大橙子网站建设,新征程启航
为企业提供网站建设、域名注册、服务器等服务
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:
10年积累的网站设计制作、网站制作经验,可以快速应对客户对网站的新想法和需求。提供各种问题对应的解决方案。让选择我们的客户得到更好、更有力的网络服务。我虽然不认识你,你也不认识我。但先网站制作后付款的网站建设流程,更有岭东免费网站建设让你可以放心的选择与我们合作。
SELECT t1.sno,t1点吸烟 o,Score
FROM SC t1
WHERE EXISTS
(SELECT COUNT(1)
FROM SC
WHERE t1点吸烟 o= cno AND t1.scorescore
HAVING COUNT(1)3)
ORDER BY t1点吸烟 o,score DESC
另外还有一种类似的写法:
SELECT t1.sno,t1点吸烟 o,Score
FROM SC t1
WHERE
(SELECT COUNT(cno)
FROM SC
WHERE t1点吸烟 o= cno AND t1.scorescore)3
ORDER BY t1点吸烟 o,score DESC
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。。。
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