PHP5变量范围和类构造

时间:2012-02-24 13:29:44

标签: php

我在从班级访问变量时遇到问题......

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;
        }
    }
}

我在全球课程中设置了这门课程'文件,稍后我将用户ID传递给以下脚本:

$u = new getuser($userid);

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

我希望这会给我用户的名字,但不是,我哪里错了?!

由于

2 个答案:

答案 0 :(得分:2)

请在您的班级中将您的用户成员定义为公开

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

答案 1 :(得分:1)

要访问类属性,您必须将其声明为public或实现getter和setter(最好是第二种解决方案)

class A {

  public $foo;

  //class methods
}

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

使用getter和setter,每个属性一个

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;
    }