数据分析师SQL试题合集
SQL数据库开发
共 4210字,需浏览 9分钟
·
2020-07-31 16:07
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简单——会考察一些group by & limit之类的用法,或者平时用的不多的函数比如rand()类;会涉及到一些表之间的关联
中等——会考察一些窗口函数的基本用法;会有表之间的关联,相对tricky的地方在于会有一些自关联的使用
困难——会有中位数或者更加复杂的取数概念,可能要求按照某特定要求生成列;一般这种题建中间表会解得清晰些
order订单表,字段为:goods_id, amount ;
pv 浏览表,字段为:goods_id,uid;
goods按照总销售金额排序,分成top10,top10~top20,其他三组
create table if not exists test.nil_goods_category as
select goods_id
,case when nn<= 10 then 'top10'
when nn<= 20 then 'top10~top20'
else 'other' end as goods_group
from
(
select goods_id
,row_number() over(partition by goods_id order by sale_sum desc) as nn
from
(
select goods_id,sum(amount) as sale_sum
from order
group by 1
) aa
) bb;
select b.goods_group,count(distinct a.uid) as num
from pv a
left join test.nil_goods_category b
on a.goods_id = b.goods_id
group by 1;
(提示:可以左右滑动代码)
1.
select count(distinct g_id) as event_goods_num
from goods_event
where (t1<=t4 and t1>=t3)
or (t2>=t3 and t2<=t4)
2.
select count(distinct g_id) as event_goods_num
from goods_event
where (t1<=t4 and t1>=t3)
union all
select a.goods_id,a.time
from event a
inner join
(
select goods_id,count(*)
from event
group by gooods_id
order by count(*) desc
limit 1
) b
on a.goods_id = b.goods_id
order by a.goods_id,a.time desc
drop table if exists koo.nil_temp0222_a2;
create table if not exists koo.nil_temp0222_a2 as
select *
,row_number() over(partition by userid order by inserttime) as nn1
from
(
select a.*
,b.inserttime as inserttime_aftr
,datediff(b.inserttime,a.inserttime) as session_diff
from
(
select userid,inserttime
,row_number() over(partition by userid order by inserttime asc) nn
from koo.nil_temp0222
where userid = 1900000169
) a
left join
(
select userid,inserttime
,row_number() over(partition by userid order by inserttime asc) nn
from koo.nil_temp0222
where userid = 1900000169
) b
on a.userid = b.userid and a.nn = b.nn-1
) aa
where session_diff >10 or nn = 1
order by userid,inserttime;
drop table if exists koo.nil_temp0222_a2_1;
create table if not exists koo.nil_temp0222_a2_1 as
select a.*
,case when b.nn is null then a.nn+3 else b.nn end as nn_end
from koo.nil_temp0222_a2 a
left join koo.nil_temp0222_a2 b
on a.userid = b.userid
and a.nn1 = b.nn1 - 1;
select a.*,b.nn1 as session_id
from
(
select userid,inserttime
,row_number() over(partition by userid order by inserttime asc) nn
from koo.nil_temp0222
where userid = 1900000169
) a
left join koo.nil_temp0222_a2_1 b
on a.userid = b.userid
and a.nn>=b.nn
and a.nn
select *
from
(
select *
,rank() over(partition by mm order by dd desc) as nn1
,row_number() over(partition by mm,dd order by inserttime desc) as nn2
from
(
select
cast(right(to_date(inserttime),2) as int) as dd,
month(inserttime) as mm,userid,inserttime
from koo.nil_temp0222
) aa
) bb
where nn1 = 1 and nn2<=3;
select a.*,b.num as num2,c.num as num3
from table a
left join table b
on a.userid = b.userid
and a.dt = date_add(b.dt,-1)
left join table c
on a.userid = c.userid
and a.dt = date_add(c.dt,-2)
where b.num>100
and a.num>100
and c.num>100
每个特征列都完全匹配的情况下
最多有一个特征列不匹配,其他19个特征列都完全匹配,但哪个列不匹配未知
1.
select aa.*
from
(
select *,concat(d1,d2,d3……d20) as mmd
from table
) aa
left join
(
select id,concat(d1,d2,d3……d20) as mmd
from table
) bb
on aa.id = bb.id
and aa.mmd = bb.mmd
2.
select a.*,sum(d1_jp,d2_jp……,d20_jp) as same_judge
from
(
select a.*
,case when a.d1 = b.d1 then 1 else 0 end as d1_jp
,case when a.d2 = b.d2 then 1 else 0 end as d2_jp
,case when a.d3 = b.d3 then 1 else 0 end as d3_jp
,case when a.d4 = b.d4 then 1 else 0 end as d4_jp
,case when a.d5 = b.d5 then 1 else 0 end as d5_jp
,case when a.d6 = b.d6 then 1 else 0 end as d6_jp
,case when a.d7 = b.d7 then 1 else 0 end as d7_jp
,case when a.d8 = b.d8 then 1 else 0 end as d8_jp
,case when a.d9 = b.d9 then 1 else 0 end as d9_jp
,case when a.d10 = b.d10 then 1 else 0 end as d10_jp
,case when a.d20 = b.d20 then 1 else 0 end as d20_jp
,case when a.d11 = b.d11 then 1 else 0 end as d11_jp
,case when a.d12 = b.d12 then 1 else 0 end as d12_jp
,case when a.d13 = b.d13 then 1 else 0 end as d13_jp
,case when a.d14 = b.d14 then 1 else 0 end as d14_jp
,case when a.d15 = b.d15 then 1 else 0 end as d15_jp
,case when a.d16 = b.d16 then 1 else 0 end as d16_jp
,case when a.d17 = b.d17 then 1 else 0 end as d17_jp
,case when a.d18 = b.d18 then 1 else 0 end as d18_jp
,case when a.d19 = b.d19 then 1 else 0 end as d19_jp
from table a
left join table b
on a.id = b.id
) aa
where sum(d1_jp,d2_jp……,d20_jp) = 19
t的字段有:uid,goods_id,star。uid是用户id
goodsid是商品id
star是用户对该商品的评分,值为1-5
U0 g1 4
U1 g0 3
U1 g1 1
select aa.uid1,aa.uid2
,sum(star_multi) as result
from
(
select a.uid as uid1
,b.uid as uid2
,a.goods_id
,a.star * b.star as star_multi
from t a
left join t b
on a.goods_id = b.goods_id
and a.udi<>b.uid
) aa
group by 1,2
select uid1,uid2,sum(multiply) as result
from
(select t.uid as uid1, t.uid as uid2, goods_id,a.star*star as multiply
from a left join b
on a.goods_id = goods_id
and a.uid<>uid) aa
group by goods
select a.*
,b.s_mid_n
,c.l_mid_n
,avg(b.s_mid_n,c.l_mid_n)
from
(
select
case when mod(count(*),2) = 0 then count(*)/2 else (count(*)+1)/2 end as s_mid
,case when mod(count(*),2) = 0 then count(*)/2+1 else (count(*)+1)/2 end as l_mid
from table
) a
left join
(
select id,num,row_number() over(partition by id order by num asc) nn
from table
) b
on a.s_mid = b.nn
left join
(
select id,num,row_number() over(partition by id order by num asc) nn
from table
) c
on a.l_mid = c.nn
select distinct credit_level
from
(
select credit_level,count(distinct nn) as number
from
(
select userid,credit_level,inserttime,month(inserttime) as mm
,weekofyear(inserttime) as week
,dense_rank() over(partition by credit_level,month(inserttime) order by weekofyear(inserttime) asc) as nn
from koo.nil_temp0222
where substring(inserttime,1,7) = '2019-12'
order by credit_level ,inserttime
) aa
group by 1
) bb
where number = (select count(distinct weekofyear(inserttime))
from koo.nil_temp0222
where substring(inserttime,1,7) = '2019-12')
作者:adorable_new
来源:简书
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