Question

Mysql 分组以后取每组的前三名数据

mysql> select * from t_subject;
| id | uid | subject |
| 1 | 1 | aa |
| 2 | 2 | bb |
| 3 | 3 | cc |
| 4 | 4 | dd |
| 5 | 2 | ee |
| 6 | 2 | rr |
| 7 | 3 | tt |
| 8 | 2 | yy |
| 9 | 3 | qq |
| 10 | 4 | oo |
| 11 | 3 | pp |
| 12 | 3 | kk |
| 13 | 1 | mm |
| 14 | 4 | nn |
| 15 | 1 | ss |
| 16 | 4 | vv |
| 17 | 1 | ff |

mysql> select uid,group_concat(a.subject) from t_subject a group by a.uid ;

| uid | group_concat(a.subject) |
| 1 | aa,ff,ss,mm |
| 2 | yy,rr,ee,bb |
| 3 | kk,pp,xx,qq,tt,cc |
| 4 | nn,dd,vv,oo |

----我想要的是如下效果的 该怎么写啊（就是取出分组后每组的前三条记录）---
| uid | group_concat(a.subject) |
| 1 | aa,ff,mm |
| 2 | yy,rr,ee |
| 3 | kk,pp,xx |
| 4 | nn,dd,vv |

要排序的啊！用substring不过必须要排序后才能截取啊

Answer 1

Try this one,should be fine
下面这个已经有排序了哦，不行么？

SELECT uid, group_concat(subject)
FROM (SELECT id, uid, subject
FROM (SELECT id, uid, subject,
(SELECT COUNT(*)
FROM t_subject
WHERE uid = t.uid
AND subject <= t.subject) RK
FROM t_subject t) t1
WHERE rk <= 3) t2
GROUP BY uid

多了个a.原来
或者你直接用个substring()得了。。。

Answer 2

不列出表结构及测试数据,只能这样大概写个思路了:

select a.*
from
(
select t1.*,(select count(*)+1 from 表 where 分组字段=t1.分组字段 and 排序字段<t1.排序字段) as group_id
from 表 t1
) a
where a.group_id<=3

Answer 3

在后面加上 limit 3
就像这样select * from table limit 3

Answer 4

你这个需求在数据库里面实现起来很麻烦，而且过于复杂效率就会很低，建议你用最简单的方式查出来之后用编程语言来实现是取的三个还是几个！