题解 | #牛客每个人最近的登录日期(二)#

# 先对时间排序
# select u.name u_n ,c.name c_n ,l.date ,
# rank()over(partition by l.user_id order by l.date desc) as rn 
# from login l,user u,client c
# where l.user_id=u.id and l.client_id= c.id ;
# 再取出时间排第一的
select u_n,c_n,date from (select u.name u_n ,c.name c_n ,l.date ,
rank()over(partition by l.user_id order by l.date desc) as rn 
from login l,user u,client c
where l.user_id=u.id and l.client_id= c.id ) as a
where a.rn=1
order by u_n;