Adapted from my answer here: Get the number of days between the Mondays of both weeks (using TRUNC( datevalue, ‘IW’ ) as an NLS_LANGUAGE independent method of finding the Monday of the week) then add the day of the week (Monday = 1, Tuesday = 2, etc., to a maximum of 5 to ignore weekends) … Read more
Here’s a two line solution: long msDiff = Calendar.getInstance().getTimeInMillis() – testCalendar.getTimeInMillis(); long daysDiff = TimeUnit.MILLISECONDS.toDays(msDiff); In this example it gets the number of days between date “testCalendar” and the current date.
To remove the ‘days’ text element, you can also make use of the dt() accessor for series: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.html So, df[[‘A’,’B’]] = df[[‘A’,’B’]].apply(pd.to_datetime) #if conversion required df[‘C’] = (df[‘B’] – df[‘A’]).dt.days which returns: A B C one 2014-01-01 2014-02-28 58 two 2014-02-03 2014-03-01 26
You have to consider the time difference as well. For example if you compare the dates 2015-01-01 10:00 and 2015-01-02 09:00, days between those dates will return as 0 (zero) since the difference between those dates is less than 24 hours (it’s 23 hours). If your purpose is to get the exact day number between … Read more
If you want logical calendar days, use DAYS.between() method from java.time.temporal.ChronoUnit: LocalDate dateBefore; LocalDate dateAfter; long daysBetween = DAYS.between(dateBefore, dateAfter); If you want literal 24 hour days, (a duration), you can use the Duration class instead: LocalDate today = LocalDate.now() LocalDate yesterday = today.minusDays(1); // Duration oneDay = Duration.between(today, yesterday); // throws an exception Duration.between(today.atStartOfDay(), … Read more
I’m not sure what format you’re looking for in your difference but here’s how to do it using DateTime $datetime1 = new DateTime(); $datetime2 = new DateTime(‘2011-01-03 17:13:00’); $interval = $datetime1->diff($datetime2); $elapsed = $interval->format(‘%y years %m months %a days %h hours %i minutes %s seconds’); echo $elapsed;
This may help 🙂 date1 = “2018-12-25”; //date1 date2 = “2019-01-01”; //date2 var one_day = 1000 * 60 * 60 * 24; var x = date1.split(“-“); var y = date2.split(“-“); var new_date1 = new Date(x[0], (x[1] – 1), x[2]); var new_date2 = new Date(y[0], (y[1] – 1), y[2]); var month1 = x[1] – 1; var … Read more