代碼如下 |
複製代碼 |
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 | 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 | 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 該怎麼寫啊(就是取出分組後每組的前三條記錄)---
代碼如下 |
複製代碼 |
| uid | group_concat(a.subject) | | 1 | aa,ff,mm | | 2 | yy,rr,ee | | 3 | kk,pp,xx | | 4 | nn,dd,vv | |
執行個體二
mysql中GROUP BY分組取前N條記錄實現
mysql分組,取記錄
GROUP BY之後如何取每組的前兩位下面我來講述mysql中GROUP BY分組取前N條記錄實現方法。
這是測試表(也不知道怎麼想的,當時表名直接敲了個aa,汗~~~~):
結果:
方法一:
代碼如下 |
複製代碼 |
SELECT a.id,a.SName,a.ClsNo,a.Score FROM aa a LEFT JOIN aa b ON a.ClsNo=b.ClsNo AND a.Score<b.Score group by a.id,a.SName,a.ClsNo,a.Score having count(b.id)<2 ORDER BY a.ClsNo,a.Score desc; |
拆開分析:
<!--[if !supportLists]-->1、 <!--[endif]-->LEFT JOIN aa b ON a.ClsNo=b.ClsNo AND a.Score<b.Score
同一個班級(每個班級四個人),分數比當前學生高的記錄,那就意味這成績墊底的學生,將會產生三條記錄
<!--[if !supportLists]-->2、 <!--[endif]-->group by a.id,a.SName,a.ClsNo,a.Score having count(b.id)<2
a.id,a.SName,a.ClsNo,a.Score可以代表一個學生(以學生分組),如果count(b.id)<2(成績超過你的人不能多於2個),那就只剩第一第二了。
方法二:
代碼如下 |
複製代碼 |
SELECT * FROM aa a WHERE 2>(SELECT COUNT(*) FROM aa WHERE ClsNo=a.ClsNo and Score>a.Score) ORDER BY a.ClsNo,a.Score DESC; |
這個我覺得是比較有意思的,取每一條記錄,判斷同一個班級,大於當前成績的同學是不是小於2個人。
方法三:
代碼如下 |
複製代碼 |
SELECT * FROM aa WHERE id IN (SELECT id FROM aa WHERE ClsNo=a.ClsNo ORDER BY Score DESC LIMIT 2) ORDER BY a.ClsNo,a.Score DESC; |
這種方式進過測試不通過,ERROR 1235 (42000): This version of MySQL doesn't yet support 'LIMIT & IN/ALL/ANY/SOME subquery' ,不能在這幾個子查詢中使用limit。