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

最强最全面的大数据SQL经典面试题完整PDF版

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

最强最全面的大数据SQL经典面试题完整PDF版

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