laravel - Laravel - 如何从休假请求中提取假期和周末

Question

我正在使用 Laravel-5.8 开发员工休假平台。

目前，我正在申请休假。每年年初，管理员设置所有假期并存储在 hr_holiday_dates 表中，如下所示：

class HrHolidayDate extends Model
{
  protected $table = 'hr_holiday_dates';

  protected $primaryKey = 'id';

  protected $fillable = [
              'holiday_name',
              'date_from',
              'date_to',
          ];
}

员工登录应用程序提出请假请求。

型号：HrLeaveRequest：

class HrLeaveRequest extends Model
{
  public $timestamps = false;
  protected $table = 'hr_leave_requests';

  protected $primaryKey = 'id';
  protected $fillable = [
              'id',
              'employee_id',
              'reason',
              'leave_type_id',
              'commencement_date',
              'resumption_date',
              'no_of_days',
          ];

  public function employee()
  {
    return $this->belongsTo('App\Models\Hr\HrEmployee','employee_id');
  } 

  public function leaveType()
  {
    return $this->belongsTo('App\Models\Hr\HrLeaveType','leave_type_id');
  }
}

当员工提出请假时，他选择leave_type、beginning_date和resumption_date。no_of_days 未选择，而是基于应用程序智能。

控制器

public function create()
{       
    return view('hr.leave_requests.create');
}

public function store(StoreLeaveRequest $request)
{
  try {
    $commencementDate = Carbon::parse($request->commencement_date);
    $resumptionDate = Carbon::parse($request->resumption_date);        
    $employeeId = Auth::user()->employee_id;

        $leave = new HrLeaveRequest();
        $leave->leave_type_id               = $request->leave_type_id;
        $leave->commencement_date       = $commencementDate ->toDateTimeString();
        $leave->resumption_date             = resumptionDate ->toDateTimeString();
        $leave->no_of_days                  = $request->no_of_days;

            Session::flash('success', 'Leave Request is created successfully');
            return redirect()->route('hr.leave_requests.index');
    } catch (Exception $exception) {
            Session::flash('error', 'Action failed! Please try again');
            return redirect()->route('hr.leave_requests.index');
    }
}

对于休假 no_of_days，我想要的是这个，应用程序在开始日期和恢复日期之间得到它。但在此之前，它会从选定日期中删除周末（周六和周日）。此外，它会进入所选日期范围内的 hr_holiday_days 并提取假期。它还应该考虑到假期中的周末。

如何得到：

$leave->no_of_days

从上面的解释？

谢谢你。

标签： laravel

Answer 1

Enter the starting and ending dates, along with an array of any holidays that might be in between, and it returns the working days as an integer:

<?php
//The function returns the no. of business days between two dates and it skips the holidays
function getWorkingDays($startDate,$endDate,$holidays){
    // do strtotime calculations just once
    $endDate = strtotime($endDate);
    $startDate = strtotime($startDate);


    //The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24
    //We add one to inlude both dates in the interval.
    $days = ($endDate - $startDate) / 86400 + 1;

    $no_full_weeks = floor($days / 7);
    $no_remaining_days = fmod($days, 7);

    //It will return 1 if it's Monday,.. ,7 for Sunday
    $the_first_day_of_week = date("N", $startDate);
    $the_last_day_of_week = date("N", $endDate);

    //---->The two can be equal in leap years when february has 29 days, the equal sign is added here
    //In the first case the whole interval is within a week, in the second case the interval falls in two weeks.
    if ($the_first_day_of_week <= $the_last_day_of_week) {
        if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week) $no_remaining_days--;
        if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week) $no_remaining_days--;
    }
    else {
        // (edit by Tokes to fix an edge case where the start day was a Sunday
        // and the end day was NOT a Saturday)

        // the day of the week for start is later than the day of the week for end
        if ($the_first_day_of_week == 7) {
            // if the start date is a Sunday, then we definitely subtract 1 day
            $no_remaining_days--;

            if ($the_last_day_of_week == 6) {
                // if the end date is a Saturday, then we subtract another day
                $no_remaining_days--;
            }
        }
        else {
            // the start date was a Saturday (or earlier), and the end date was (Mon..Fri)
            // so we skip an entire weekend and subtract 2 days
            $no_remaining_days -= 2;
        }
    }

    //The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder
//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it
   $workingDays = $no_full_weeks * 5;
    if ($no_remaining_days > 0 )
    {
      $workingDays += $no_remaining_days;
    }

    //We subtract the holidays
    foreach($holidays as $holiday){
        $time_stamp=strtotime($holiday);
        //If the holiday doesn't fall in weekend
        if ($startDate <= $time_stamp && $time_stamp <= $endDate && date("N",$time_stamp) != 6 && date("N",$time_stamp) != 7)
            $workingDays--;
    }

    return $workingDays;
}

//Example:

$holidays=array("2008-12-25","2008-12-26","2009-01-01");

echo getWorkingDays("2008-12-22","2009-01-02",$holidays)
// => will return 7
?>

Referenced from : https://stackoverflow.com/a/336175/10309265