILLIS.between(twoDaysAgo, now));
System.out.println(ChronoUnit.MILLIS.between(now, inTwoDays));
Output:
169200000
172800000
So how many milliseconds are in two days depends on which two days you mean. And in which time zone.
So what to do?
If for your purpose you can safely define a day as 24 hours always, for example because your days are counted in UTC or your users are fine with the inaccuracy, use either Duration or TimeUnit. Since Java 9 the Duration class will additionally tell you how many hours, minutes and seconds there are in addition to the whole days. See the answer by Arvind Kumar Avinash. For the TimeUnit enum see the answers by whaley and Dev Parzival. In any case the good news is that it doesn’t matter if you suck at math because the math is taken care of for you.
If you know a time zone and a starting point, use ZonedDateTime and ChronoUnit.DAYS. In this case too the math is taken care of for you.
ZonedDateTime start = LocalDate.of(2021, Month.MARCH, 28).atStartOfDay(myTimeZone);
long millisToConvert = 170_000_000;
ZonedDateTime end = start.plus(millisToConvert, ChronoUnit.MILLIS);
long days = ChronoUnit.DAYS.between(start, end);
System.out.format("%d days%n", days);
2 days
If you additionally want the hours, minutes and seconds:
Duration remainingTime = Duration.between(start.plusDays(days), end);
System.out.format(" - and an additional %s hours %d minutes %d seconds%n",
remainingTime.toHours(),
remainingTime.toMinutesPart(),
remainingTime.toSecondsPart());
- and an additional 0 hours 13 minutes 20 seconds
If instead you had got an endpoint, subtract your milliseconds from the endpoint using the minus method (instead of the plus method used in the above code) to get the start point.
Under no circumstances do the math yourself as in the question and in the currently accepted answer. It’s error-prone and results in code that is hard to read. And if your reader sucks at math, he or she can spend much precious developer time trying to verify that you have done it correctly. Leave the math to proven library methods, and it will be much easier for your reader to trust that your code is correct.
In case you solve a more complex task of logging execution statistics in your code:
public void logExecutionMillis(LocalDateTime start, String callerMethodName) {
LocalDateTime end = getNow();
long difference = Duration.between(start, end).toMillis();
Logger logger = LoggerFactory.getLogger(ProfilerInterceptor.class);
long millisInDay = 1000 * 60 * 60 * 24;
long millisInHour = 1000 * 60 * 60;
long millisInMinute = 1000 * 60;
long millisInSecond = 1000;
long days = difference / millisInDay;
long daysDivisionResidueMillis = difference - days * millisInDay;
long hours = daysDivisionResidueMillis / millisInHour;
long hoursDivisionResidueMillis = daysDivisionResidueMillis - hours * millisInHour;
long minutes = hoursDivisionResidueMillis / millisInMinute;
long minutesDivisionResidueMillis = hoursDivisionResidueMillis - minutes * millisInMinute;
long seconds = minutesDivisionResidueMillis / millisInSecond;
long secondsDivisionResidueMillis = minutesDivisionResidueMillis - seconds * millisInSecond;
logger.info(
"\n************************************************************************\n"
+ callerMethodName
+ "() - "
+ difference
+ " millis ("
+ days
+ " d. "
+ hours
+ " h. "
+ minutes
+ " min. "
+ seconds
+ " sec."
+ secondsDivisionResidueMillis
+ " millis).");
}
P.S. Logger can be replaced with simple System.out.println() if you like.