重庆分公司,新征程启航
为企业提供网站建设、域名注册、服务器等服务
Try this one,should be fine
成都一家集口碑和实力的网站建设服务商,拥有专业的企业建站团队和靠谱的建站技术,十载企业及个人网站建设经验 ,为成都超过千家客户提供网页设计制作,网站开发,企业网站制作建设等服务,包括成都营销型网站建设,成都品牌网站建设,同时也为不同行业的客户提供做网站、网站设计的服务,包括成都电商型网站制作建设,装修行业网站制作建设,传统机械行业网站建设,传统农业行业网站制作建设。在成都做网站,选网站制作建设服务商就选成都创新互联公司。
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。。。
select * from
(select Student.S#,Sname,Sage,Ssex,SC.C#,score,Cname
from SC inner join Student on SC.S#=Student.S#
inner join Course on Course.C#=SC.C#) as t
where exists(
select count(1) from t as d
where C#=d.C# and score d.score
having count(1)3
)
select class,total,name from (select *,ywsc+sxsc as total from st ORDER BY total DESC) b
where
not EXISTS(select * from (select *,ywsc+sxsc as total from st ORDER BY total DESC) c where c.class=b.class and b.total c.total GROUP BY c.class HAVING COUNT(*)2 )
ORDER BY b.class,b.total DESC
上面那条答案应该是可以的啊,可能是看起来复杂了点吧,我自己试过是没问题的:
SELECT t1.sno,t1.cno,Score
FROM SC t1
WHERE EXISTS
(SELECT COUNT(1)
FROM SC
WHERE t1.cno= cno AND t1.scorescore
HAVING COUNT(1)3)
ORDER BY t1.cno,score DESC
另外还有一种类似的写法:
SELECT t1.sno,t1.cno,Score
FROM SC t1
WHERE
(SELECT COUNT(cno)
FROM SC
WHERE t1.cno= cno AND t1.scorescore)3
ORDER BY t1.cno,score DESC