如何在php中声明扩展类的变量

Question

我试图声明一个变量,特别是$ publisher变量,但它不起作用.我不确定我是否正确地扩展了课程.

书籍是主要的类,而Novel是扩展类.

PHP课程

class Books {

    /* Member variables */
    public $price;
    public $title;

    /* Member functions */
    public function setPrice($par){
        $this->price = $par;    
    }

    public function getPrice(){
        echo $this->price.'<br/>';
    }

    public function setTitle($par){
        $this->title = $par;
    }

    public function getTitle(){
        echo $this->title.'<br/>';
    }

    public function __construct($par1,$par2){
        $this->title = $par1;
        $this->price = $par2;
    }

}

class Novel extends Books {
    public  $publisher;

    public function setPublisher($par){
        $this->publisher = $par;
    }

    public function getPublisher(){
        echo $this->publisher;
    }

}

**打电话给**

include 'includes/book.inc.php';

$physics = new Books("Physics for High School" , 2000);
$chemistry = new Books("Advanced Chemistry" , 1200);
$maths = new Books("Algebra", 3400);


$physics->getTitle();
$chemistry->getTitle();
$maths->getTitle();

$physics->getPrice();
$chemistry->getPrice();
$maths->getPrice();

if (class_exists('Novel')) {
    echo 'yes';
}

$pN = new Novel();
$pN->setPublisher("Barnes and Noble");
$pN->getPublisher();

**结果**

Physics for High School
Advanced Chemistry
Algebra
2000
1200
3400
yes

正如您所看到的,它没有声明$ Publisher()值.

我的问题是我错过了什么？

Answer 1

如果启用错误报告,您将看到错误,代码实际上并未运行.

致命错误:未捕获的ArgumentCountError:函数Books :: __ construct()的参数太少,0在第62行传入......正好是2个预期的

因为您正在扩展类并且父类具有构造,所以您需要将值传递给它$pN = new Novel('To Kill a Mockingbird', 2.99);,或者将它们定义为默认值.

public function __construct($par1 = '',$par2 = ''){
    $this->title = $par1;
    $this->price = $par2;
}

然后它会正常工作:

高中物理
高等化学
代数
2000
1200
3400
yesBarnes and Noble

https://3v4l.org/Ydthj