CODEWITHSUNDEEP
phpdatedatetimedifferencedatestamp
HenkB
HenkB

Reputation: 43

Difference between 2 dates in minutes excluding weekends and holidays

I'm relative new to PHP programming and run into an interesting problem. I've tried multiple searches and found different solutions, but none of them fit my precise problem.

I have 2 datestamps in mysql format (2014-04-10 09:00:00 for example). I need to know the difference in minutes between those 2 timestamps, but must exclude the out-of-office hours, weekends and holidays.

For example, a timestamp of today (2014-04-11 14:00:00) and a timestamp of monday (2014-04-14 11:00:00) must show as a result of 390 minutes (workdays are 08.30 to 18.00).

All of the solutions on stackexchange show the results as hours or days, but I need more accuracy.

Thanks in advance and apologies of there is something unclear.

Upvotes: 4

Views: 2896

Answers (3)

Glavić
Glavić

Reputation: 43552

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 ;)

If you wish to exclude holidays, see this example: https://stackoverflow.com/a/19221403/67332

Upvotes: 3

Vitthal
Vitthal

Reputation: 327

I found this and converted it in to minutes,

$start = new DateTime('2014-03-03 09:21:30');
$end = new DateTime('2014-03-11 17:23:15');
// otherwise the  end date is excluded (bug?)
$end->modify('+1 day');

$interval = $end->diff($start);

// total days
$days = $interval->days;
$days_inMin = ($interval->d*24*60) + ($interval->h*60) + $interval->i;

// create an iterateable period of date (P1D equates to 1 day)
$period = new DatePeriod($start, new DateInterval('P1D'), $end);

// best stored as array, so you can add more than one
$holidays = array('2014-03-07');

foreach($period as $dt) {
    $curr = $dt->format('D');

    // for the updated question
    if (in_array($dt->format('Y-m-d'), $holidays)) {
       $days--;
       $days_inMin -= (24*60);
    }

    // substract if Saturday or Sunday
    if ($curr == 'Sat' || $curr == 'Sun') {
        $days--;
        $days_inMin -= (24*60);
    }
}

echo 'Days: ' . $days; 
echo '<br>Days in Minutes: ' . $days_inMin . ' min = ' . $days_inMin/(24*60) . ' days';

Edit: 

$office_hrs_min = $days_inMin - ($days * (14.5*60));
// as out of 24 only 8.5 hrs are working

Upvotes: 2

Guns
Guns

Reputation: 2728

I got the answer from here

For PHP >= 5.3.0, use the DatePeriod class. It's unfortunately barely documented.

$start = new DateTime('6/30/2010');
$end = new DateTime('7/6/2010');
$oneday = new DateInterval("P1D");

$days = array();
$data = "7.5";

/* Iterate from $start up to $end+1 day, one day in each iteration.
   We add one day to the $end date, because the DatePeriod only iterates up to,
   not including, the end date. */
foreach(new DatePeriod($start, $oneday, $end->add($oneday)) as $day) {
    $day_num = $day->format("N"); /* 'N' number days 1 (mon) to 7 (sun) */
    if($day_num < 6) { /* weekday */
        $days[$day->format("Y-m-d")] = $data;
    } 
}    
print_r($days);

Upvotes: 1

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