Calculate number of days,hours,minutes,seconds only working hours

Question

I have a php function to calculate timestamp (start date) to current (end) date that will return 1hr 30min 2s format. What I want to achieve is to calculate only from 8AM to 5PM of a working day. Anything that goes beyond will not be counted. Here is the php code I have.

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);

This would be so much easier with a datetime object. http://de2.php.net/manual/en/book.datetime.php — cptnk, Sep 12 '13 at 07:29
possible duplicate of [PHP - subtract dates on business days and working hours](http://stackoverflow.com/questions/13655722/php-subtract-dates-on-business-days-and-working-hours) — Bora, Sep 12 '13 at 07:33
@Bora: that junk of code? Come on guys, make some nice DateTime examples ;) — Glavić, Sep 12 '13 at 07:37
@Glavić Questioner didnt search and efford. He can do this ownself and i just show him alternative. The function that on question already copied, not trying. — Bora, Sep 12 '13 at 07:42
@Bore: still, it would be nice to see DateTime example for future OP's ;) — Glavić, Sep 12 '13 at 07:43
i'm not finish modifying so i can't tell yet. little more time. :) — Je Bor, Sep 23 '13 at 01:23

score 1 · Answer 1 · answered Sep 12 '13 at 10:38

Forget about date(), strtotime(), time(), etc. function, use DateTime :

Use example :

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

Output :

1 day, 22 hours, 14 minutes, 28 seconds

Function :

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);
}

This function can be shorter/better; but that is your job now ;)

I've been searching for an answer to this problem and this is by far the cleanest solution out there! The only thing it doesn't seem to be able to handle is working hours of night shift workers, ie that span two days like 22:00 - 06:00. My current strategy is to breakup the hours into 2 days and run this function twice, however I was wondering if you had a cleaner solution? — Tim, Jun 18 '15 at 06:49

Calculate number of days,hours,minutes,seconds only working hours

1 Answers1

Use example :

Output :

Function :

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