PHP5變量范圍和類構造

Question

我在從類中訪問變量時遇到問題...

class getuser {
    public function __construct($id) {
        $userquery = "SELECT * FROM users WHERE id = ".$id."";
        $userresult = mysql_query($userquery);
        $this->user = array();
        $idx = 0;
        while($user = mysql_fetch_object($userresult)){
           $this->user[$idx] = $user;
           ++$idx;
        }
    }
}

我在全局“類”文件中設置此類，稍后我將用戶ID傳遞給以下腳本：

$u = new getuser($userid);

    foreach($u->user as $user){
        echo $user->username;
    }

我希望這會給我用戶的名字，但不是，我哪里錯了？！

謝謝

Answer 1

請像這樣在您的班級中將您的用戶成員定義為公開

class getuser {
    public $user = null;
    //...
}

Answer 2

為了訪問類屬性，您必須將其聲明為公共屬性或實現getter和setter方法（最好使用第二種解決方案）

class A {

  public $foo;

  //class methods
}

$a = new A();
$a->foo = 'whatever';

有吸氣劑和二傳手，每個屬性一個

class B {

  private $foo2;

  public function getFoo2() {
    return $this->foo2;
  }

  public function setFoo2($value) {
    $this->foo2 = $value;
  }

}

$b = new B();
$b->setFoo2('whatever');  
echo $b->getFoo2();

在你的例子中：

class getuser {
    private $user;

    public function __construct($id) {
        $userquery = "SELECT * FROM users WHERE id = ".$id."";
        $userresult = mysql_query($userquery);
        $this->user = array();
        $idx = 0;
        while($user = mysql_fetch_object($userresult)){
           $this->user[$idx] = $user;
           ++$idx;
        }
    }

    /* returns the property value */
    public function getUser() {
      return $this->user;
    }

    /* sets the property value */
    public function setUser($value) {
      $this->user = $value;
    }

}


$u = new getuser($userid);
$users_list = $u->getUser();

    foreach($users_list as $user) {
        echo $user->username;
    }