Php 通过对象调用公共变量以

Php 通过对象调用公共变量以,php,function,object,constructor,pdo,Php,Function,Object,Constructor,Pdo,$sql=$pardConfig->prepare(“从“$this->”菜单中选择*)我想动态调用db表。在这里,它只选择了菜单。我想从对象调用它 <?php $dbhost=null; $dbname=null; $dbuser=null; $dbpass=null; $file = __DIR__ ."/config.json"; $array = file_get_contents($file); $dbConfig=

$sql=$pardConfig->prepare(“从“$this->”菜单中选择*)我想动态调用db表。在这里,它只选择了菜单。我想从对象调用它

 <?php

    $dbhost=null;
    $dbname=null;
    $dbuser=null;
    $dbpass=null;

    $file = __DIR__ ."/config.json";

    $array = file_get_contents($file);
    $dbConfig=json_decode($array);

    $pardConfig=new PDO('mysql:host='.$dbConfig[0].';'.'dbname='.$dbConfig[1],$dbConfig[2],$dbConfig[3]);

    class pardDb
    {
        public $config = "pard_admin_config";
        public $article = "pard_article";
        public $menu = "pard_menu";
        public $user = "pard_user";
        public $images = "pard_images";

        function pardTemplate($pardConfig,$pardDbTable){
            $sql=$pardConfig->prepare("SELECT * FROM ".$this->menu);
            $sql->execute();
            $result=$sql->fetchALL(PDO::FETCH_OBJ); 
            $item = array_reverse($result);
            return $item;
        }
    }

    $pardDbTable = new pardDb();
    $pardDbTable->pardTemplate($config,$pardConfig);

    ?>

尝试在一个设计模式中重构此模式(通过活动记录和数据映射器开始学习)添加继承:

在看到神奇的方法uuu get和uuu set之后

    echo $obj->menu; 
echo $obj->article;