注：以下参考答案都经过简单数据场景进行测试通过，但并未测试其他复杂情况。本文档的SQL主要使用Hive SQL。
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
这种累积相关的表，常做桥接表。
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
表t23_1和表t23_2通过a和b关联时，有相等的取相等的值匹配，不相等时每一个a的值在b中找差值最小的来匹配。
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
