首先来造一部分数据,表mygoods为商品表,cat_id为分类id,goods_id为商品id,status为商品当前的状态位(1:有效,0:无效)。
CREATE TABLE `mygoods` ( `goods_id` int(11) unsigned NOT NULL AUTO_INCREMENT, `cat_id` int(11) NOT NULL DEFAULT '0', `price` tinyint(3) NOT NULL DEFAULT '0', `status` tinyint(3) DEFAULT '1', PRIMARY KEY (`goods_id`), KEY `icatid` (`cat_id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; INSERT INTO `mygoods` VALUES (1, 101, 90, 0); INSERT INTO `mygoods` VALUES (2, 101, 99, 1); INSERT INTO `mygoods` VALUES (3, 102, 98, 0); INSERT INTO `mygoods` VALUES (4, 103, 96, 0); INSERT INTO `mygoods` VALUES (5, 102, 95, 0); INSERT INTO `mygoods` VALUES (6, 102, 94, 1); INSERT INTO `mygoods` VALUES (7, 102, 93, 1); INSERT INTO `mygoods` VALUES (8, 103, 99, 1); INSERT INTO `mygoods` VALUES (9, 103, 98, 1); INSERT INTO `mygoods` VALUES (10, 103, 97, 1); INSERT INTO `mygoods` VALUES (11, 104, 96, 1); INSERT INTO `mygoods` VALUES (12, 104, 95, 1); INSERT INTO `mygoods` VALUES (13, 104, 94, 1); INSERT INTO `mygoods` VALUES (15, 101, 92, 1); INSERT INTO `mygoods` VALUES (16, 101, 93, 1); INSERT INTO `mygoods` VALUES (17, 101, 94, 0); INSERT INTO `mygoods` VALUES (18, 102, 99, 1); INSERT INTO `mygoods` VALUES (19, 105, 85, 1); INSERT INTO `mygoods` VALUES (20, 105, 89, 0); INSERT INTO `mygoods` VALUES (21, 105, 99, 1);
需求:每个分类下,找出两个价格最高的有效的商品。
1. 每个分类找出价格最高的两个商品
mysql> select a.* -> from mygoods a -> where (select count(*) -> from mygoods -> where cat_id = a.cat_id and price > a.price ) <2 -> order by a.cat_id,a.price desc; +----------+--------+-------+--------+ | goods_id | cat_id | price | status | +----------+--------+-------+--------+ | 2 | 101 | 99 | 1 | | 17 | 101 | 94 | 0 | | 18 | 102 | 99 | 1 | | 3 | 102 | 98 | 0 | | 8 | 103 | 99 | 1 | | 9 | 103 | 98 | 1 | | 11 | 104 | 96 | 1 | | 12 | 104 | 95 | 1 | | 21 | 105 | 99 | 1 | | 20 | 105 | 89 | 0 | +----------+--------+-------+--------+ 10 rows in set (0.00 sec)
2. 每个分类找出价格最高的有效的两个商品(正确)
mysql> select a.* -> from mygoods a -> where (select count(*) from mygoods -> where cat_id = a.cat_id and price > a.price and status=1 ) <2 -> and status=1 -> order by a.cat_id,a.price desc ; +----------+--------+-------+--------+ | goods_id | cat_id | price | status | +----------+--------+-------+--------+ | 2 | 101 | 99 | 1 | | 16 | 101 | 93 | 1 | | 18 | 102 | 99 | 1 | | 6 | 102 | 94 | 1 | | 8 | 103 | 99 | 1 | | 9 | 103 | 98 | 1 | | 11 | 104 | 96 | 1 | | 12 | 104 | 95 | 1 | | 21 | 105 | 99 | 1 | | 19 | 105 | 85 | 1 | +----------+--------+-------+--------+ 10 rows in set (0.00 sec)
3. 每个分类找出价格最高的有效的两个商品(正确)
mysql> select a.* -> from mygoods a -> left join mygoods b -> on a.cat_id = b.cat_id and a.price < b.price and b.status=1 -> where a.status=1 -> group by a.goods_id,a.cat_id,a.price -> having count(b.goods_id) < 2 -> order by a.cat_id,a.price desc; +----------+--------+-------+--------+ | goods_id | cat_id | price | status | +----------+--------+-------+--------+ | 2 | 101 | 99 | 1 | | 16 | 101 | 93 | 1 | | 18 | 102 | 99 | 1 | | 6 | 102 | 94 | 1 | | 8 | 103 | 99 | 1 | | 9 | 103 | 98 | 1 | | 11 | 104 | 96 | 1 | | 12 | 104 | 95 | 1 | | 21 | 105 | 99 | 1 | | 19 | 105 | 85 | 1 | +----------+--------+-------+--------+ 10 rows in set (0.00 sec)
4.每个分类找出价格最高的有效的两个商品(错误)
mysql> select a.* -> from mygoods a -> where (select count(*) from mygoods -> where cat_id = a.cat_id and price > a.price ) <2 and status=1 -> order by a.cat_id,a.price desc; +----------+--------+-------+--------+ | goods_id | cat_id | price | status | +----------+--------+-------+--------+ | 2 | 101 | 99 | 1 | | 18 | 102 | 99 | 1 | | 8 | 103 | 99 | 1 | | 9 | 103 | 98 | 1 | | 11 | 104 | 96 | 1 | | 12 | 104 | 95 | 1 | | 21 | 105 | 99 | 1 | +----------+--------+-------+--------+ 7 rows in set (0.00 sec)
由上可知,如果需要增加条件的话,需要在两处增加条件。
mysql> select cat_id,GROUP_CONCAT(goods_id) from mygoods group by cat_id; +--------+------------------------+ | cat_id | GROUP_CONCAT(goods_id) | +--------+------------------------+ | 101 | 1,2,15,16,17 | | 102 | 3,5,6,7,18 | | 103 | 4,8,9,10 | | 104 | 11,12,13 | | 105 | 19,20,21 | +--------+------------------------+ 5 rows in set (0.00 sec)