Maruf scribbled this post.
Here’s how you add and subtract dates from one another. For example, this simple code can calculate what the date will be 2 weeks before or after 1998-08-14 (yyyy-mm-dd).
Subtracting days from a date
The following example will subtract 3 days from 1998-08-14. The result will be 1998-08-11.
1 2 3 4 5 | $date = "1998-08-14"; $newdate = strtotime ( '-3 day' , strtotime ( $date ) ) ; $newdate = date ( 'Y-m-j' , $newdate ); echo $newdate; |
Subtracting Weeks from a date
The following example will subtract 3 weeks from 1998-08-14. The result will be 1998-07-24. Notice that the only difference in the code is the week statement.
1 2 3 4 5 | $date = "1998-08-14"; $newdate = strtotime ( '-3 week' , strtotime ( $date ) ) ; $newdate = date ( 'Y-m-j' , $newdate ); echo $newdate; |
Subtracting Months from a date
The following example will subtract 3 months from 1998-08-14. The result will be 1998-05-14. Notice that the only difference in the code is the month statement.
1 2 3 4 5 | $date = "1998-08-14"; $newdate = strtotime ( '-3 month' , strtotime ( $date ) ) ; $newdate = date ( 'Y-m-j' , $newdate ); echo $newdate; |
Subtracting Years from a date
The following example will subtract 3 years from 1998-08-14. The result will be 1995-08-14. Notice that the only difference in the code is the year statement.
1 2 3 4 5 | $date = "1998-08-14"; $newdate = strtotime ( '-3 year' , strtotime ( $date ) ) ; $newdate = date ( 'Y-m-j' , $newdate ); echo $newdate; |
Adding days, months, weeks and years from a date
There isn’t really much difference from subtracting and adding dates. To add dates, just use any of the examples above and replace the negative (-) with a positive (+) e.g. ‘+3 weeks’
comments
You could also just do:
echo date(“Y-m-j”, strtotime(“1998-08-14 -3 days”));
(sub days, weeks, months, as needed)
Thanks from Turkey…
HI,Thanx for the hint,but what about subtracting a date from another date.e.g check in date from check out date to get the number of days to spend.Thanx once more
Man! This is great, thank you so much.It helps a lot.
Thank you! Exactly what I was looking for!
Thank you! it’s good.
i got correct coding…it’s useful for me…
Hey Guys , thanks for this very helpful, but how can I add $variable days?
inteh above example we – 3 days.
say i need to add a $varible numbe rof days ? how can i do that ? anybody ?
cheers
Thanks!
Thanks this is exactly what I was looking for.
Hey thanks Maruf. Lovely piece of code!
hi, i’m having problem with this function.
it works fine until when i tried to add 40 year to the date, the result turns out to be “1970-01-1″.
It’s weird, coz it works fine with +39 year, but not +40 & above. has anyone face the same problem?
hi what can i do for variable days?
Excelente script, gracias!
i have problem when substract date to februari month.
$date = “2010-05-30″;
$newdate = strtotime ( ‘-3 month’ , strtotime ( $date ) ) ;
$newdate = date ( ‘Y-m-j’ , $newdate );
echo $newdate;
The codes gives result 2010-03-2, the correct result should be 2010-02-28
can you fix it and mail me the fix? thks
chill, here is how you can add substract days as variables:
$diff = 16;
$newdate = strtotime( ‘+’ . $diff .’ day’ , strtotime ( $date1 ) ) ;
Muchas gracias!!!
It works!
Hey, the reson that your having trouble with adding 40 and not 39 to the year is in the way that php codes 2 digit years. just change your code into a 4 digit format and that will fix it for you.
No. The reason you are having trouble adding 40 years is because of the limitation of the unix timestamp. It can’t show dates beyond 2038, until they fix this, I don’t know of a workaround.