CODEWITHSUNDEEP
phpmysql
Kim Carlo
Kim Carlo

Reputation: 1223

calculating time attendance in PHP MySQL

I am working with laravel and i have a table with all the users attendances. each row has a flag stating if the user logs in or out. I want to calculate the total working hours of user.

this is my table and sample data

id | user_id |      date     |   timestamp   | status |
 1 |    1    |   2018-05-10  |    17:15:31   |   out  |
 1 |    1    |   2018-05-10  |    13:15:31   |   in   |
 1 |    1    |   2018-05-10  |    12:15:31   |   out  |
 1 |    1    |   2018-05-10  |    08:01:31   |   in   |

I want to calculate the total working hours

$logs = DB::table('attendances as at')
        ->join('infos as in','in.user_id','=','at.user_id')
        ->select('in.name','in.avatar','at.*')
        ->orderBy('id','desc')
        ->where('at.user_id',$id)
        ->get();

 $total_hours = [];

    for($i=0; $i < count($logs)-1; $i++ ){

         if($logs[$i]->status == 'out'){
           $dattime1 = new DateTime($logs[$i]->dt.' '. $logs[$i]->time);
           $dattime2 = new DateTime($logs[$i+1]->dt.' '. $logs[$i+1]->time);

            $total_hours[] = $dattime1 ->diff($dattime2);
         }

    }

 $working_hours = array_sum($total_hours);

Is this the best way to achieve accurate results? Please help. Thanks

Upvotes: 0

Views: 501

Answers (1)

Bits Please
Bits Please

Reputation: 897

Can you try like this?

$time1 = "17:15:00";
$time2 = "00:30:00";

$strtotime1 = strtotime($time1);
$strtotime2 = strtotime($time2);
 $o = ($strtotime1) + ($strtotime2);
echo $time = date("h:i:s A T",$o);

Output will be like this:

05:45:00 PM UTC

Upvotes: 1

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