If I\'ve got a $date
YYYY-mm-dd
and want to get a specific $day
(specified by 0 (sunday) to 6 (saturday)) of the week that YYYY-
Just:
2012-10-11 as $date and 5 as $day
<?php
$day=5;
$w = date("w", strtotime("2011-01-11")) + 1; // you must add 1 to for Sunday
echo $w;
$sunday = date("Y-m-d", strtotime("2011-01-11")-strtotime("+$w day"));
$result = date("Y-m-d", strtotime($sunday)+strtotime("+$day day"));
echo $result;
?>
The $result = '2012-10-12' is what you want.
Try
$date = '2012-10-11';
$day = 1;
$days = array('Sunday', 'Monday', 'Tuesday', 'Wednesday','Thursday','Friday', 'Saturday');
echo date('Y-m-d', strtotime($days[$day], strtotime($date)));
PHP Manual said :
w Numeric representation of the day of the week
You can therefore construct a date with mktime, and use in it date("w", $yourTime);
I had to use a similar solution for Portuguese (Brazil):
<?php
$scheduled_day = '2018-07-28';
$days = ['Dom','Seg','Ter','Qua','Qui','Sex','Sáb'];
$day = date('w',strtotime($scheduled_day));
$scheduled_day = date('d-m-Y', strtotime($scheduled_day))." ($days[$day])";
// provides 28-07-2018 (Sáb)
If your date is already a DateTime
or DateTimeImmutable
you can use the format
method.
$day_of_week = intval($date_time->format('w'));
The format string is identical to the one used by the date function.
To answer the intended question:
$date_time->modify($target_day_of_week - $day_of_week . ' days');
I think this is what you want.
$dayofweek = date('w', strtotime($date));
$result = date('Y-m-d', strtotime(($day - $dayofweek).' day', strtotime($date)));