访问PHP中的另一个类保护变量

Question

我正在用PHP写一个足球经理模拟器，[HARD AlgorithMS！]

我有3个课程：

播放器

class Player {

protected $name;

public function addAttr($name) {
    $this->name = $name;
}
}

团队

class Team {

protected $name;

protected $players = array();


public function setName($name) {
    $this->name = $name;
}

public function getName() {
    return $this->name;
}

public function addPlayer($player) {
    $this->players[] = $player;
}

public function getPlayers() {
    print_r($this->players);
}

public function getOpponetsPosition() {
    GAME::getOpponetPlayersPosition();
}

和游戏

class Game {

protected $t1;
protected $t2;

function setTeams($team1,$team2) {
    $this->t1 = $team1;
    $this->t2 = $team2;
}

function getOpponetPlayersPosition() {
        $this->t1->getPlayers();
}

}

和主要脚本

require_once 'classes/CPlayer.php';
require_once 'classes/CTeam.php';
require_once 'classes/CGame.php';

$game = new Game;

$team1 = new Team;
$team1->setName("PO-1");
$team2 = new Team;
$team2->setName("PO-2");

$p1 = new Player;
$p2 = new Player;


$p1->addAttr("payam babaiy");
$p2->addAttr("parsa babaiy");

$team1->addPlayer($p1);
$team2->addplayer($p2);

$game->setTeams($team1,$team2);

$team1->getOpponetsPosition();

我需要在团队类

中使用 getOpponetsPosition（）功能获得游戏中的所有玩家位置

但它不会返回我在主脚本中输入的值。 我这样做对吗？这是app im writing的好方法吗？

Answer 1

你的方法很好，有几点：

使用构造函数，它们可以让您的生活更轻松：

class Player {

    protected $name;

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

    public function addAttr($name) {
        $this->name = $name;
    }
}

然后

new Player("Lionel Messi");

构造商还确保您不会让玩家/团队空名！您可以更好地控制课程中的内容！

不要混用静态代码和普通代码

你的getOpponentsPosition函数不正确

public function getOpponetsPosition() {
    GAME::getOpponetPlayersPosition();
}

事实上，它甚至不应该在这里，获取其他球队的位置并不是Team的工作，因为它包含两者。

<强> See this gist for how I would have accomplished your goal.

Answer 2

不，您正在调用静态函数GAME::getOpponetPlayersPosition();，因此未定义该函数内的$this。