PHP: Get the sum of a different dates in array - php

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.

Answer 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

Answer 2

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