I would recommend to use date->diff function, as in example below: $dStart = new DateTime(‘2012-07-26’); $dEnd = new DateTime(‘2012-08-26’); $dDiff = $dStart->diff($dEnd); echo $dDiff->format(‘%r%a’); // use for point out relation: smaller/greater see http://www.php.net/manual/en/datetime.diff.php
if you want to compute differences between two known dates, use total_seconds like this: import datetime as dt a = dt.datetime(2013,12,30,23,59,59) b = dt.datetime(2013,12,31,23,59,59) (b-a).total_seconds() 86400.0 #note that seconds doesn’t give you what you want: (b-a).seconds 0
When you need something like that on server side, you usually create a table which contains all possible dates between two points in time, and then left join this table with query results. Something like this: create procedure sp1(d1 date, d2 date) declare d datetime; create temporary table foo (d date not null); set d … Read more
I suggest to use DateTime and DateInterval objects. $date1 = new DateTime(“2007-03-24”); $date2 = new DateTime(“2009-06-26”); $interval = $date1->diff($date2); echo “difference ” . $interval->y . ” years, ” . $interval->m.” months, “.$interval->d.” days “; // shows the total amount of days (not divided into years, months and days like above) echo “difference ” . $interval->days … Read more
Take a look at the PrettyTime library. It’s quite simple to use: import org.ocpsoft.prettytime.PrettyTime; PrettyTime p = new PrettyTime(); System.out.println(p.format(new Date())); // prints “moments ago” You can also pass in a locale for internationalized messages: PrettyTime p = new PrettyTime(new Locale(“fr”)); System.out.println(p.format(new Date())); // prints “à l’instant” As noted in the comments, Android has this … Read more
The definition of “the number of months in the difference” is subject to a lot of interpretation. 🙂 You can get the year, month, and day of month from a JavaScript date object. Depending on what information you’re looking for, you can use those to figure out how many months are between two points in … Read more
By using the Date object and its milliseconds value, differences can be calculated: var a = new Date(); // Current date now. var b = new Date(2010, 0, 1, 0, 0, 0, 0); // Start of 2010. var d = (b-a); // Difference in milliseconds. You can get the number of seconds (as a integer/whole … Read more
I would suggest you use the excellent Joda Time library instead of the flawed java.util.Date and friends. You could simply write import java.util.Date; import org.joda.time.DateTime; import org.joda.time.Days; Date past = new Date(110, 5, 20); // June 20th, 2010 Date today = new Date(110, 6, 24); // July 24th int days = Days.daysBetween(new DateTime(past), new DateTime(today)).getDays(); … Read more
Here is some T-SQL that gives you the number of years, months, and days since the day specified in @date. It takes into account the fact that DATEDIFF() computes the difference without considering what month or day it is (so the month diff between 8/31 and 9/1 is 1 month) and handles that with a … Read more
How precise do you need to be? If you do need to take into account common years and leap years, and the exact difference in days between months then you’ll have to write something more advanced but for a basic and rough calculation this should do the trick: today = new Date() past = new … Read more