无法将$ _POST ['variable']分配给对象属性

Question

我正在与代码作斗争：

$worker = new Worker(); // :|
$worker->addWorker;
//testing                   
//testing
// $_SESSION['worker']->addWorker();
echo $_POST['name'];
echo $worker->name;

哪两个最后一行应显示相同的值。类中的assign方法如下所示：

public function addWorker()
{
      $this->name = $_POST['name'];
      $this->surname = $_POST['surname'];
      $this->dob = $_POST['dob'];
      $this->skills = $_POST['skills'];
      $this->postcode = $_POST['postcode'];
      $this->street = $_POST['street'];
      $this->email = $_POST['email'];
      $this->tel = $_POST['tel'];
      $this->erefnumber = $_POST['erefnumber'];
}

我已经声明了类Worker属性。问题是我没有任何错误，$ worker-＆gt; name只提供任何内容。 echo $ _POST ['name']工作正常。

Answer 1

您的代码几乎没问题，但您确实：

$worker->addWorker;

虽然它应该是：

$worker->addWorker();

另外，为了提高代码的可读性，我考虑将addWorker()更改为要求数组参数：

public function addWorker( array $postArray ) {
    $this->name = $postArray['name'];
    ....

并在调用时传递$_POST：

$worker->addWorker( $_POST );