PHP: Get the sum of a different dates in array

Question

Hello Im just a beginner programmer but Im now building an app which gets the total hours of work.

Here I have a time_start array which the threshold for the customer to in and same as out for time_end

The logs here is where the person checks in.

            $time_start = [
                [0] => '22:00:00',
                [1] => '03:00:00',
                ];
            $time_end = [
                [0] => '02:00:00',
                [1] => '08:00:00',
                ];
            $logs = [
                [0] => '2019-07-09 22:00:00',
                [1] => '2019-07-10 02:00:00',
                [2] => '2019-07-10 03:00:00',
                [3] => '2019-07-10 08:00:00',
            ];

So I'm trying to get the sum of their hours in night shift which has different date.

Im trying to get it in:

            foreach( $logs as $check_time ){

                $attendance_date_time = date("Y-m-d",strtotime($check_time));
                $time_starts = date('Y-m-d H:i:s', strtotime("$attendance_date_time $time_start"));
                $time_ends = date('Y-m-d H:i:s', strtotime("$attendance_date_time $time_end"));
                if ( $check_time >= $time_starts && $check_time <= $time_ends )
                {
                    $time[] = $check_time;
                }else{
                    $time = null
                }

            }

Supposedly I will get all logs because it is between time_start and time_end and store it to $time

And summing the time i get.

Since I'm new at php is their any easiest idea to get the total work hours? Cuz I get a null because the time_end it gets was the date of check_in.

How it will be 9 hours, can you explain? Because I see its 5 hours! — Rahul
– Rahul, Commented Jul 10, 2019 at 6:02
2019-09-09 22:00 to 2019-09-10 02:00 equals to 4hours. 2019-07-10 03:00:00 to 2019-07-10 08:00:00 equals to 5hours — Rubics
– Rubics, Commented Jul 10, 2019 at 6:07
Since the $logs are beetween the time_in and out it will add to $time array. And add each array. equals to 9hours — Rubics
– Rubics, Commented Jul 10, 2019 at 6:08

Rahul · Accepted Answer · 2019-07-10 07:00:43Z

1

Here is your snippet,

$time_start = [
    0 => '22:00:00',
    1 => '03:00:00',
];
$time_end = [
    0 => '02:00:00',
    1 => '08:00:00',
];
$logs = [
    0 => '2019-07-09 22:00:00',
    1 => '2019-07-10 02:00:00',
    2 => '2019-07-10 03:00:00',
    3 => '2019-07-10 08:00:00',
];
$time = [];
$diff = 0;
for ($j = 0; $j < count($logs); $j = $j + 2) {
    $attendance_start = date("Y-m-d", strtotime($logs[$j]));
    $attendance_end   = date("Y-m-d", strtotime($logs[$j + 1]));
    for ($i = 0; $i < count($time_start); $i++) {
        $time_starts = date('Y-m-d H:i:s', strtotime("$attendance_start " . $time_start[$i] . ""));
        $time_ends   = date('Y-m-d H:i:s', strtotime("$attendance_end " . $time_end[$i] . ""));
        if ($logs[$j] >= $time_starts && $logs[$j + 1] <= $time_ends) {
            array_push($time, $logs[$j], $logs[$j + 1]);
            // calculating diff for two alternate days
            $diff += (round((strtotime($logs[$j + 1]) - strtotime($logs[$j])) / 3600, 1));
            break;
        }
    }
}
echo $diff . " Hours";die;

Demo

Output:-

9 Hours

edited Jul 10, 2019 at 7:00

answered Jul 10, 2019 at 6:05

Rahul

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3 Comments

Rubics Over a year ago

2019-09-09 22:00 to 2019-09-10 02:00 equals to 4hours. 2019-07-10 03:00:00 to 2019-07-10 08:00:00 equals to 5hours We have a date 2019-09-09 and 2019-09-10

Rahul Over a year ago

@ElliotA I have made changes as per your requirement. PLease have a look

Rubics Over a year ago

thank you. link in case you can help me with this one?

GrenierJ · Accepted Answer · 2019-07-10 06:15:36Z

0

The for loop is the easiest way to do this

$time_start = [
  '22:00:00',
  '03:00:00',
];
$time_end = [
  '02:00:00',
  '08:00:00',
];
$logs = [
  '2019-07-09 22:00:00',
  '2019-07-10 02:00:00',
  '2019-07-10 03:00:00',
  '2019-07-10 08:00:00',
];

$totalDiff = 0;

for($i =0 ; $i < count($logs); $i = $i + 2)
{
  $startDate = strtotime($logs[$i]);
  $endDate = strtotime($logs[$i+1]);

  $diff = $endDate-$startDate;

  $totalDiff += $diff;
}

$hourDiff = $totalDiff / 3600;
echo $hourDiff . ' hours';

Demo

answered Jul 10, 2019 at 6:15

GrenierJ

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