[英]Accessing parent variables in child method
I currently have two classes, one called Dog, one called Poodle. 我目前有两堂课,一堂叫狗,一堂叫贵宾犬。 Now how can I use a variable defined in Dog from the Poodle class.
现在如何使用Poodle类中Dog中定义的变量。 My code is as follows:
我的代码如下:
class dog {
protected static $name = '';
function __construct($name) {
$this->name = $name
}
}
class Poodle extends dog {
function __construct($name) {
parent::__construct($name)
}
function getName(){
return parent::$name;
}
}
$poodle = new Poodle("Benjy");
print $poodle->getName();
I get this error 我得到这个错误
Notice: Undefined variable: name
注意:未定义的变量:名称
i guess 'name' is an attribute of the concrete Dog, so it shouldn't be static in the first place. 我想'name'是具体Dog的属性,因此它首先不应该是静态的。 To access non-static parent class attributes from within an inherited class, just use "$this".
要从继承的类中访问非静态父类属性,只需使用“ $ this”。
class dog {
protected $name = '';
function __construct($name) {
$this->name = $name;
}
}
class Poodle extends dog {
function getName(){
return $this->name;
}
}
The problem is in your Dog
constructor. 问题出在您的
Dog
构造函数中。 You wrote: 你写了:
$this->name = $name;
But using $this
implies that name
is an instance variable, when in fact it's a static variable. 但是使用
$this
表示name
是一个实例变量,而实际上它是一个静态变量。 Change it to this: 更改为此:
self::$name = $name;
That should work fine. 那应该工作正常。
In your dog class you have declared the variable $name as static , you have to declare the variable without the static word 在您的dog类中,您已将变量$ name声明为static ,您必须声明不带静态词的变量
class dog {
protected $name = '';
function __construct($name) {
$this->name = $name
}
}
class Poodle extends dog {
function __construct($name) {
parent::__construct($name)
}
function getName(){
return $this->name;
}
}
$poodle = new Poodle("Benjy");
print $poodle->getName();
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