From: http://blog.chinaunix.net/uid-26729093-id-4294287.html
请参考:http://bbs.csdn.net/topics/330021260
create table t2 (
id int primary key,gid char,
col1 int,
col2 int
) engine=myisam;
insert into t2 values
(1,'A',31,6),
(2,'B',25,83),
(3,'C',76,21),
(4,'D',63,56),
(5,'E',3,17),
(6,'A',29,97),
(7,'B',88,63),
(8,'C',16,22),
(9,'D',25,43),
(10,'E',45,28),
(11,'A',2,78),
(12,'B',30,79),
(13,'C',96,73),
(14,'D',37,40),
(15,'E',14,86),
(16,'A',32,67),
(17,'B',84,38),
(18,'C',27,9),
(19,'D',31,21),
(20,'E',80,63),
(21,'A',89,9),
(22,'B',15,22),
(23,'C',46,84),
(24,'D',54,79),
(25,'E',85,64),
(26,'A',87,13),
(27,'B',40,45),
(28,'C',34,90),
(29,'D',63,8),
(30,'E',66,40),
(31,'A',83,49),
(32,'B',4,90),
(33,'C',81,7),
(34,'D',11,12),
(35,'E',85,10),
(36,'A',39,75),
(37,'B',22,39),
(38,'C',76,67),
(39,'D',20,11),
(40,'E',81,36);
期望结果
1) N=1 取GID每组 COL2最大的记录
+----+------+------+------+
| id | gid | col1 | col2 |
+----+------+------+------+
| 6 | A | 29 | 97 |
| 15 | E | 14 | 86 |
| 24 | D | 54 | 79 |
| 28 | C | 34 | 90 |
| 32 | B | 4 | 90 |
+----+------+------+------+
2) N=3 取GID每组 COL2最大的3条记录
+----+------+------+------+
| id | gid | col1 | col2 |
+----+------+------+------+
| 6 | A | 29 | 97 |
| 11 | A | 2 | 78 |
| 36 | A | 39 | 75 |
| 32 | B | 4 | 90 |
| 2 | B | 25 | 83 |
| 12 | B | 30 | 79 |
| 28 | C | 34 | 90 |
| 23 | C | 46 | 84 |
| 13 | C | 96 | 73 |
| 24 | D | 54 | 79 |
| 4 | D | 63 | 56 |
| 9 | D | 25 | 43 |
| 15 | E | 14 | 86 |
| 25 | E | 85 | 64 |
| 20 | E | 80 | 63 |
+----+------+------+------+
问题(1)的sql:
SELECT a.id,a.gid,a.col1,a.col2 FROM t2 a
LEFT JOIN t2 b
ON a.gid=b.gid AND a.col2<=b.col2
GROUP BY a.id,a.gid,a.col1,a.col2
HAVING COUNT(*)<=1
ORDER BY a.gid,a.col2 desc
或者
SELECT a.id,a.gid,a.col1,a.col2 FROM t2 a
WHERE 1>=(
SELECT COUNT(*) FROM t2 b
WHERE a.gid=b.gid AND a.col2<=b.col2)
ORDER BY a.gid,a.col2 desc
问题(2)的sql:
SELECT a.id,a.gid,a.col1,a.col2 FROM t2 aLEFT JOIN t2 b
ON a.gid=b.gid AND a.col2<=b.col2
GROUP BY a.id,a.gid,a.col1,a.col2
HAVING COUNT(*)<=3
ORDER BY a.gid,a.col2 desc
或者
SELECT a.id,a.gid,a.col1,a.col2 FROM t2 a
WHERE 3>=(
SELECT COUNT(*) FROM t2 b
WHERE a.gid=b.gid AND a.col2<=b.col2)
ORDER BY a.gid,a.col2 desc
实际上,这两种解法都是不准确的,如果最大值重复的话,就会出现错误
比如再插入一条数据:
INSERT INTO t2 VALUES
(41,'A',31,97);
然后用该解答方式解答步骤一,得到下面的答案:
| id | gid | col1 | col2 |
+----+------+------+------+
| 32 | B | 4 | 90 |
| 28 | C | 34 | 90 |
| 24 | D | 54 | 79 |
| 15 | E | 14 | 86 |
+----+------+------+------+