在PHP中将值推送到数组中
我在接收我要推入卡阵列的值时遇到问题。我不知道是我没有调用正确的属性,还是我只是没有正确地添加到数组中在PHP中将值推送到数组中,php,arrays,Php,Arrays,我在接收我要推入卡阵列的值时遇到问题。我不知道是我没有调用正确的属性,还是我只是没有正确地添加到数组中 <?php class Deck{ public function __construct(){ $values =array('2','3','4','5','6','7','8','9','10','J','Q','K','A'); $suits =array('Diamond','Club','Heart','Spade');
<?php
class Deck{
public function __construct(){
$values =array('2','3','4','5','6','7','8','9','10','J','Q','K','A');
$suits =array('Diamond','Club','Heart','Spade');
$cards = array();
foreach ($suits as $suit) {
foreach($values as $value){
$cards[] = "$value of $suit's";
}
}
}
}
$deck = new Deck();
var_dump($deck);
$cards
是\u构造的局部变量:一旦该函数结束,该变量就会消失。相反,您可能希望使卡成为该类的成员:
class Deck {
public $cards = [];
public function __construct() {
$values =array('2','3','4','5','6','7','8','9','10','J','Q','K','A');
$suits =array('Diamond','Club','Heart','Spade');
$cards = array();
foreach ($suits as $suit) {
foreach($values as $value){
$this->cards[] = "$value of $suit's";
}
}
}
}
然后,您可以在对象内部使用$this->cards
,或在对象外部使用$deck->cards
。在$cards[]=“$suit's的$value”。。是打字错误吗?$cards[]=$value.“..$suites<代码>$deck
是您的对象。在构造函数中有一个局部变量$cards
保存创建的数组。只需打印($cards)代码>在构造函数内部?!然后做var_转储($deck->cards)?