如何使用PHP查找日期中某一天的日期？

Question

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-mm-dd is in.

For example, if I got 2012-10-11 as $date and 5 as $day , I want to get 2012-10-12 , if I've got 0 as $day , 2012-10-14

EDIT:
Most of you misunderstood it. I got some date, $date and want to get a day specified by 0-6 of the same week $date is in.

So no, I don't want the day of $date ...

Answer 1

I think this is what you want.

$dayofweek = date('w', strtotime($date));
$result    = date('Y-m-d', strtotime(($day - $dayofweek).' day', strtotime($date)));

Answer 2

You can use the date() function:

date('w'); // day of week

or

date('l'); // dayname

Example function to get the day nr.:

function getWeekday($date) {
    return date('w', strtotime($date));
}

echo getWeekday('2012-10-11'); // returns 4

Answer 3

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)));

Answer 4

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.

Answer 5

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');

Answer 6

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);

Answer 7

I'm afraid you have to do it manually. Get the date's current day of week, calculate the offset and add the offset to the date.

$current = date("w", $date)
$offset = $day - $current
$new_date = new DateTime($date)
    ->add(
        new DateInterval($offset."D")
    )->format('Y-m-d')

Answer 8

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)

Answer 9

<?php echo date("H:i", time()); ?>
<?php echo $days[date("l", time())] . date(", d.m.Y", time()); ?>

Simple, this should do the trick