PHP calc difference between time greater than 24 hours

Question

What is the best way to calculate the difference in time, when time is greater than 24 hours.

Example

$time1 = '76:00:00';
$time2 = '30:00:00';

// result should be 46:00:00
echo date('H:i:s', strtotime($time1) - strtotime($time2));

But this could not be done with this because its greater then 24 hours.

Also in a database i've stored a time like this: 33:30:00 How in php could i format it to: 33:30

If you use normal timestamps its a lot easier to calculate the difference — Bolli, Nov 22 '16 at 12:32
In the database the field type is Time. Because i only need the time and not te date. — Bham, Nov 22 '16 at 12:38

Bartosz Zasada · Accepted Answer · 2016-11-22T14:56:19.680

3

Use \DateTime and \DateInterval to perform calculations:

$date1 = new \DateTime('now', new DateTimeZone('UTC'));
$date2 = new \DateTime('now', new DateTimeZone('UTC'));
$time1 = new \DateInterval('PT76H');
$time2 = new \DateInterval('PT30H');

$date1->add($time1);
$date2->add($time2);

$diff = $date1->diff($date2);
echo ($diff->days * 24 + $diff->h) . $diff->format(':%I:%S');

Explanation: It's not possible to perform calculations directly on DateIntervals, so you have to create dates as a basis for calculations. Then add two different intervals to current dates, and calculate a difference between them. diff() returns \DateInterval that contains total number of days, that you have to multiply by 24 to get hours, and hours that don't make full days.

EDIT: The timezone should be specified as UTC to avoid daylight saving time issues.

edited Nov 22 '16 at 14:56

answered Nov 22 '16 at 12:42

Bartosz Zasada

3,762
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What are the values: 'PT76H' and 'PT30H' – Bham Nov 22 '16 at 12:51
These are time interval definitions. See examples here: http://php.net/manual/en/dateinterval.construct.php – Bartosz Zasada Nov 22 '16 at 12:53
Is it possible to show a negative value when one time is greather then the other. – Bham Nov 22 '16 at 13:42
Yes it is. There is an `invert` field in `$diff`, which is `1` if the difference is negative. You can check this field and prepend `-` to your result. – Bartosz Zasada Nov 22 '16 at 14:52

score 0 · Answer 2 · answered Jan 19 '23 at 10:26

You can try this:

/**
 * Common method to diff between two time
 * 
 * @param string $start
 * @param string $end
 * @return string
 */
public function diffTime($start, $end)
{
    /* $start = '76:00:00'; */
    /* $end = '30:00:00'; */
    /* Get total time in Seconds */
    $totalStartTimeSec = self::getTotalSeconds($start);
    $totalEndTimeSec = self::getTotalSeconds($end);
    /* Get difference time in Seconds */
    $difference = abs($totalStartTimeSec - $totalEndTimeSec);
    $hours = floor($difference / 3600);
    $minutes = floor(($difference % 3600) / 60);
    $seconds = $difference % 60;
    $diffFormat = str_pad($hours, 2, '0', STR_PAD_LEFT). ":" . str_pad($minutes, 2, '0', STR_PAD_LEFT). ":" . str_pad($seconds, 2, '0', STR_PAD_LEFT);
    
    /* result $diffFormat is '46:00:00'; */
    return $diffFormat;
}

/**
 * Get total time in Seconds
 * 
 * @param string $time
 * @return array
 */
public function getTotalSeconds($time) {
    $timeArr = explode(':', $time);
    $allTime = [];
    if (isset($timeArr[0]) && $timeArr[0] > 24) {
        $greaterDuration = $timeArr[0];
        while ($greaterDuration > 24) {
            $greaterDuration -= 24;
            $allTime[] = strtotime('24:00:00') - strtotime('00:00');
        }
        if ($greaterDuration == 24 && ($timeArr[1] != '00' || $timeArr[2] != '00')) {
            $greaterDuration -= 24;
            $allTime[] = strtotime('24:00:00') - strtotime('00:00');
        }
        $remainingTime = str_pad($greaterDuration, 2, '0', STR_PAD_LEFT) . ':' . $timeArr[1] . ':' . $timeArr[2];
        $allTime[] = strtotime($remainingTime) - strtotime('00:00');
    } else {
        $allTime[] = strtotime($time) - strtotime('00:00');
    }
    
    return array_sum($allTime);
}

Thanks!

PHP calc difference between time greater than 24 hours

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