php - 为什么类扩展没有获得私有(private)变量

Question

以下代码不会生成名为 Jock 的输出。我怀疑是因为在 class Animal 中 $name 是 private，但构造是 public 所以子类应该无法从构造中获取 $name 。我不想将 $name public。

class Animal{
    private $name;
    public function __construct($name) {
        $this->name = $name;
    }
    public function Greet(){
        echo "Hello, I'm some sort of animal and my name is ", $this->name ;
    }
}

 class Dog extends Animal{
     private $type;

     public function __construct($name,$type) {
         $this->type = $type;
           parent::__construct($name);

     }
     public function Greet(){
         echo "Hello, I'm a ", $this->type, " and my name is ", $this->name;
     }
 }
   $dog2 = new Dog('Jock','dog');
   $dog2->Greet();

Answer 1

你是对的:删除 private 变量或在 animal 类的第一行使用 protected 就可以了。

class Animal{
    protected $name; //see here!
    public function __construct($name) {
        $this->name = $name;
    }
    public function Greet(){
        echo "Hello, I'm some sort of animal and my name is ".$this->name ;
    }
}

$animal = new Animal("Gizmo");
$animal->greet(); //produces the desired result.
echo $animal->name; //this will throw an error - unable to access protected variable $name

$name 不会是公共(public)的，因为它是公共(public)构造函数中使用的参数，因此仅限于该函数的范围。但是，除非您使用protected，否则狗的属性name 将是公开的。

点用于连接字符串。但是 echo 允许逗号输出多个表达式。

 public function Greet(){
     echo "Hello, I'm a ".$this->type." and my name is ".$this->name;
 }

同样当使用双引号时；您可以将变量放入字符串中:

 public function Greet(){
     echo "Hello, I'm a $this->type and my name is $this->name";
 }