大数据培训经典 SQL 面试题解析

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以下文章来源于大数据技术与架构

一、提要

作为一名数据工作人员，SQL 是日常工作中最常用的数据提取 &简单预处理语言。因为其使用的广泛性和易学程度也被其他岗位比如产品经理、研发广泛学习使用，本篇文章主要结合经典面试题，给出通过数据开发面试的 SQL 方法与实战。

二、解题思路

• 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_大数据培训
  

• 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

三、SQL 真题

第一题

• 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

求每组商品的浏览用户数（同组内同一用户只能算一次）

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;

第二题

商品活动表 goods_event，g_id（有可能重复），t1（开始时间）,t2（结束时间）

给定时间段（t3,t4)，求在时间段内做活动的商品数

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

第三题

商品活动流水表，表名为 event，字段：goods_id， time；

求参加活动次数最多的商品的最近一次参加活动的时间

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

第四题

用户登录的 log 数据，划定 session，同一个用户一个小时之内的登录算一个 session；

生成 session 列

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<b.nn_end

第五题

订单表，字段有订单编号和时间；

取每月最后一天的最后三笔订单

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;

第六题

数据库表 Tourists，记录了某个景点 7 月份每天来访游客的数量如下：

id date visits 1 2017-07-01 100 …… 非常巧，id 字段刚好等于日期里面的几号。

现在请筛选出连续三天都有大于 100 天的日期。

上面例子的输出为：date 2017-07-01 ……

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

第七题

现有 A 表，有 21 个列，第一列 id，剩余列为特征字段，列名从 d1-d20，共 10W 条数据！

另外一个表 B 称为模式表，和 A 表结构一样，共 5W 条数据

请找到 A 表中的特征符合 B 表中模式的数据，并记录下相对应的 id

有两种情况满足要求：

• 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

第八题

我们把用户对商品的评分用稀疏向量表示，保存在数据库表 t 里面：

• 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

现在我们想要计算向量两两之间的内积，内积在这里的语义为：

对于两个不同的用户，如果他们都对同样的一批商品打了分，那么对于这里面的每个人的分数乘起来，并对这些乘积求和。

例子，数据库表里有以下的数据：U0 g0 2U0 g1 4U1 g0 3U1 g1 1

计算后的结果为：U0 U1 23+41=10 ……

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

第九题

给出一堆数和频数的表格，统计这一堆数中位数

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

第十题

表 order 有三个字段，店铺 ID，订单时间，订单金额

查询一个月内每周都有销量的店铺

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')

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