Question

我试图在扩展类中使用显示函数，该类首先在父类中获取显示函数。但是，它不会在echo语句中显示变量。游戏类型（在本例中为“一日”）不会显示。

<?php
class Cricket
{
    protected $gameType;

    function __construct($gameType)
    {
        $this->gameType=$gameType;
    }

    function display()
    {
        echo 'The cricket match is a ' . $this->gameType . " match";
    }
}

class Bowler extends Cricket
{
    public $type;
    public $number;

    function __construct($type,$number)
    {
        $this->type=$type;
        $this->number=$number;

        parent::__construct($this->gameType);
    }

    function display()
    {
        parent:: display();
        echo " with " . $this->number . " " . $this->type . " bowler";
    }
}   

$one = new Cricket("day-night");
$one->display();

echo'<br>';

$two  = new Cricket("day-night");
$two = new Bowler("left-hand","2");
$two->display();
?>

Answer 1

实例化保龄球类的过程实际上将通过调用父级构造函数parent::__construct();来暗示，创建一个崭新的Cricket类以及Bowler类。

因此，尝试访问此新创建的Cricket类的属性没有任何意义。

因此，当实例化Bowler类时，您还必须传递Cricket类成功构建所需的任何数据。

例如

<?php
class Cricket
{
    protected $gameType;

    function __construct($gameType)
    {
        $this->gameType=$gameType;
    }

    function display()
    {
        echo 'The cricket match is a ' . $this->gameType . " match";
    }
}

class Bowler extends Cricket
{
    public $type;
    public $number;

    function __construct($gameType, $type, $number)
    {
        $this->type=$type;
        $this->number=$number;

        parent::__construct($gameType);
    }

    function display()
    {
        parent:: display();
        echo " with " . $this->number . " " . $this->type . " bowler";
    }
}   

$two = new Bowler('day-night', "left-hand","2");
$two->display();

扩展类无法从父类获取数据

1 个答案: