Question

我的数据库中有一个用户列表，如果用户被禁止，我还有另一个表...

我需要帮助来调整我的sql查询，这样如果用户在禁止表中，他们将被排除在sql结果之外

tbl_ban_user    
banID user_id nickname isBan date_ban   ban_cause
98    32      Michael  Y     2016-03-11 whatever reason
99    84      Joe      Y     2016-03-11 whatever reason



tbl_users
user_id nickname is_private verified etc..
32      Michael  N          Y    
84      Joe      N          Y

这是我的sql查询，然后输出JSON，

public function fetchUsers($keyword, $user_id, $who = 0, $who_nickname = '') {
        $sql = "SELECT
                A.*,
                SUM(IF(D.grader_id='$user_id', 0, 1)) AS ungraded_count,
                ROUND(AVG(D.grader_value),0) as avg_grade,
                COUNT(DISTINCT B.grader_id) as grading,
                COUNT(DISTINCT E.grade_id) as graders,
                COUNT(DISTINCT C.photo_id) as photos,
                SUM(IF(B.grader_id='$user_id', 1, 0)) AS is_graded,
                SUM(IF(B.grader_id='$user_id', B.notification, 0)) AS notification
                FROM
                tbl_users A
                LEFT JOIN
                tbl_grade B ON (A.user_id = B.grade_id AND B.grade_type=2)
                LEFT JOIN
                tbl_photo C ON (A.user_id = C.user_id)
                LEFT JOIN
                tbl_grade D ON (C.photo_id=D.grade_id AND D.grade_type=1)
                LEFT JOIN
                tbl_grade E ON (A.user_id = E.grader_id AND E.grade_type=2)
                WHERE
                A.verified = 'Y' AND 
                A.user_name LIKE '%$keyword%'
                 AND A.user_id <> '$user_id' " . ($who == 0 ? '' : " AND A.user_id = '$who'") . ($who_nickname == '' ? '' : " AND A.nickname = '$who_nickname'") . "

                GROUP BY A.user_id
                ORDER BY A.user_name";

        return $this->db->query($sql)->result();
    }

所以基本上我想从结果中排除tbl_ban_user中列出的任何用户我将如何编码？

感谢您的帮助

Answer 1

最快的方法可能是：

public function fetchUsers($keyword, $user_id, $who = 0, $who_nickname = '') {
        $sql = "SELECT
                A.*,
                SUM(IF(D.grader_id='$user_id', 0, 1)) AS ungraded_count,
                ROUND(AVG(D.grader_value),0) as avg_grade,
                COUNT(DISTINCT B.grader_id) as grading,
                COUNT(DISTINCT E.grade_id) as graders,
                COUNT(DISTINCT C.photo_id) as photos,
                SUM(IF(B.grader_id='$user_id', 1, 0)) AS is_graded,
                SUM(IF(B.grader_id='$user_id', B.notification, 0)) AS notification
                FROM
                tbl_users A
                LEFT JOIN
                tbl_grade B ON (A.user_id = B.grade_id AND B.grade_type=2)
                LEFT JOIN
                tbl_photo C ON (A.user_id = C.user_id)
                LEFT JOIN
                tbl_grade D ON (C.photo_id=D.grade_id AND D.grade_type=1)
                LEFT JOIN
                tbl_grade E ON (A.user_id = E.grader_id AND E.grade_type=2)
                WHERE
                A.verified = 'Y' AND 
                A.user_name LIKE '%$keyword%'
                 AND A.user_id <> '$user_id' " . ($who == 0 ? '' : " AND A.user_id = '$who'") . ($who_nickname == '' ? '' : " AND A.nickname = '$who_nickname'") . "
                AND A.user_id NOT IN(SELECT DISTINCT(user_id) FROM tbl_ban_user)

                GROUP BY A.user_id
                ORDER BY A.user_name";

        return $this->db->query($sql)->result();
    }

另一种方式是：

public function fetchUsers($keyword, $user_id, $who = 0, $who_nickname = '') {
        $sql = "SELECT
                A.*,
                SUM(IF(D.grader_id='$user_id', 0, 1)) AS ungraded_count,
                ROUND(AVG(D.grader_value),0) as avg_grade,
                COUNT(DISTINCT B.grader_id) as grading,
                COUNT(DISTINCT E.grade_id) as graders,
                COUNT(DISTINCT C.photo_id) as photos,
                SUM(IF(B.grader_id='$user_id', 1, 0)) AS is_graded,
                SUM(IF(B.grader_id='$user_id', B.notification, 0)) AS notification
                FROM
                tbl_users A
                LEFT JOIN
                tbl_grade B ON (A.user_id = B.grade_id AND B.grade_type=2)
                LEFT JOIN
                tbl_photo C ON (A.user_id = C.user_id)
                LEFT JOIN
                tbl_grade D ON (C.photo_id=D.grade_id AND D.grade_type=1)
                LEFT JOIN
                tbl_grade E ON (A.user_id = E.grader_id AND E.grade_type=2)
                LEFT JOIN
                tbl_bar_user F ON F.user_id = A.user_id
                WHERE
                A.verified = 'Y' AND 
                A.user_name LIKE '%$keyword%'
                 AND A.user_id <> '$user_id' " . ($who == 0 ? '' : " AND A.user_id = '$who'") . ($who_nickname == '' ? '' : " AND A.nickname = '$who_nickname'") . "
                AND F.isBan <> 'Y'

                GROUP BY A.user_id
                ORDER BY A.user_name";

        return $this->db->query($sql)->result();
    }

您可以测试两者并查看更适合您的数据集/编码首选项。

Answer 2

永远不要在生产中运行此功能，因为这些行易受攻击：

A.nickname ='$ who_nickname'

尝试使用PDO来避免sql注入

使用Josh的第二个请求示例（JOIN变体）。子查询“SELECT DISTINCT（user_id）FROM tbl_ban_user”在大型表中运行速度非常慢。

如果在其他表

2 个答案: