我是PHP编程的新手,遇到了一个有趣的问题.我尝试了多次搜索并找到了不同的解决方案,但没有一个适合我的确切问题.

我有2个采用mysql格式的datestamps(例如2014-04-10 09:00:00).我需要知道这两个时间戳之间以分钟为单位的差异,但必须排除非办公时间,周末和节假日.

例如,必须显示今天(2014-04-11 14:00:00)的时间戳记和星期一(2014-04-14 11:00:00)的时间戳记,作为390分钟的结果(工作日为08.30到18.00) ).

stackexchange上的所有解决方案都将结果显示为小时或几天,但是我需要更高的准确性.

在此先感谢您,并表示歉意.

解决方法:

使用示例：

$from = '2013-09-06 15:45:32';
$to   = '2013-09-14 21:00:00';
echo some_func_name($from, $to);

输出：

1 day, 22 hours, 14 minutes, 28 seconds

功能说明

function some_func_name($from, $to) {
    $workingDays = [1, 2, 3, 4, 5]; # date format = N
    $workingHours = ['from' => ['08', '00'], 'to' => ['17', '00']];

    $start = new DateTime($from);
    $end = new DateTime($to);

    $startP = clone $start;
    $startP->setTime(0, 0, 0);
    $endP = clone $end;
    $endP->setTime(23, 59, 59);
    $interval = new DateInterval('P1D');
    $periods = new DatePeriod($startP, $interval, $endP);

    $sum = [];
    foreach ($periods as $i => $period) {
        if (!in_array($period->format('N'), $workingDays)) continue;

        $startT = clone $period;
        $startT->setTime($workingHours['from'][0], $workingHours['from'][1]);
        if (!$i && $start->diff($startT)->invert) $startT = $start;

        $endT = clone $period;
        $endT->setTime($workingHours['to'][0], $workingHours['to'][1]);
        if (!$end->diff($endT)->invert) $endT = $end;

        #echo $startT->format('Y-m-d H:i') . ' - ' . $endT->format('Y-m-d H:i') . "\n"; # debug

        $diff = $startT->diff($endT);
        if ($diff->invert) continue;
        foreach ($diff as $k => $v) {
            if (!isset($sum[$k])) $sum[$k] = 0;
            $sum[$k] += $v;
        }
    }

    if (!$sum) return 'ccc, no time on job?';

    $spec = "P{$sum['y']}Y{$sum['m']}M{$sum['d']}DT{$sum['h']}H{$sum['i']}M{$sum['s']}S";
    $interval = new DateInterval($spec);
    $startS = new DateTime;
    $endS = clone $startS;
    $endS->sub($interval);
    $diff = $endS->diff($startS);

    $labels = [
        'y' => 'year',
        'm' => 'month',
        'd' => 'day',
        'h' => 'hour',
        'i' => 'minute',
        's' => 'second',
    ];
    $return = [];
    foreach ($labels as $k => $v) {
        if ($diff->$k) {
            $return[] = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
        }
    }

    return implode(', ', $return);
}

此功能可以更短/更好；但这就是你现在的工作;)

如果要排除假期,请参见以下示例：https://*.com/a/19221403/67332