大厂的SQL面试题都是怎样的(附赠答案)
年底了,你们有没有开始写年终总结呢?
可乐最近也在写年终总结了,过两天给你们分享一下年终总结漂亮的写法,今天先给大家带来SQL试题的分享,SQL吧上手很容易,想要精通也没那么简单。
正 文
作为一名互联网商业数据分析师,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,给出通过数据分析师面试的SQL方法论。以下题目均来与笔者经历&网上分享的中高难度SQL题。
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
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 asselect goods_id,case when nn<= 10 then 'top10'when nn<= 20 then 'top10~top20'else 'other' end as goods_groupfrom(select goods_id,row_number() over(partition by goods_id order by sale_sum desc) as nnfrom(select goods_id,sum(amount) as sale_sumfrom ordergroup by 1) aa) bb;select b.goods_group,count(distinct a.uid) as numfrom pv aleft join test.nil_goods_category bon a.goods_id = b.goods_idgroup 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_numfrom goods_eventwhere (t1<=t4 and t1>=t3)or (t2>=t3 and t2<=t4)
2.select count(distinct g_id) as event_goods_numfrom goods_eventwhere (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.timefrom event ainner join(select goods_id,count(*)from eventgroup by gooods_idorder by count(*) desclimit 1) bon a.goods_id = b.goods_idorder 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 asselect *,row_number() over(partition by userid order by inserttime) as nn1from(select a.*,b.inserttime as inserttime_aftr,datediff(b.inserttime,a.inserttime) as session_difffrom(select userid,inserttime,row_number() over(partition by userid order by inserttime asc) nnfrom koo.nil_temp0222where userid = 1900000169) aleft join(select userid,inserttime,row_number() over(partition by userid order by inserttime asc) nnfrom koo.nil_temp0222where userid = 1900000169) bon a.userid = b.userid and a.nn = b.nn-1) aawhere session_diff >10 or nn = 1order by userid,inserttime;
drop table if exists koo.nil_temp0222_a2_1;create table if not exists koo.nil_temp0222_a2_1 asselect a.*,case when b.nn is null then a.nn+3 else b.nn end as nn_endfrom koo.nil_temp0222_a2 aleft join koo.nil_temp0222_a2 bon a.userid = b.useridand a.nn1 = b.nn1 - 1;
select a.*,b.nn1 as session_idfrom(select userid,inserttime,row_number() over(partition by userid order by inserttime asc) nnfrom koo.nil_temp0222where userid = 1900000169) aleft join koo.nil_temp0222_a2_1 bon a.userid = b.useridand a.nn>=b.nnand a.nn
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 nn2from(select cast(right(to_date(inserttime),2) as int) as dd,month(inserttime) as mm,userid,inserttimefrom koo.nil_temp0222) aa) bbwhere 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 num3from table aleft join table bon a.userid = b.useridand a.dt = date_add(b.dt,-1)left join table con a.userid = c.useridand a.dt = date_add(c.dt,-2)where b.num>100and a.num>100and 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 mmdfrom table) aaleft join(select id,concat(d1,d2,d3……d20) as mmdfrom table) bbon aa.id = bb.idand aa.mmd = bb.mmd
2.select a.*,sum(d1_jp,d2_jp……,d20_jp) as same_judgefrom(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_jpfrom table aleft join table bon a.id = b.id) aawhere 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 2
U0 g1 4
U1 g0 3
U1 g1 1
计算后的结果为:
U0 U1 23+41=10 ……
select aa.uid1,aa.uid2,sum(star_multi) as resultfrom(select a.uid as uid1,b.uid as uid2,a.goods_id,a.star * b.star as star_multifrom t aleft join t bon a.goods_id = b.goods_idand a.udi<>b.uid) aagroup by 1,2
select uid1,uid2,sum(multiply) as resultfrom(select t.uid as uid1, t.uid as uid2, goods_id,a.star*star as multiplyfrom a left join bon a.goods_id = goods_idand a.uid<>uid) aagroup 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(selectcase 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_midfrom table) aleft join(select id,num,row_number() over(partition by id order by num asc) nnfrom table) bon a.s_mid = b.nnleft join(select id,num,row_number() over(partition by id order by num asc) nnfrom table) con 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_levelfrom(select credit_level,count(distinct nn) as numberfrom(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 nnfrom koo.nil_temp0222where substring(inserttime,1,7) = '2019-12'order by credit_level ,inserttime) aagroup by 1) bbwhere number = (select count(distinct weekofyear(inserttime))from koo.nil_temp0222where substring(inserttime,1,7) = '2019-12')

