mysql分组取每组前几条记录(排序)

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。

 

可以将每个分组下的goods_id合并。

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)

下面是我的:
select a.* from t1 a where (select count(*)from t1 where grouper = a.grouper and score>a.score)<2 order by a.grouper,a.score desc