Question

我有两种90％相同的方法，即90％的重复代码。我正在尝试扩展第二种方法。

第一种方法：

public function getResultsByID($userID = null){
    $sqlParams = array();
    if (!$userID)
            {
                throw new Exception("No User ID Provided");
            }   

    $sqlParams['userID'] = $userID;
    $sql = "SELECT t.user_id,
                   t.owner_id,
                   t.store_id
            FROM users t
            LEFT JOIN store s
            ON s.store_id = t.store_id
            WHERE t.user_id = :userID";

    $db = $this->dbWrite : $this->dbRead;       
    $results = $db->getRow($sql, $sqlParams);

    return $results;

}

我的第二种方法非常相同，只是我将多个表连接到此。

public function getMoreResultsByID($userID = null){
    $sqlParams = array();
    if (!$userID)
            {
                throw new Exception("No User ID Provided");
            }   

    $sqlParams['userID'] = $userID;
    $sql = "SELECT t.user_id,
                   t.owner_id,
                   t.store_id
            FROM users t
            LEFT JOIN store s ON s.store_id = t.store_id
            LEFT JOIN owner_contact oc ON oc.owner_id = t.owner_id
            LEFT JOIN owner_detail od ON od.owner_id = t.owner_id
            WHERE t.user_id = :userID";

    $db = $this->dbWrite : $this->dbRead;       
    $results = $db->getRow($sql, $sqlParams);

    return $results;

}

我无法弄清楚如何从getResultsByID（）扩展我的getMoreResultsByID（），以便我可以摆脱相同的代码

提前致谢

Answer 1

您可以创建另一个私有方法，将SQL查询和userId作为参数传递给它。

<?php

private function queryResults($sql, $userID = null){
    $sqlParams = array();
     if (!$userID)
        {
            throw new Exception("No User ID Provided");
        }   

    $sqlParams['userID'] = $userID;
    $db = $this->dbWrite : $this->dbRead;       
    $results = $db->getRow($sql, $sqlParams);

    return $results;

}
?>

然后，您可以使用这样的方法。

<?php
public function getResultsByID($userID = null){
    $sql = "SELECT t.user_id,
                   t.owner_id,
                   t.store_id
            FROM users t
            LEFT JOIN store s
            ON s.store_id = t.store_id
            WHERE t.user_id = :userID";

    return $this->queryResults($sql, $userID);
}

同样，另一个。您还可以进一步修改queryResults()方法，以便为您的班级提供其他方法。

Answer 2

你不需要添加两个函数来完成相同的工作，你需要做的就是将参数传递给函数

public function getResultsByID($userID = null,$flag){
    $sqlParams = array();
    if (!$userID)
        {
            throw new Exception("No User ID Provided");
        }   

    $sqlParams['userID'] = $userID;
    if($flag=value1){
       $sql = "SELECT t.user_id,t.owner_id,t.store_id
        FROM users t
        LEFT JOIN store s
        ON s.store_id = t.store_id
        WHERE t.user_id = :userID";
    }elseif($flag=value2){
        $sql = "SELECT t.user_id,
               t.owner_id,
               t.store_id
        FROM users t
        LEFT JOIN store s ON s.store_id = t.store_id
        LEFT JOIN owner_contact oc ON oc.owner_id = t.owner_id
        LEFT JOIN owner_detail od ON od.owner_id = t.owner_id
        WHERE t.user_id = :userID";
    }

    $db = $this->dbWrite : $this->dbRead;       
    $results = $db->getRow($sql, $sqlParams);

    return $results;
}

Answer 3

您可以将两个参数传递给函数getResultsById

<?php

public function getMoreResultsByID($userID = null, join = TRUE){
    $sqlParams = array();
    if (!$userID)
            {
                throw new Exception("No User ID Provided");
            }   

    $sqlParams['userID'] = $userID;
    $sql = "SELECT t.user_id,
                   t.owner_id,
                   t.store_id
            FROM users t";
    if (join){
        $sql.="LEFT JOIN store s ON s.store_id = t.store_id
                LEFT JOIN owner_contact oc ON oc.owner_id = t.owner_id
                LEFT JOIN owner_detail od ON od.owner_id = t.owner_id";
    }else{
        $sql.="LEFT JOIN store s
                ON s.store_id = t.store_id";
    }
    $sql.="WHERE t.user_id = :userID";

    $db = $this->dbWrite : $this->dbRead;       
    $results = $db->getRow($sql, $sqlParams);

    return $results;

}

在PHP中扩展方法

3 个答案: