最强最全面的大数据SQL经典面试题（由31位大佬共同协作完成）
Posted 园陌
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本套SQL题的答案是由许多小伙伴共同贡献的，1+1的力量是远远大于2的，有不少题目都采用了非常巧妙的解法，也有不少题目有多种解法。本套大数据SQL题不仅题目丰富多样，答案更是精彩绝伦！
因内容较多，带目录的PDF查看是比较方便的：
一、行列转换
描述：表中记录了各年份各部门的平均绩效考核成绩。
\\
表名：t1
\\
表结构：
a -- 年份
b -- 部门
c -- 绩效得分表内容：
a b c
2014 B 9
2015 A 8
2014 A 10
2015 B 7问题一：多行转多列
问题描述：将上述表内容转为如下输出结果所示：
a col_A col_B
2014 10 9
2015 8 7参考答案：
select
a,
max(case when b="A" then c end) col_A,
max(case when b="B" then c end) col_B
from t1
group by a;问题二：如何将结果转成源表？（多列转多行）
问题描述：将问题一的结果转成源表，问题一结果表名为t1_2。
参考答案：
select
a,
b,
c
from (
select a,"A" as b,col_a as c from t1_2
union all
select a,"B" as b,col_b as c from t1_2
)tmp; 问题三：同一部门会有多个绩效，求多行转多列结果
问题描述：2014年公司组织架构调整，导致部门出现多个绩效，业务及人员不同，无法合并算绩效，源表内容如下：
2014 B 9
2015 A 8
2014 A 10
2015 B 7
2014 B 6输出结果如下所示：
a col_A col_B
2014 10 6,9
2015 8 7参考答案:
select
a,
max(case when b="A" then c end) col_A,
max(case when b="B" then c end) col_B
from (
select
a,
b,
concat_ws(",",collect_set(cast(c as string))) as c
from t1
group by a,b
)tmp
group by a;二、排名中取他值
表名：t2
\\
表字段及内容：
a b c
2014 A 3
2014 B 1
2014 C 2
2015 A 4
2015 D 3问题一：按a分组取b字段最小时对应的c字段
输出结果如下所示：
a min_c
2014 3
2015 4参考答案:
select
a,
c as min_c
from
(
select
a,
b,
c,
row_number() over(partition by a order by b) as rn
from t2
)a
where rn = 1;问题二：按a分组取b字段排第二时对应的c字段
输出结果如下所示：
a second_c
2014 1
2015 3参考答案：
select
a,
c as second_c
from
(
select
a,
b,
c,
row_number() over(partition by a order by b) as rn
from t2
)a
where rn = 2;问题三：按a分组取b字段最小和最大时对应的c字段
输出结果如下所示：
a min_c max_c
2014 3 2
2015 4 3参考答案:
select
a,
min(if(asc_rn = 1, c, null)) as min_c,
max(if(desc_rn = 1, c, null)) as max_c
from
(
select
a,
b,
c,
row_number() over(partition by a order by b) as asc_rn,
row_number() over(partition by a order by b desc) as desc_rn
from t2
)a
where asc_rn = 1 or desc_rn = 1
group by a; 问题四：按a分组取b字段第二小和第二大时对应的c字段
输出结果如下所示：
a min_c max_c
2014 1 1
2015 3 4参考答案：
select
ret.a
,max(case when ret.rn_min = 2 then ret.c else null end) as min_c
,max(case when ret.rn_max = 2 then ret.c else null end) as max_c
from (
select
*
,row_number() over(partition by t2.a order by t2.b) as rn_min
,row_number() over(partition by t2.a order by t2.b desc) as rn_max
from t2
) as ret
where ret.rn_min = 2
or ret.rn_max = 2
group by ret.a;问题五：按a分组取b字段前两小和前两大时对应的c字段
注意：需保持b字段最小、最大排首位
输出结果如下所示：
a min_c max_c
2014 3,1 2,1
2015 4,3 3,4参考答案：
select
tmp1.a as a,
min_c,
max_c
from
(
select
a,
concat_ws(,, collect_list(c)) as min_c
from
(
select
a,
b,
c,
row_number() over(partition by a order by b) as asc_rn
from t2
)a
where asc_rn <= 2
group by a
)tmp1
join
(
select
a,
concat_ws(,, collect_list(c)) as max_c
from
(
select
a,
b,
c,
row_number() over(partition by a order by b desc) as desc_rn
from t2
)a
where desc_rn <= 2
group by a
)tmp2
on tmp1.a = tmp2.a; 三、累计求值
表名：t3
\\
表字段及内容：
a b c
2014 A 3
2014 B 1
2014 C 2
2015 A 4
2015 D 3问题一：按a分组按b字段排序，对c累计求和
输出结果如下所示：
a b sum_c
2014 A 3
2014 B 4
2014 C 6
2015 A 4
2015 D 7参考答案：
select
a,
b,
c,
sum(c) over(partition by a order by b) as sum_c
from t3; 问题二：按a分组按b字段排序，对c取累计平均值
输出结果如下所示：
a b avg_c
2014 A 3
2014 B 2
2014 C 2
2015 A 4
2015 D 3.5参考答案：
select
a,
b,
c,
avg(c) over(partition by a order by b) as avg_c
from t3;问题三：按a分组按b字段排序，对b取累计排名比例
输出结果如下所示：
a b ratio_c
2014 A 0.33
2014 B 0.67
2014 C 1.00
2015 A 0.50
2015 D 1.00参考答案：
select
a,
b,
c,
round(row_number() over(partition by a order by b) / (count(c) over(partition by a)),2) as ratio_c
from t3
order by a,b;问题四：按a分组按b字段排序，对b取累计求和比例
输出结果如下所示：
a b ratio_c
2014 A 0.50
2014 B 0.67
2014 C 1.00
2015 A 0.57
2015 D 1.00参考答案：
select
a,
b,
c,
round(sum(c) over(partition by a order by b) / (sum(c) over(partition by a)),2) as ratio_c
from t3
order by a,b;四、窗口大小控制
表名：t4
\\
表字段及内容：
a b c
2014 A 3
2014 B 1
2014 C 2
2015 A 4
2015 D 3问题一：按a分组按b字段排序，对c取前后各一行的和
输出结果如下所示：
a b sum_c
2014 A 1
2014 B 5
2014 C 1
2015 A 3
2015 D 4参考答案：
select
a,
b,
lag(c,1,0) over(partition by a order by b)+lead(c,1,0) over(partition by a order by b) as sum_c
from t4;问题二：按a分组按b字段排序，对c取平均值
问题描述：前一行与当前行的均值！
输出结果如下所示：
a b avg_c
2014 A 3
2014 B 2
2014 C 1.5
2015 A 4
2015 D 3.5参考答案：
select
a,
b,
case when lag_c is null then c
else (c+lag_c)/2 end as avg_c
from
(
select
a,
b,
c,
lag(c,1) over(partition by a order by b) as lag_c
from t4
)temp;五、产生连续数值
输出结果如下所示：
1
2
3
4
5
...
100参考答案：
\\
不借助其他任何外表，实现产生连续数值
\\
此处给出两种解法，其一：
select
id_start+pos as id
from(
select
1 as id_start,
1000000 as id_end
) m lateral view posexplode(split(space(id_end-id_start), )) t as pos, val其二：
select
row_number() over() as id
from
(select split(space(99), ) as x) t
lateral view
explode(x) ex;那如何产生1至1000000连续数值？
参考答案：
select
row_number() over() as id
from
(select split(space(999999), ) as x) t
lateral view
explode(x) ex;六、数据扩充与收缩
表名：t6
\\
表字段及内容：
a
3
2
4问题一：数据扩充
输出结果如下所示：
a b
3 3、2、1
2 2、1
4 4、3、2、1参考答案：
select
t.a,
concat_ws(、,collect_set(cast(t.rn as string))) as b
from
(
select
t6.a,
b.rn
from t6
left join
(
select
row_number() over() as rn
from
(select split(space(5), ) as x) t -- space(5)可根据t6表的最大值灵活调整
lateral view
explode(x) pe
) b
on 1 = 1
where t6.a >= b.rn
order by t6.a, b.rn desc
) t
group by t.a;问题二：数据扩充，排除偶数
输出结果如下所示：
a b
3 3、1
2 1
4 3、1参考答案：
select
t.a,
concat_ws(、,collect_set(cast(t.rn as string))) as b
from
(
select
t6.a,
b.rn
from t6
left join
(
select
row_number() over() as rn
from
(select split(space(5), ) as x) t
lateral view
explode(x) pe
) b
on 1 = 1
where t6.a >= b.rn and b.rn % 2 = 1
order by t6.a, b.rn desc
) t
group by t.a;问题三：如何处理字符串累计拼接
问题描述：将小于等于a字段的值聚合拼接起来
输出结果如下所示：
a b
3 2、3
2 2
4 2、3、4参考答案：
select
t.a,
concat_ws(、,collect_set(cast(t.a1 as string))) as b
from
(
select
t6.a,
b.a1
from t6
left join
(
select a as a1
from t6
) b
on 1 = 1
where t6.a >= b.a1
order by t6.a, b.a1
) t
group by t.a;问题四：如果a字段有重复，如何实现字符串累计拼接
输出结果如下所示：
a b
2 2
3 2、3
3 2、3、3
4 2、3、3、4参考答案：
select
a,
b
from
(
select
t.a,
t.rn,
concat_ws(、,collect_list(cast(t.a1 as string))) as b
from
(
select
a.a,
a.rn,
b.a1
from
(
select
a,
row_number() over(order by a ) as rn
from t6
) a
left join
(
select a as a1,
row_number() over(order by a ) as rn
from t6
) b
on 1 = 1
where a.a >= b.a1 and a.rn >= b.rn
order by a.a, b.a1
) t
group by t.a,t.rn
order by t.a,t.rn
) tt; 问题五：数据展开
问题描述：如何将字符串"1-5,16,11-13,9"扩展成"1,2,3,4,5,16,11,12,13,9"？注意顺序不变。
参考答案：
select
concat_ws(,,collect_list(cast(rn as string)))
from
(
select
a.rn,
b.num,
b.pos
from
(
select
row_number() over() as rn
from (select split(space(20), ) as x) t -- space(20)可灵活调整
lateral view
explode(x) pe
) a lateral view outer
posexplode(split(1-5,16,11-13,9, ,)) b as pos, num
where a.rn between cast(split(num, -)[0] as int) and cast(split(num, -)[1] as int) or a.rn = num
order by pos, rn
) t;七、合并与拆分
表名：t7
\\
表字段及内容：
a b
2014 A
2014 B
2015 B
2015 D问题一：合并
输出结果如下所示：
2014 A、B
2015 B、D参考答案:
select
a,
concat_ws(、, collect_set(t.b)) b
from t7
group by a;问题二：拆分
问题描述：将分组合并的结果拆分出来
参考答案：
select
t.a,
d
from
(
select
a,
concat_ws(、, collect_set(t7.b)) b
from t7
group by a
)t
lateral view
explode(split(t.b, 、)) table_tmp as d;八、模拟循环操作
表名：t8
\\
表字段及内容：
a
1011
0101问题一：如何将字符1的位置提取出来
输出结果如下所示:
1,3,4
2,4参考答案：
select
a,
concat_ws(",",collect_list(cast(index as string))) as res
from (
select
a,
index+1 as index,
chr
from (
select
a,
concat_ws(",",substr(a,1,1),substr(a,2,1),substr(a,3,1),substr(a,-1)) str
from t8
) tmp1
lateral view posexplode(split(str,",")) t as index,chr
where chr = "1"
) tmp2
group by a;九、不使用distinct或group by去重
表名：t9
\\
表字段及内容：
a b c d
2014 2016 2014 A
2014 2015 2015 B问题一：不使用distinct或group by去重
输出结果如下所示：
2014 A
2016 A
2014 B
2015 B参考答案：
select
t2.year
,t2.num
from
(
select
*
,row_number() over (partition by t1.year,t1.num) as rank_1
from
(
select
a as year,
d as num
from t9
union all
select
b as year,
d as num
from t9
union all
select
c as year,
d as num
from t9
)t1
)t2
where rank_1=1
order by num;十、容器--反转内容
表名：t10
\\
表字段及内容：
a
AB,CA,BAD
BD,EA问题一：反转逗号分隔的数据：改变顺序，内容不变
输出结果如下所示：
BAD,CA,AB
EA,BD参考答案：
select
a,
concat_ws(",",collect_list(reverse(str)))
from
(
select
a,
str
from t10
lateral view explode(split(reverse(a),",")) t as str
) tmp1
group by a;问题二：反转逗号分隔的数据：改变内容，顺序不变
输出结果如下所示：
BA,AC,DAB
DB,AE参考答案：
select
a,
concat_ws(",",collect_list(reverse(str)))
from
(
select
a,
str
from t10
lateral view explode(split(a,",")) t as str
) tmp1
group by a;十一、多容器--成对提取数据
表名：t11
\\
表字段及内容：
a b
A/B 1/3
B/C/D 4/5/2问题一：成对提取数据，字段一一对应
输出结果如下所示：
a b
A 1
B 3
B 4
C 5
D 2参考答案:
select
a_inx,
b_inx
from
(
select
a,
b,
a_id,
a_inx,
b_id,
b_inx
from t11
lateral view posexplode(split(a,/)) t as a_id,a_inx
lateral view posexplode(split(b,/)) t as b_id,b_inx
) tmp
where a_id=b_id;十二、多容器--转多行
表名：t12
\\
表字段及内容：
a b c
001 A/B 1/3/5
002 B/C/D 4/5问题一：转多行
输出结果如下所示：
a d e
001 type_b A
001 type_b B
001 type_c 1
001 type_c 3
001 type_c 5
002 type_b B
002 type_b C
002 type_b D
002 type_c 4
002 type_c 5参考答案:
select
a,
d,
e
from
(
select
a,
"type_b" as d,
str as e
from t12
lateral view explode(split(b,"/")) t as str
union all
select
a,
"type_c" as d,
str as e
from t12
lateral view explode(split(c,"/")) t as str
) tmp
order by a,d;十三、抽象分组--断点排序
表名：t13
\\
表字段及内容：
a b
2014 1
2015 1
2016 1
2017 0
2018 0
2019 -1
2020 -1
2021 -1
2022 1
2023 1问题一：断点排序
输出结果如下所示：
a b c
2014 1 1
2015 1 2
2016 1 3
2017 0 1
2018 0 2
2019 -1 1
2020 -1 2
2021 -1 3
2022 1 1
2023 1 2参考答案:
select
a,
b,
row_number() over( partition by b,repair_a order by a asc) as c--按照b列和[b的组首]分组，排序
from
(
select
a,
b,
a-b_rn as repair_a--根据b列值出现的次序,修复a列值为b首次出现的a列值,称为b的[组首]
from
(
select
a,
b,
row_number() over( partition by b order by a asc ) as b_rn--按b列分组,按a列排序,得到b列各值出现的次序
from t13
)tmp1
)tmp2--注意，如果不同的b列值，可能出现同样的组首值，但组首值需要和a列值 一并参与分组，故并不影响排序。
order by a asc; 十四、业务逻辑的分类与抽象--时效
日期表：d_date
\\
表字段及内容：
date_id is_work
2017-04-13 1
2017-04-14 1
2017-04-15 0
2017-04-16 0
2017-04-17 1工作日：周一至周五09:30-18:30
客户申请表：t14
\\
表字段及内容：
a b c
1 申请 2017-04-14 18:03:00
1 通过 2017-04-17 09:43:00
2 申请 2017-04-13 17:02:00
2 通过 2017-04-15 09:42:00问题一：计算上表中从申请到通过占用的工作时长
输出结果如下所示：
a d
1 0.67h
2 10.67h 参考答案:
select
a,
round(sum(diff)/3600,2) as d
from (
select
a,
apply_time,
pass_time,
dates,
rn,
ct,
is_work,
case when is_work=1 and rn=1 then unix_timestamp(concat(dates, 18:30:00),yyyy-MM-dd HH:mm:ss)-unix_timestamp(apply_time,yyyy-MM-dd HH:mm:ss)
when is_work=0 then 0
when is_work=1 and rn=ct then unix_timestamp(pass_time,yyyy-MM-dd HH:mm:ss)-unix_timestamp(concat(dates, 09:30:00),yyyy-MM-dd HH:mm:ss)
when is_work=1 and rn!=ct then 9*3600
end diff
from (
select
a,
apply_time,
pass_time,
time_diff,
day_diff,
rn,
ct,
date_add(start,rn-1) dates
from (
select
a,
apply_time,
pass_time,
time_diff,
day_diff,
strs,
start,
row_number() over(partition by a) as rn,
count(*) over(partition by a) as ct
from (
select
a,
apply_time,
pass_time,
time_diff,
day_diff,
substr(repeat(concat(substr(apply_time,1,10),,),day_diff+1),1,11*(day_diff+1)-1) strs
from (
select
a,
apply_time,
pass_time,
unix_timestamp(pass_time,yyyy-MM-dd HH:mm:ss)-unix_timestamp(apply_time,yyyy-MM-dd HH:mm:ss) time_diff,
datediff(substr(pass_time,1,10),substr(apply_time,1,10)) day_diff
from (
select
a,
max(case when b=申请 then c end) apply_time,
max(case when b=通过 then c end) pass_time
from t14
group by a
) tmp1
) tmp2
) tmp3
lateral view explode(split(strs,",")) t as start
) tmp4
) tmp5
join d_date
on tmp5.dates = d_date.date_id
) tmp6
group by a;十五、时间序列--进度及剩余
表名：t15
\\
表字段及内容：
date_id is_work
2017-07-30 0
2017-07-31 1
2017-08-01 1
2017-08-02 1
2017-08-03 1
2017-08-04 1
2017-08-05 0
2017-08-06 0
2017-08-07 1问题一：求每天的累计周工作日，剩余周工作日
输出结果如下所示：
date_id week_to_work week_left_work
2017-07-31 1 4
2017-08-01 2 3
2017-08-02 3 2
2017-08-03 4 1
2017-08-04 5 0
2017-08-05 5 0
2017-08-06 5 0参考答案:
\\
此处给出两种解法，其一：
select
date_id
,case date_format(date_id,u)
when 1 then 1
when 2 then 2
when 3 then 3
when 4 then 4
when 5 then 5
when 6 then 5
when 7 then 5
end as week_to_work
,case date_format(date_id,u)
when 1 then 4
when 2 then 3
when 3 then 2
when 4 then 1
when 5 then 0
when 6 then 0
when 7 then 0
end as week_to_work
from t15其二：
select
date_id,
week_to_work,
week_sum_work-week_to_work as week_left_work
from(
select
date_id,
sum(is_work) over(partition by year,week order by date_id) as week_to_work,
sum(is_work) over(partition by year,week) as week_sum_work
from(
select
date_id,
is_work,
year(date_id) as year,
weekofyear(date_id) as week
from t15
) ta
) tb order by date_id;十六、时间序列--构造日期
问题一：直接使用SQL实现一张日期维度表，包含以下字段：
date string 日期
d_week string 年内第几周
weeks int 周几
w_start string 周开始日
w_end string 周结束日
d_month int 第几月
m_start string 月开始日
m_end string 月结束日
d_quarter int 第几季
q_start string 季开始日
q_end string 季结束日
d_year int 年份
y_start string 年开始日
y_end string 年结束日参考答案：
drop table if exists dim_date;
create table if not exists dim_date(
`date` string comment 日期,
d_week string comment 年内第几周,
weeks string comment 周几,
w_start string comment 周开始日,
w_end string comment 周结束日,
d_month string comment 第几月,
m_start string comment 月开始日,
m_end string comment 月结束日,
d_quarter int comment 第几季,
q_start string comment 季开始日,
q_end string comment 季结束日,
d_year int comment 年份,
y_start string comment 年开始日,
y_end string comment 年结束日
);
--自然月: 指每月的1号到那个月的月底，它是按照阳历来计算的。就是从每月1号到月底，不管这个月有30天，31天，29天或者28天，都算是一个自然月。
insert overwrite table dim_date
select `date`
, d_week --年内第几周
, case weekid
when 0 then 周日
when 1 then 周一
when 2 then 周二
when 3 then 周三
when 4 then 周四
when 5 then 周五
when 6 then 周六
end as weeks -- 周
, date_add(next_day(`date`,MO),-7) as w_start --周一
, date_add(next_day(`date`,MO),-1) as w_end -- 周日_end
-- 月份日期
, concat(第, monthid, 月) as d_month
, m_start
, m_end
-- 季节
, quarterid as d_quart
, concat(d_year, -, substr(concat(0, (quarterid - 1) * 3 + 1), -2), -01) as q_start --季开始日
, date_sub(concat(d_year, -, substr(concat(0, (quarterid) * 3 + 1), -2), -01), 1) as q_end --季结束日
-- 年
, d_year
, y_start
, y_end
from (
select `date`
, pmod(datediff(`date`, 2012-01-01), 7) as weekid --获取周几
, cast(substr(`date`, 6, 2) as int) as monthid --获取月份
, case
when cast(substr(`date`, 6, 2) as int) <= 3 then 1
when cast(substr(`date`, 6, 2) as int) <= 6 then 2
when cast(substr(`date`, 6, 2) as int) <= 9 then 3
when cast(substr(`date`, 6, 2) as int) <= 12 then 4
end as quarterid --获取季节 可以直接使用 quarter(`date`)
, substr(`date`, 1, 4) as d_year -- 获取年份
, trunc(`date`, YYYY) as y_start --年开始日
, date_sub(trunc(add_months(`date`, 12), YYYY), 1) as y_end --年结束日
, date_sub(`date`, dayofmonth(`date`) - 1) as m_start --当月第一天
, last_day(date_sub(`date`, dayofmonth(`date`) - 1)) m_end --当月最后一天
, weekofyear(`date`) as d_week --年内第几周
from (
-- 2021-04-01是开始日期, 2022-03-31是截止日期
select date_add(2021-04-01, t0.pos) as `date`
from (
select posexplode(
split(
repeat(o, datediff(
from_unixtime(unix_timestamp(2022-03-31, yyyy-mm-dd),
yyyy-mm-dd),
2021-04-01)), o
)
)
) t0
) t1
) t2;十七、时间序列--构造累积日期
表名：t17
\\
表字段及内容：
date_id
2017-08-01
2017-08-02
2017-08-03问题一：每一日期，都扩展成月初至当天
输出结果如下所示：
date_id date_to_day
2017-08-01 2017-08-01
2017-08-02 2017-08-01
2017-08-02 2017-08-02
2017-08-03 2017-08-01
2017-08-03 2017-08-02
2017-08-03 2017-08-03参考答案:
select
date_id,
date_add(date_start_id,pos) as date_to_day
from
(
select
date_id,
date_sub(date_id,dayofmonth(date_id)-1) as date_start_id
from t17
) m lateral view
posexplode(split(space(datediff(from_unixtime(unix_timestamp(date_id,yyyy-MM-dd)),from_unixtime(unix_timestamp(date_start_id,yyyy-MM-dd)))), )) t as pos, val;十八、时间序列--构造连续日期
表名：t18
\\
表字段及内容：
a b c
101 2018-01-01 10
101 2018-01-03 20
101 2018-01-06 40
102 2018-01-02 20
102 2018-01-04 30
102 2018-01-07 60问题一：构造连续日期
问题描述：将表中数据的b字段扩充至范围[2018-01-01, 2018-01-07]，并累积对c求和。
\\
b字段的值是较稀疏的。
输出结果如下所示：
a b c d
101 2018-01-01 10 10
101 2018-01-02 0 10
101 2018-01-03 20 30
101 2018-01-04 0 30
101 2018-01-05 0 30
101 2018-01-06 40 70
101 2018-01-07 0 70
102 2018-01-01 0 0
102 2018-01-02 20 20
102 2018-01-03 0 20
102 2018-01-04 30 50
102 2018-01-05 0 50
102 2018-01-06 0 50
102 2018-01-07 60 110参考答案:
select
a,
b,
c,
sum(c) over(partition by a order by b) as d
from
(
select
t1.a,
t1.b,
case
when t18.b is not null then t18.c
else 0
end as c
from
(
select
a,
date_add(s,pos) as b
from
(
select
a,
2018-01-01 as s,
2018-01-07 as r
from (select a from t18 group by a) ta
) m lateral view
posexplode(split(space(datediff(from_unixtime(unix_timestamp(r,yyyy-MM-dd)),from_unixtime(unix_timestamp(s,yyyy-MM-dd)))), )) t as pos, val
) t1
left join t18
on t1.a = t18.a and t1.b = t18.b
) ts;十九、时间序列--取多个字段最新的值
表名：t19
\\
表字段及内容：
date_id a b c
2014 AB 12 bc
2015 23
2016 d
2017 BC 问题一：如何一并取出最新日期
输出结果如下所示：
date_a a date_b b date_c c
2017 BC 2015 23 2016 d参考答案:
\\
此处给出三种解法，其一：
SELECT max(CASE WHEN rn_a = 1 THEN date_id else 0 END) AS date_a
,max(CASE WHEN rn_a = 1 THEN a else null END) AS a
,max(CASE WHEN rn_b = 1 THEN date_id else 0 END) AS date_b
,max(CASE WHEN rn_b = 1 THEN b else NULL END) AS b
,max(CASE WHEN rn_c = 1 THEN date_id else 0 END) AS date_c
,max(CASE WHEN rn_c = 1 THEN c else null END) AS c
FROM (
SELECT date_id
,a
,b
,c
--对每列上不为null的值 的 日期 进行排序
,row_number()OVER( PARTITION BY 1 ORDER BY CASE WHEN a IS NULL THEN 0 ELSE date_id END DESC) AS rn_a
,row_number()OVER(PARTITION BY 1 ORDER BY CASE WHEN b IS NULL THEN 0 ELSE date_id END DESC) AS rn_b
,row_number()OVER(PARTITION BY 1 ORDER BY CASE WHEN c IS NULL THEN 0 ELSE date_id END DESC) AS rn_c
FROM t19
) t
WHERE t.rn_a = 1
OR t.rn_b = 1
OR t.rn_c = 1;其二：
SELECT
a.date_id
,a.a
,b.date_id
,b.b
,c.date_id
,c.c
FROM
(
SELECT
t.date_id,
t.a
FROM
(
SELECT
t.date_id
,t.a
,t.b
,t.c
FROM t19 t INNER JOIN t19 t1 ON t.date_id = t1.date_id AND t.a IS NOT NULL
) t
ORDER BY t.date_id DESC
LIMIT 1
) a
LEFT JOIN
(
SELECT
t.date_id
,t.b
FROM
(
SELECT
t.date_id
,t.b
FROM t19 t INNER JOIN t19 t1 ON t.date_id = t1.date_id AND t.b IS NOT NULL
) t
ORDER BY t.date_id DESC
LIMIT 1
) b ON 1 = 1
LEFT JOIN
(
SELECT
t.date_id
,t.c
FROM
(
SELECT
t.date_id
,t.c
FROM t19 t INNER JOIN t19 t1 ON t.date_id = t1.date_id AND t.c IS NOT NULL
) t
ORDER BY t.date_id DESC
LIMIT 1
) c
ON 1 = 1;其三：
select
*
from
(
select t1.date_id as date_a,t1.a from (select t1.date_id,t1.a from t19 t1 where t1.a is not null) t1
inner join (select max(t1.date_id) as date_id from t19 t1 where t1.a is not null) t2
on t1.date_id=t2.date_id
) t1
cross join
(
select t1.date_b,t1.b from (select t1.date_id as date_b,t1.b from t19 t1 where t1.b is not null) t1
inner join (select max(t1.date_id) as date_id from t19 t1 where t1.b is not null)t2
on t1.date_b=t2.date_id
) t2
cross join
(
select t1.date_c,t1.c from (select t1.date_id as date_c,t1.c from t19 t1 where t1.c is not null) t1
inner join (select max(t1.date_id) as date_id from t19 t1 where t1.c is not null)t2
on t1.date_c=t2.date_id
) t3;二十、时间序列--补全数据
表名：t20
\\
表字段及内容：
date_id a b c
2014 AB 12 bc
2015 23
2016 d
2017 BC 问题一：如何使用最新数据补全表格
输出结果如下所示：
date_id a b c
2014 AB 12 bc
2015 AB 23 bc
2016 AB 23 d
2017 BC 23 d参考答案:
select
date_id,
first_value(a) over(partition by aa order by date_id) as a,
first_value(b) over(partition by bb order by date_id) as b,
first_value(c) over(partition by cc order by date_id) as c
from
(
select
date_id,
a,
b,
c,
count(a) over(order by date_id) as aa,
count(b) over(order by date_id) as bb,
count(c) over(order by date_id) as cc
from t20
)tmp1;二十一、时间序列--取最新完成状态的前一个状态
表名：t21
\\
表字段及内容：
date_id a b
2014 1 A
2015 1 B
2016 1 A
2017 1 B
2013 2 A
2014 2 B
2015 2 A
2014 3 A
2015 3 A
2016 3 B
2017 3 A上表中B为完成状态。
问题一：取最新完成状态的前一个状态
输出结果如下所示：
date_id a b
2016 1 A
2013 2 A
2015 3 A参考答案:
\\
此处给出两种解法，其一：
select
t21.date_id,
t21.a,
t21.b
from
(
select
max(date_id) date_id,
a
from
t21
where
b = B
group by
a
) t1
inner join t21 on t1.date_id -1 = t21.date_id
and t1.a = t21.a;其二：
select
next_date_id as date_id
,a
,next_b as b
from(
select
*,min(nk) over(partition by a,b) as minb
from(
select
*,row_number() over(partition by a order by date_id desc) nk
,lead(date_id) over(partition by a order by date_id desc) next_date_id
,lead(b) over(partition by a order by date_id desc) next_b
from(
select * from t21
) t
) t
) t
where minb = nk and b = B;问题二：如何将完成状态的过程合并
输出结果如下所示:
a b_merge
1 A、B、A、B
2 A、B
3 A、A、B参考答案：
select
a
,collect_list(b) as b
from(
select
*
,min(if(b = B,nk,null)) over(partition by a) as minb
from(
select
*,row_number() over(partition by a order by date_id desc) nk
from(
select * from t21
) t
) t
) t
where nk >= minb
group by a;二十二、非等值连接--范围匹配
表f是事实表，表d是匹配表，在hive中如何将匹配表中的值关联到事实表中？
表d相当于拉链过的变化维，但日期范围可能是不全的。
表f：
date_id p_id
2017 C
2018 B
2019 A
2013 C表d：
d_start d_end p_id p_value
2016 2018 A 1
2016 2018 B 2
2008 2009 C 4
2010 2015 C 3问题一：范围匹配
输出结果如下所示：
date_id p_id p_value
2017 C null
2018 B 2
2019 A null
2013 C 3**参考答案：
\\
此处给出两种解法，其一：
select
f.date_id,
f.p_id,
A.p_value
from f
left join
(
select
date_id,
p_id,
p_value
from
(
select
f.date_id,
f.p_id,
d.p_value
from f
left join d on f.p_id = d.p_id
where f.date_id >= d.d_start and f.date_id <= d.d_end
)A
)A
ON f.date_id = A.date_id;其二：
select
date_id,
p_id,
flag as p_value
from (
select
f.date_id,
f.p_id,
d.d_start,
d.d_end,
d.p_value,
if(f.date_id between d.d_start and d.d_end,d.p_value,null) flag,
max(d.d_end) over(partition by date_id) max_end
from f
left join d
on f.p_id = d.p_id
) tmp
where d_end = max_end;二十三、非等值连接--最近匹配
t23_1和t23_2为两个班的成绩单，t23_1班的每个学生成绩在t23_2班中找出成绩最接近的成绩。
表t23_1：a中无重复值
a
1
2
4
5
8
10表t23_2：b中无重复值
b
2
3
7
11
13问题一：单向最近匹配
输出结果如下所示：
\\
注意：b的值可能会被丢弃
a b
1 2
2 2
4 3
5 3
5 7
8 7
10 11参考答案：
select
*
from
(
select
ttt1.a,
ttt1.b
from
(
select
tt1.a,
t23_2.b,
dense_rank() over(partition by tt1.a order by abs(tt1.a-t23_2.b)) as dr
from
(
select
t23_1.a
from t23_1
left join t23_2 on t23_1.a=t23_2.b
where t23_2.b is null
) tt1
cross join t23_2
) ttt1
where ttt1.dr=1
union all
select
t23_1.a,
t23_2.b
from t23_1
inner join t23_2 on t23_1.a=t23_2.b
) result_t
order by result_t.a;二十四、N指标--累计去重
假设表A为事件流水表，客户当天有一条记录则视为当天活跃。
表A：
time_id user_id
2018-01-01 10:00:00 001
2018-01-01 11:03:00 002
2018-01-01 13:18:00 001
2018-01-02 08:34:00 004
2018-01-02 10:08:00 002
2018-01-02 10:40:00 003
2018-01-02 14:21:00 002
2018-01-02 15:39:00 004
2018-01-03 08:34:00 005
2018-01-03 10:08:00 003
2018-01-03 10:40:00 001
2018-01-03 14:21:00 005假设客户活跃非常，一天产生的事件记录平均达千条。
问题一：累计去重
输出结果如下所示：
日期 当日活跃人数 月累计活跃人数_截至当日
date_id user_cnt_act user_cnt_act_month
2018-01-01 2 2
2018-01-02 3 4
2018-01-03 3 5参考答案：
SELECT tt1.date_id
,tt2.user_cnt_act
,tt1.user_cnt_act_month
FROM
( -- ④ 按照t.date_id分组求出user_cnt_act_month，得到tt1
SELECT t.date_id
,COUNT(user_id) AS user_cnt_act_month
FROM
( -- ③ 表a和表b进行笛卡尔积，按照a.date_id,b.user_id分组，保证截止到当日的用户唯一，得出表t。
SELECT a.date_id
,b.user_id
FROM
( -- ① 按照日期分组，取出date_id字段当主表的维度字段 得出表a
SELECT from_unixtime(unix_timestamp(time_id),yyyy-MM-dd) AS date_id
FROM test.temp_tanhaidi_20211213_1
GROUP BY from_unixtime(unix_timestamp(time_id),yyyy-MM-dd)
) a
INNER JOIN
( -- ② 按照date_id、user_id分组，保证每天每个用户只有一条记录，得出表b
SELECT from_unixtime(unix_timestamp(time_id),yyyy-MM-dd) AS date_id
,user_id
FROM test.temp_tanhaidi_20211213_1
GROUP BY from_unixtime(unix_timestamp(time_id),yyyy-MM-dd)
,user_id
) b
ON 1 = 1
WHERE a.date_id >= b.date_id
GROUP BY a.date_id
,b.user_id
) t
GROUP BY t.date_id
) tt1
LEFT JOIN
( -- ⑥ 按照date_id分组求出user_cnt_act，得到tt2
SELECT date_id
,COUNT(user_id) AS user_cnt_act
FROM
( -- ⑤ 按照日期分组，取出date_id字段当主表的维度字段 得出表a
SELECT from_unixtime(unix_timestamp(time_id),yyyy-MM-dd) AS date_id
,user_id
FROM test.temp_tanhaidi_20211213_1
GROUP BY from_unixtime(unix_timestamp(time_id),yyyy-MM-dd)
,user_id
) a
GROUP BY date_id
) tt2
ON tt2.date_id = tt1.date_id参考：
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