LARAVEL Invalid argument supplied for foreach() loop

Question

I have a Session::get('employee_leave_tag'); that contains the payroll_leave_employee_id since data is not only one i used a foreach loop. But im having an error

Invalid argument supplied for foreach()

Can someone help me what's the cause why i am getting this error? Thanks.

     public function get_session_leave_tagv2()
     {
          $employee = [0 => 0];
          if(Session::has('employee_leave_tag'))
          {
               $employee = Session::get('employee_leave_tag');
          }

          $empdata = array();
          foreach($employee as $emp)
          {

                 $employee_id = Tbl_payroll_leave_employeev2::select('payroll_employee_id')
                                                          ->join('tbl_payroll_leave_schedulev2','tbl_payroll_leave_employee_v2.payroll_leave_employee_id','=','tbl_payroll_leave_schedulev2.payroll_leave_employee_id')
                                                          ->where('tbl_payroll_leave_schedulev2.payroll_leave_employee_id',$emp)
                                                          ->distinct()
                                                          ->get();

                    if(count($employee_id) == 0)
                    {
                         $empa = Tbl_payroll_employee_basic::join('tbl_payroll_leave_employee_v2','tbl_payroll_leave_employee_v2.payroll_employee_id','=','tbl_payroll_employee_basic.payroll_employee_id')->whereIn('tbl_payroll_leave_employee_v2.payroll_leave_employee_id',$emp)->get();
                         array_push($empdata,$empa);
                    }
                    else
                    {
                         $empb = Tbl_payroll_leave_schedulev2::getallemployeeleavedata($employee_id)->get();
                         array_push($empdata,$empb);
                    }                            

          }

          $data['new_record'] = $empdata;

          return json_encode($data);
     }

Answer 1

Try this one...

if (isset($employee) && count($employee) > 0) {
    if (is_array($employee)) {
        foreach($employee as $emp) {
            //display Array employee
        }
    } else {
        //display for NON Array employee
    }
}

Answer 2

You need to make two changes in your code.

1st change: Your array declaration $employee = array(); or $employee[];

2nd change: Its always a good practice to check if the variable is an array and not empty using if condition before we write foreach loop. It will save us from the error which you are facing.

if(is_array($employee) && !empty($employee))
{
    foreach($employee as $emp)
    {
    }
}