2019-08-31 18:25 已编辑 招银网络_零售信息应用开发_数据研发工程师
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什么时候才能有offer啊 楼主
楼主 招银网络_零售信息应用开发_数据研发工程师
欢迎各位大佬指正 1. select a.type, sum(a.vnum)/count(a.uid) as avg,
round(
sum(case when a.date='20180711'then a.vnum else 0 end)/count(distinct case when a.date='20180711'then a.uid else null end)
-sum(case when a.date='20180710' then a.vnum else 0 end)/count(distinct case when a.date='20180710'then a.uid else null end),
4)as increase
from(
select v.date as date, v.uid as uid, v.vid as vid,
info.channel_type as type, count(vid) as vnum from
user_video as v left join new_user_info as info on v.uid=info.uid
where v.date in ('20180710','20180711')group by date,uid,type
)a
group by a.type; 2. select uid from(
select a.date,a.uid,
sum(if(date>='20190801'and date<='20190807',1,0))as num1,
sum(if(date='20190808',1,0))as num2
from(
select v.date,v.uid from user_video v join user_info i on
v.uid=i.uid
where v.tag='美食' and i.age<=25 and i.gender='女'
)a group by a.uid
)b where b.num1>10 and b.num2<3;
3. select a.tag,max(a.num)-min(a.num)from(
select tag,uid,count(distinct vid) as num from user_video group by tag,uid
) a
group by a.tag;
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