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。以下题目均来与笔者经历&网上分享的中高难度SQL题。
二、解题思路
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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
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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
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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
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order订单表，字段为：goods_id， amount ；
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pv 浏览表，字段为：goods_id，uid；
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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 nn from ( select goods_id,sum(amount) as sale_sum from order group by 1 ) aa) bb;select b.goods_group,count(distinct ) as numfrom pv a left join test.nil_goods_category b on 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,from event a inner join( select goods_id,count(*) from event group by gooods_id order by count(*) desc limit 1) bon a.goods_id = b.goods_idorder by a.goods_id, 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.* , as inserttime_aftr ,datediff(,) 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 = and = -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 is null then +3 else end as nn_endfrom koo.nil_temp0222_a2 a left join koo.nil_temp0222_a2 b on = and 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) nn from koo.nil_temp0222 where userid = 1900000169) aleft join koo.nil_temp0222_a2_1 b on = and >=and <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.*, as num2, as num3from table a left join table bon = and = date_add(,-1)left join table con = and = date_add(,-2)where >100and >100and >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：
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每个特征列都完全匹配的情况下
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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 = and =
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_jp from table a left join table b on = ) 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：
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t的字段有：uid，goods_id，star。uid是用户id
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goodsid是商品id
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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 as uid1 , as uid2 ,a.goods_id , * as star_multi from t a left join t b on a.goods_id = b.goods_id and <> ) aa group by 1,2
select uid1,uid2,sum(multiply) as resultfrom(select as uid1, as uid2, goods_id,*star as multiplyfrom a left join b on a.goods_id = goods_idand <>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( 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 = left join( select id,num,row_number() over(partition by id order by num asc) nn from table) c on a.l_mid =
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 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 ) bbwhere number = (select count(distinct weekofyear(inserttime))from koo.nil_temp0222 where substring(inserttime,1,7) = '2019-12')