Reputation: 8633
I have month in number (i.e. 1,2,3...12) I have year in number (2015,2014...2010)
What I try to do is that using strtotime() to find
strtotime("last day of 2 month 2021");
in this case, 2 and 2012 will be variable in a number listed in 1st and 2nd sentence.
I wonder is that possible to do? Because I tried the one I wrote here, its not getting the right date after I test the result of $at = strtotime("last day of 2 month 2021");
with $time = date('Y-m-d', $at);
it always seems sending me the date of the current year. in the above example, result of date('Y-m-d', $at);
will be
2015-04-30
what I expect to get is
2021-02-28
Anyone?
Upvotes: 0
Views: 943
Reputation: 4880
Its really easy to find a leap year. Which really seems to be what you're trying to do.
if(2012%4 == 0) {
// it's a leap year
}
Upvotes: 0
Reputation: 326
Try DateTime object
$date = new DateTime('2021-02');
$date->modify('last day of this month');
echo $date->format('Y-m-d');
Also, the last day of 2021 Feb is 28...
Upvotes: 4