Question

我有这个查询，很慢。我得到每个选择的计数，并将它们定义为free_cnt，plus_cnt列，以便可以在父选择查询中进行操作。

$query = $this->db->query("SELECT movie_title,movie_id,MAX(x.free_cnt) as free_cnt, MAX(x.plus_cnt) as plus_cnt, ((MAX(x.free_cnt)*1) + (MAX(x.plus_cnt)*3)) AS score, (MAX(x.free_cnt) + MAX(x.plus_cnt)) AS total
            FROM (
                SELECT b.id as movie_id, b.movie_title as movie_title, COUNT(*) AS free_cnt, 0 as plus_cnt
                FROM preview_movie_request a1
                LEFT JOIN movies b on a1.movie_id=b.id
                JOIN users c on c.email=a1.email
                WHERE c.subsc_status='0' AND c.package_type='' AND b.movie_type=2 $where1
                GROUP BY b.movie_title
                UNION ALL

                SELECT d.id as movie_id, d.movie_title as movie_title,  0 as free_cnt, COUNT(*) AS plus_cnt
                FROM preview_movie_request a2
                LEFT JOIN movies d on a2.movie_id=d.id
                JOIN users e on e.email=a2.email
                WHERE e.subsc_status='1' AND e.package_type!='' AND d.movie_type=2 $where2
                GROUP BY d.movie_title
                UNION ALL

            ) AS x
            GROUP BY movie_title
            $orderby
            $limit");

反正有没有简化查询并使其更快？

Answer 1

您可以消除一个子查询来移动计数条件，以防万一

SELECT movie_title,movie_id,MAX(x.free_cnt) as free_cnt, MAX(x.plus_cnt) as plus_cnt, ((MAX(x.free_cnt)*1) + (MAX(x.plus_cnt)*3)) AS score, (MAX(x.free_cnt) + MAX(x.plus_cnt)) AS total FROM
(
SELECT b.id as movie_id, b.movie_title as movie_title,
 SUM(CASE WHEN c.subsc_status='0' AND c.package_type='' AND b.movie_type=2 $where1 THEN 1 END ) AS free_cnt,
 SUM(CASE WHEN c.subsc_status='1' AND c.package_type!='' AND b.movie_type=2  $where2 THEN 1 END ) AS  plus_cnt,

                FROM preview_movie_request a1
                LEFT JOIN movies b on a1.movie_id=b.id
                JOIN users c on c.email=a1.email

                GROUP BY b.movie_title)x    
GROUP BY movie_title
$orderby
$limit

简化mysql联合查询

1 个答案: