计算两个工作日之间的中间工作日 In PHP

Calculate intermediate week days between two week days In PHP

给定两个工作日:星期一、星期三。

你如何获得该数组中的中间天数? 答:周一、周二、周三。

案例2:周一到周一 答:周一、周二、周三、周四、周五、周六、周日、周一

谢谢!

我已经创建了这个函数来执行此操作,评论会引导您了解它是如何完成的。

function list_days($start, $end) {

    //convert day "word" to it's given number
    //Monday = 1, Tuesday = 2 ... Sunday = 7
    $start_n = date('N', strtotime($start));
    $end_n = $end_range = date('N', strtotime($end));

    //we also set $end_range above, by default it's set to the $end day number, 
    //but if $start and $end are the same it will be changed later.

    //create an empty output array
    $output = [];

    //determine end_range for the for loop
    //if $start and $end are not the same, the $end_range is simply the number of $end (set earlier)
    if($start_n == $end_n) {

        //if $start and $end ARE the same, we know there is always 7 days between the days
        //So we just add 7 to the start day number.
        $end_range = $start_n + 7;
    }

    //loop through, generate a list of days
    for ($x = $start_n; $x <= $end_range; $x++) {

        //convert day number back to the readable text, and put it in the output array
        $output[] = date('l', strtotime("Sunday +{$x} days"));
    }

    //return a string with commas separating the words.
    return implode(', ', $output);
}

用法:

示例 1:

echo list_days('Monday', 'Wednesday');
//output: Monday, Tuesday, Wednesday

示例 2:

echo list_days('Monday', 'Monday');
//output: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday, Monday

我的解决方案更多是移动数组的内部指针,直到找到边距,然后将边距之间的元素推入另一个结果数组。无论初始数组中有什么数据都可以使用。

function getDaysInBetween($start, $end)
    {
        $weekdays = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"];

        $start_found = false;

        $days = [];

        while(true) {
            $next = next($weekdays);
            $day = (empty($day) || !$next)?reset($weekdays):$next;

            if($day === $start) $start_found = true;

            if($start_found) {
                $days[] = $day;
                if($day===$end && count($days)>1) return implode(", ",$days);
            }

        }

    }

现场演示:https://3v4l.org/D5063