扩展类无法从父类获取数据

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（在这种情况下为“一日”）不会显示。<?phpclass 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

实际上，实例化保龄球类的过程，正如调用父级构造函数所暗示的那样parent::__construct();，将创建一个全新的板球类以及保龄球类。

因此，尝试访问此新创建的Cricket类的属性没有任何意义。

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

所以举个例子

<?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();