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我有一个 php 函数来计算时间戳（开始日期）到当前（结束）日期，它将返回 1hr 30min 2s 格式。我想要实现的是只计算一个工作日的上午 8 点到下午 5 点。任何超出的都不会被计算在内。这是我拥有的php代码。

class duration_computation {
   function duration( $time ) {
      $d[0] = array(1, "s");
      $d[1] = array(60, "min");
      $d[2] = array(3600, "hr");
      $d[3] = array(86400, "dy");
      $d[4] = array(604800, "wk");
      $d[5] = array(2592000, "mth");
      $d[6] = array(31104000, "yr");

      $numbers = array();

      $result = "";
      $now = time();
      $time_difference = ( $now - $time );
      $seconds_left = $time_difference;

      for ( $i = 6; $i > -1; $i-- ) {
          $numbers[$i] = intval( $seconds_left / $d[$i][0] );
          $seconds_left -= ( $numbers[$i] * $d[$i][0] );
          if ( $numbers[$i] != 0 ) {
          $result.= abs($numbers[$i]) . "" . $d[$i][1] . (($numbers[$i]>1)?'':'') ." ";
          }
      }
      return $result;
   }
}

$duration = new duration_computation();
echo $duration->duration($trail->duration);
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1 回答 1

1

忘记date(), strtotime(), time(), etc.功能，使用DateTime：

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 ：

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);
}

这个函数可以更短/更好；但那是你现在的工作；）

于 2013-09-12T10:38:59.190 回答