Where to check user login?

Question

I am developing a new application using an object-oriented approach involving some REST, I do not use any frameworks.

I have a question where it is best to check the user login to the setter, as shown below:

public function setSalary($salary)
{
    if (Validator::money($salary))
        $this->salary = $salary;
    else
        return 'Error that is an invalid number';
}

Or in the controller?

public function updateSalary()
{
    $errors = array();

    if (Validator::money($_POST['salary']))
        $salary = $_POST['salary'];
        else
            $errors ['salary']  = 'Error that is an invalid number';

    if(count($errors))
        return $errors;

    $employee = new Employee($_POST['e_Id']);
    $employee->setSalary($salary);

    $employee->save();
}

If I were to install the installer, what should my controller look like and return verification errors?

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Answer 1

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  :

  $employee = new Entity\Employee;
  $employee->setID($id);
  $employee->setSalary($money);    
  if ($employee->isValid()) {
      $mapper = new Mapper\Employee($dbConn);
      $mapper->store($emplyee);
  }
  

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Answer 2

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public function updateSalary()
{
    $employee = new Employee($_POST['e_Id']);
    $employee->setSalary($_POST['salary']));
    if ($employee->validate()) {
        $employee->save();             
    } else {
        return $employee->getErrors();
    }
}

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Answer 3

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Answer 4

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Answer 5

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$dbh = new PDO($dsn, $user, $password);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

, :

public function setSalary($salary) {
    if (!Validator::money($salary)) {
        throw new Exception('Invalid value provided as salary.');
    }
    $this->salary = $salary;
}

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Employee , :

public function updateSalary($emp_id, $salary) {
    try {    
        // Note that any of the following statements could trigger exceptions:
        $employee = $this->$model->getEmployee($emp_id);
        $employee->setSalary($salary);
        $employee->save();
    } catch(Exception $e) {
        $this->$model->logError('salary', $e->getMessage());
    }
}

, , . PHP :

$model = new Model();
$controller = new Controller($model);
$view = new View($controller, $model);

$controller->updateSalary($_POST['e_Id'], $_POST['salary']);

echo $view->output();

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