doubeijian2257 2013-02-22 11:38
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PHP未定义的变量错误

I have this class ( it is simple card class):

    class Card{
private $suit;
private $rank;

public function __construct($suit, $rank){
    $this->$suit = $suit;
    $this->$rank = $rank;
}

public function get_suit(){
    return $this->$suit;
}

public function get_rank(){
    return $this->$rank;
}
    }

I instance every card ( with suit and rank) for deck :

        $tmp_deck = array();
    foreach ($SUITS as $suit){
        foreach($RANKS as $rank){
            array_push( $tmp_deck, new Card($suit, $rank) );
        }
    }
    echo $tmp_deck[0]->get_suit();

And error it gives me :

Notice: Undefined variable: suit in card.php on line 13

I really cant get what is wrong. Can any one help me ?

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2条回答 默认 最新

  • dtfpznrbn503027700 2013-02-22 11:39
    关注

    Class variable access like $this->suit not like $this->$suit

    change this

    public function __construct($suit, $rank){
    $this->$suit = $suit;
    $this->$rank = $rank;
    }
    

    to

    public function __construct($suit, $rank){
       $this->suit = $suit;
       $this->rank = $rank;
    }
    

    Change others as well.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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