如何使用多维数组以及重叠检查从开始时间和结束时间获得总小时数？

Question

我有以下情况，我需要仅在特定日期数组中检查 重叠 并获取总参加时间。

array (
  '2020-07-14' => 
  array (
    'total_attended_hours' => 0,
    0 => 
    array (
      'start_time' => '09:00:00',
      'end_time' => '13:00:00',
      'hours' => '4 hours 0 mins',
    ),
    1 => 
    array (
      'start_time' => '13:30:00',
      'end_time' => '16:30:00',
      'hours' => '3 hours 0 mins',
    ),
    2 => 
    array (
      'start_time' => '09:00:00',
      'end_time' => '14:00:00',
      'hours' => '5 hours 0 mins',
    ),
  ),
  '2020-07-15' => 
  array (
    'total_attended_hours' => 0,
    0 => 
    array (
      'start_time' => '13:30:00',
      'end_time' => '17:00:00',
      'hours' => '3 hours 30 mins',
    ),
    1 => 
    array (
      'start_time' => '09:00:00',
      'end_time' => '14:00:00',
      'hours' => '5 hours 0 mins',
    ),
  ),
)

在上面的例子中，日期为2020-07-14，我们有start_time and end_time:- total_attended_hoursshould be equals to7 小时 30 分钟

对于下一个2020-07-15，它应该是total_attended_hours = 8 hours 0 mins

Playground

以下数组的新问题

$data = [
  '2020-07-14' => 
  [
    [
      'start_time' => '14:15:00',
      'end_time' => '17:45:00',
    ],[
      'start_time' => '14:30:00',
      'end_time' => '17:30:00',
    ],[
      'start_time' => '14:30:00',
      'end_time' => '17:30:00',
    ],
  ],
  '2020-07-15' => [
    [
      'start_time' => '13:30:00',
      'end_time' => '17:00:00',
    ],[
      'start_time' => '09:00:00',
      'end_time' => '14:00:00',
    ],
  ],
];

结果：-

Array
(
    [2020-07-14] => Array
        (
            [0] => Array
                (
                    [start_time] => 14:15:00
                    [end_time] => 17:45:00
                )

            [1] => Array
                (
                    [start_time] => 14:30:00
                    [end_time] => 17:30:00
                )

            [2] => Array
                (
                    [start_time] => 14:30:00
                    [end_time] => 17:30:00
                )

            [total_attended_hours] => 03:15:00
        )

[total_attended_hours] => 03:15:00 应该是 [total_attended_hours] => 03:30:00

Answer 1

这里有算法：

1. 对于每组时间预订，请执行以下操作
2. 找到最小的start_time
3. 将start_time 和end_time 之间的duration 添加到sum
4. 通过start_time查找下一个最小的时间预订
5. IF current_end_time previous_end_time 跳转到4 END IF
6. IF start_time previous_end_time 从sum END IF 中减去差异
7. 在start_time 和end_time 之间添加duration
8. 跳到 4 直到没有匹配的元素。

编码愉快:)

编辑 - 添加更干净的实现

function getSortedDays(array $days): array {
    return array_map(function (array $day) {
       array_multisort(array_column($day, 'start_time'), SORT_ASC, $day);
       
       return $day;
    }, $days);
}

function addTotalAttendedHours(array $days): array {
    $sortedDays = getSortedDays($days);
    
    $days = array_map(function (array $day) {
        $sum = (new DateTime())->setTimestamp(0);
        $previousEnd = null;
        
        foreach ($day as $time) {
            $currentStart = new DateTimeImmutable($time['start_time']);
            $currentEnd = new DateTimeImmutable($time['end_time']);

            if ($currentEnd < $previousEnd) continue; // this has been added
            
            $sum->add($currentStart->diff($currentEnd));
            
            if ($previousEnd !== null && $currentStart < $previousEnd) {
                $sum->sub($currentStart->diff($previousEnd));
            }
            
            $previousEnd = $currentEnd;
        }
        
        $attendedSeconds = $sum->getTimestamp();
        $day['total_attended_hours'] = sprintf(
            '%02u:%02u:%02u',
            $attendedSeconds / 60 / 60,
            ($attendedSeconds / 60) % 60,
            $attendedSeconds % 60
        );
        
        return $day;
    }, $sortedDays);
    
    return $days;
}

工作example。