选择首先出现的日期或日期

Question

I am trying to pick either Friday / Saturday or 3rd, 18th, 30th, 31st... which ever comes first.

Real example: Current date is Tuesday, January 10th, 2017.

Options:

1. Friday, January 13th 2017
2. Wednesday, January 18th 2017

My code should pick the Friday, January 13th 2017.

PHP

$dayOfWeekArray  = ['friday', 'saturday'];
$dayOfMonthArray = [3, 18, 30, 31];

foreach ($dayOfWeekArray as $dayOfWeek) {
    $nextDayDate = new DateTime("next $dayOfWeek");

    foreach ($dayOfMonthArray as $dayOfMonth) {
        # Continue if $nextDayDate is before, or equals, whichever date $dayOfMonth turns out to be.

        # Otherwise, determine which comes first, $nextDayDate or $dayOfMonth
    }
}

Question: How can I turn $dayOfMonth into a date?

Answer 1

I would use DatePeriod to loop through days at an interval of 1 day until the formatted date matches one of the values you are looking for —

$start = new \DateTime("2017-01-10");

// You can probably do some math to determine the minimum number needed here

// php version 7.1.5 
// $end = (clone $start)->modify("+31 days");

// php version 5.6
$end = clone $start;
$end->modify("+31 days");

$interval = new \DateInterval("P1D"); // An interval of 1 day
$period = new \DatePeriod($start, $interval, $end);

foreach ($period as $date) {
    if (in_array($date->format('l'), ['Friday', 'Saturday']) || in_array((int) $date->format('d'), [3, 18, 30, 31])) {
        break;
    }
}

echo $date->format('l, Y-m-d') ;

// outputs 'Friday, 2017-01-13'

I think you’re asking that the final date not be 2017-01-13 if the current day is already 2017-01-13 — in which case you can just add a day to the start day, like $start = (new \DateTime("today"))->modify("+1 day")