搜索 OOP - 模型设计（MVC）[关闭]

Question

我想在模型（MVC/框架）中创建一个“搜索”类，我想知道该方法应该如何构造。

有两种类型的搜索和SQL查询包含一些JOIN。

这将允许客户从一个搜索文本框中输入邮政编码、公司名称或城市：

SELECT * FROM Shop as S
   JOIN shop_options as O on O.ShopID = S.ShopID
   JOIN shop_openhours as OH on OH.ShopID = S.ShopID
WHERE (O.postcode = :postcode  OR S.company LIKE :companyName OR S.city = :townName)

如果客户通过 url 访问特定城镇，例如：/town/London

那么查询将如下所示：

SELECT * FROM Shop as S
   JOIN shop_options as O on O.ShopID = S.ShopID
   JOIN shop_openhours as OH on OH.ShopID = S.ShopID
WHERE S.city = :townName

需要在模型中对数据进行操作，以确定商店是开门还是关门。

这是我想出的快速设计：

class modelSearch extends Model {

    public $id;
    public $shopName;
    public $isOpen = false;

    public static function findAll($searchType) {
        $SQL = "SELECT * FROM Shop as S
                 JOIN shop_options as O on O.ShopID = S.ShopID
                 JOIN shop_openhours as OH on OH.ShopID = S.ShopID
               WHERE (O.postcode = :postcode  OR S.company LIKE :companyName OR S.town = :townName)";

         //fetch data into $data

        $search = array();
        foreach ($Dbdata as $data) {
            $searchModel = new modelSearch();
            $searchModel->isOpen = modelSearch::isShopOpen($data['opentime'], $data['closetime']);
            $searchModel->id = $data['id'];
            $searchModel->shopName = $data['shopName'];

            $search[] = $searchModel;
        }
        return $search;
    }

    public static function findByTown($search) {
        //Same code as findAll() apart from SQL query (WHERE)
        //WHERE S.town = :townName
    }

    public static function isShopOpen($open, $close) {
        $min = ($open > 123) ? $open : 345; // Earliest allow time
        if ($close < $min)
        //some block code here
            return true;
        return false;
    }
}

我走在正确的道路上还是可以改进什么？

Answer 1

抱歉，这还不是正确的方法。静态函数是包装在类中的过程代码。去掉所有的静态，它会看起来更像 OOP，尽管 OOP 真正是关于你创建的对象的交互。一个接一个地调用一个方法来实现你想要的，只是写函数的一种奇特方式。

这是我要做的：

class modelSearch extends Model {

   /**** public variables should not be common in OOP.  We want to control
    **** the interface.  Someone could change these if they were left public.
    ****/
   private $id;
   private $shopName;
   private $isOpen = false;

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

    public function findAll($searchType) {
        $SQL = "SELECT * FROM Shop as S
                 JOIN shop_options as O on O.ShopID = S.ShopID
                 JOIN shop_openhours as OH on OH.ShopID = S.ShopID
               WHERE (O.postcode = :postcode  OR S.company LIKE :companyName OR S.town = :townName)";


        /**** There is no need to instantiate any more objects in the class to
     **** do the work. I am replacing the rest of your function with
     **** pseudo-code.
     ****/
    $data = execution_of_sql();

    // I am not clear on what you want.
        foreach ($data as $record) {
       if ($this->isShopOpen($record['opentime'], $record['closetime']))
       {
          $openShops[] = array('id' => $record['id'],
                   'name' => $record['shopName']);
       }
    }

    return $openShops;

    // **** Notice how there is no need to instantiate another object
    // **** because we are the object.
    // $searchModel = new modelSearch();

    }

    // **** These could possibly be joined into 1 function with more parameters.
    public function findByTown($search) {
        //Same code as findAll() apart from SQL query (WHERE)
        //WHERE S.town = :townName
    }


    // **** protected and private hide information so that you can ignore it
    // **** outside the class.  The interface to the class is now less
    // **** cluttered.
    protected function isShopOpen($open, $close) {
        $min = ($open > 123) ? $open : 345; // Earliest allow time
        if ($close < $min)
        //some block code here
            return true;
        return false;
    }
}

Answer 2

是的，你的方法是正确的。遵循一个 MVC 框架。我认为你正在为此遵循一个框架。你创建了一个类 Search as Model 并且还具有控制器和视图。