在PHP中扩展方法

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这个中.

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,您将SQL查询和userId作为参数传递给此类.

<?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queryResults()进一步修改方法,以便为您的课程中的其他方法提供服务.