本套SQL题的答案是由许多小伙伴独特奉献的，1+1的力量是远远大于2的，有不少题目都采纳了十分奇妙的解法，也有不少题目有多种解法。本套大数据SQL题不仅题目丰盛多样，答案更是精彩绝伦！
注：以下参考答案都通过简略数据场景进行测试通过，但并未测试其余简单状况。本文档的SQL次要应用Hive SQL。
因内容较多，带目录的PDF查看是比拟不便的：
最强最全面的大数据SQL经典面试题残缺PDF版
一、行列转换
形容：表中记录了各年份各部门的均匀绩效考核问题。
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表名：t1
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表构造：
a -- 年份b -- 部门c -- 绩效得分表内容：
a b c2014 B 92015 A 82014 A 102015 B 7问题一：多行转多列
问题形容：将上述表内容转为如下输入后果所示：
a col_A col_B2014 10 92015 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_Bfrom t1group by a;问题二：如何将后果转成源表？（多列转多行）
问题形容：将问题一的后果转成源表，问题一后果表名为t1_2。
参考答案：
select a, b, cfrom ( 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，求多行转多列后果
问题形容：2014年公司组织架构调整，导致部门呈现多个绩效，业务及人员不同，无奈合并算绩效，源表内容如下：
2014 B 92015 A 82014 A 102015 B 72014 B 6输入后果如下所示：
a col_A col_B2014 10 6,92015 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_Bfrom ( select a, b, concat_ws(",",collect_set(cast(c as string))) as c from t1 group by a,b)tmpgroup by a;二、排名中取他值
表名：t2
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表字段及内容：
a b c2014 A 32014 B 12014 C 22015 A 42015 D 3问题一：按a分组取b字段最小时对应的c字段
输入后果如下所示：
a min_c2014 32015 4参考答案:
select a, c as min_cfrom( select a, b, c, row_number() over(partition by a order by b) as rn from t2 )awhere rn = 1;问题二：按a分组取b字段排第二时对应的c字段
输入后果如下所示：
a second_c2014 12015 3参考答案：
select a, c as second_cfrom( select a, b, c, row_number() over(partition by a order by b) as rn from t2 )awhere 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
输入后果如下所示：
a min_c max_c2014 3 22015 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_cfrom( 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 )awhere asc_rn = 1 or desc_rn = 1group 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
输入后果如下所示：
a min_c max_c2014 1 12015 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_cfrom ( 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 retwhere ret.rn_min = 2or ret.rn_max = 2group 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
留神：需放弃b字段最小、最大排首位
输入后果如下所示：
a min_c max_c2014 3,1 2,12015 4,3 3,4参考答案：
select tmp1.a as a, min_c, max_cfrom ( 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
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表字段及内容：
a b c2014 A 32014 B 12014 C 22015 A 42015 D 3问题一：按a分组按b字段排序，对c累计求和
输入后果如下所示：
a b sum_c2014 A 32014 B 42014 C 62015 A 42015 D 7参考答案：
select a, b, c, sum(c) over(partition by a order by b) as sum_cfrom t3; 问题二：按a分组按b字段排序，对c取累计平均值
输入后果如下所示：
a b avg_c2014 A 32014 B 22014 C 22015 A 42015 D 3.5参考答案：
select a, b, c, avg(c) over(partition by a order by b) as avg_cfrom t3;问题三：按a分组按b字段排序，对b取累计排名比例
输入后果如下所示：
a b ratio_c2014 A 0.332014 B 0.672014 C 1.002015 A 0.502015 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_cfrom 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
输入后果如下所示：
a b ratio_c2014 A 0.502014 B 0.672014 C 1.002015 A 0.572015 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_cfrom t3 order by a,b;四、窗口大小管制
表名：t4
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表字段及内容：
a b c2014 A 32014 B 12014 C 22015 A 42015 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
输入后果如下所示：
a b sum_c2014 A 12014 B 52014 C 12015 A 32015 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_cfrom t4;问题二：按a分组按b字段排序，对c取平均值
问题形容：前一行与以后行的均值！
输入后果如下所示：
a b avg_c2014 A 32014 B 22014 C 1.52015 A 42015 D 3.5参考答案：
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此处给出两种解法，其一：
select a, b, avg(c) over(partition by a order by b rows between 1 preceding and current row )fromt4;其二：
select a, b, case when lag_c is null then c else (c+lag_c)/2 end as avg_cfrom ( select a, b, c, lag(c,1) over(partition by a order by b) as lag_c from t4 )temp;五、产生间断数值
输入后果如下所示：
12345...100参考答案：
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不借助其余任何表面，实现产生间断数值
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此处给出两种解法，其一：
selectid_start+pos as idfrom( 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 idfrom (select split(space(99), ' ') as x) tlateral viewexplode(x) ex;那如何产生1至1000000间断数值？
参考答案：
select row_number() over() as idfrom (select split(space(999999), ' ') as x) tlateral viewexplode(x) ex;六、数据裁减与膨胀
表名：t6
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表字段及内容：
a324问题一：数据裁减
输入后果如下所示：
a b3 3、2、12 2、14 4、3、2、1参考答案：
select t.a, concat_ws('、',collect_set(cast(t.rn as string))) as bfrom( 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)可依据t6表的最大值灵便调整 lateral view explode(x) pe ) b on 1 = 1 where t6.a >= b.rn order by t6.a, b.rn desc ) tgroup by t.a;问题二：数据裁减，排除偶数
输入后果如下所示：
a b3 3、12 14 3、1参考答案：
select t.a, concat_ws('、',collect_set(cast(t.rn as string))) as bfrom( 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 ) tgroup by t.a;问题三：如何解决字符串累计拼接
问题形容：将小于等于a字段的值聚合拼接起来
输入后果如下所示：
a b3 2、32 24 2、3、4参考答案：
select t.a, concat_ws('、',collect_set(cast(t.a1 as string))) as bfrom( 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 ) tgroup by t.a;问题四：如果a字段有反复，如何实现字符串累计拼接
输入后果如下所示：
a b2 23 2、33 2、3、34 2、3、3、4参考答案：
select a, bfrom ( 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：数据开展
问题形容：如何将字符串"1-5,16,11-13,9"扩大成"1,2,3,4,5,16,11,12,13,9"？留神程序不变。
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)可灵便调整 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
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表字段及内容：
a b2014 A2014 B2015 B2015 D问题一：合并
输入后果如下所示：
2014 A、B2015 B、D参考答案:
select a, concat_ws('、', collect_set(t.b)) bfrom t7group by a;问题二：拆分
问题形容：将分组合并的后果拆分进去
参考答案：
select t.a, dfrom( select a, concat_ws('、', collect_set(t7.b)) b from t7 group by a)tlateral view explode(split(t.b, '、')) table_tmp as d;八、模仿循环操作
表名：t8
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表字段及内容：
a10110101问题一：如何将字符'1'的地位提取进去
输入后果如下所示:
1,3,42,4参考答案：
select a, concat_ws(",",collect_list(cast(index as string))) as resfrom ( 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") tmp2group by a;九、不应用distinct或group by去重
表名：t9
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表字段及内容：
a b c d2014 2016 2014 A2014 2015 2015 B问题一：不应用distinct或group by去重
输入后果如下所示：
2014 A2016 A2014 B2015 B参考答案：
select t2.year ,t2.numfrom ( 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)t2where rank_1=1order by num;十、容器--反转内容
表名：t10
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表字段及内容：
aAB,CA,BADBD,EA问题一：反转逗号分隔的数据：扭转程序，内容不变
输入后果如下所示：
BAD,CA,ABEA,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) tmp1group by a;问题二：反转逗号分隔的数据：扭转内容，程序不变
输入后果如下所示：
BA,AC,DABDB,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) tmp1group by a;十一、多容器--成对提取数据
表名：t11
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表字段及内容：
a bA/B 1/3B/C/D 4/5/2问题一：成对提取数据，字段一一对应
输入后果如下所示：
a bA 1B 3B 4C 5D 2参考答案:
select a_inx, b_inxfrom ( 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) tmpwhere a_id=b_id;十二、多容器--转多行
表名：t12
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表字段及内容：
a b c001 A/B 1/3/5002 B/C/D 4/5问题一：转多行
输入后果如下所示：
a d e001 type_b A001 type_b B001 type_c 1001 type_c 3001 type_c 5002 type_b B002 type_b C002 type_b D002 type_c 4002 type_c 5参考答案:
select a, d, efrom ( 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) tmporder by a,d;十三、形象分组--断点排序
表名：t13
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表字段及内容：
a b2014 12015 12016 12017 02018 02019 -12020 -12021 -12022 12023 1问题一：断点排序
输入后果如下所示：
a b c 2014 1 12015 1 22016 1 32017 0 12018 0 22019 -1 12020 -1 22021 -1 32022 1 12023 1 2参考答案:
select a, b, row_number() over( partition by b,repair_a order by a asc) as c--依照b列和[b的组首]分组，排序from ( select a, b, a-b_rn as repair_a--依据b列值呈现的秩序,修复a列值为b首次呈现的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,失去b列各值呈现的秩序 from t13 )tmp1)tmp2--留神，如果不同的b列值，可能呈现同样的组首值，但组首值须要和a列值 一并参加分组，故并不影响排序。order by a asc; 十四、业务逻辑的分类与形象--时效
日期表：d_date
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表字段及内容：
date_id is_work2017-04-13 12017-04-14 12017-04-15 02017-04-16 02017-04-17 1工作日：周一至周五09:30-18:30
客户申请表：t14
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表字段及内容：
a b c1 申请 2017-04-14 18:03:001 通过 2017-04-17 09:43:002 申请 2017-04-13 17:02:002 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
输入后果如下所示：
a d1 0.67h2 10.67h 参考答案:
select a, round(sum(diff)/3600,2) as dfrom ( 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) tmp6group by a;十五、工夫序列--进度及残余
表名：t15
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表字段及内容：
date_id is_work2017-07-30 02017-07-31 12017-08-01 12017-08-02 12017-08-03 12017-08-04 12017-08-05 02017-08-06 02017-08-07 1问题一：求每天的累计周工作日，残余周工作日
输入后果如下所示：
date_id week_to_work week_left_work2017-07-31 1 42017-08-01 2 32017-08-02 3 22017-08-03 4 12017-08-04 5 02017-08-05 5 02017-08-06 5 0参考答案:
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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_workfrom t15其二：
selectdate_id,week_to_work,week_sum_work-week_to_work as week_left_workfrom( 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、工夫序列--结构日期
问题一：间接应用SQL实现一张日期维度表，蕴含以下字段：
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 周完结日d_month int 第几月m_start string 月开始日m_end string 月完结日d_quarter int 第几季q_start string 季开始日q_end string 季完结日d_year int 年份y_start string 年开始日y_end string 年完结日参考答案：
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 '周完结日', d_month string comment '第几月', m_start string comment '月开始日', m_end string comment '月完结日', d_quarter int comment '第几季', q_start string comment '季开始日', q_end string comment '季完结日', d_year int comment '年份', y_start string comment '年开始日', y_end string comment '年完结日');--天然月: 指每月的1号到那个月的月底，它是依照阳历来计算的。就是从每月1号到月底，不论这个月有30天，31天，29天或者28天，都算是一个天然月。insert overwrite table dim_dateselect `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 -- 节令 , 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 --季完结日 -- 年 , d_year , y_start , y_endfrom ( 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 --获取节令 能够间接应用 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 --年完结日 , date_sub(`date`, dayofmonth(`date`) - 1) as m_start --当月第一天 , last_day(date_sub(`date`, dayofmonth(`date`) - 1)) m_end --当月最初一天 , 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、工夫序列--结构累积日期
表名：t17
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表字段及内容：
date_id2017-08-012017-08-022017-08-03问题一：每一日期，都扩大成月初至当天
输入后果如下所示：
date_id date_to_day2017-08-01 2017-08-012017-08-02 2017-08-012017-08-02 2017-08-022017-08-03 2017-08-012017-08-03 2017-08-022017-08-03 2017-08-03这种累积相干的表，常做桥接表。
参考答案:
select date_id, date_add(date_start_id,pos) as date_to_dayfrom( 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、工夫序列--结构间断日期
表名：t18
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表字段及内容：
a b c101 2018-01-01 10101 2018-01-03 20101 2018-01-06 40102 2018-01-02 20102 2018-01-04 30102 2018-01-07 60问题一：结构间断日期
问题形容：将表中数据的b字段裁减至范畴[2018-01-01, 2018-01-07]，并累积对c求和。
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b字段的值是较稠密的。
输入后果如下所示：
a b c d101 2018-01-01 10 10101 2018-01-02 0 10101 2018-01-03 20 30101 2018-01-04 0 30101 2018-01-05 0 30101 2018-01-06 40 70101 2018-01-07 0 70102 2018-01-01 0 0102 2018-01-02 20 20102 2018-01-03 0 20102 2018-01-04 30 50102 2018-01-05 0 50102 2018-01-06 0 50102 2018-01-07 60 110参考答案:
select a, b, c, sum(c) over(partition by a order by b) as dfrom( 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、工夫序列--取多个字段最新的值
表名：t19
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表字段及内容：
date_id a b c2014 AB 12 bc2015 23 2016 d2017 BC 问题一：如何一并取出最新日期
输入后果如下所示：
date_a a date_b b date_c c2017 BC 2015 23 2016 d参考答案:
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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 cFROM ( 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 ) tWHERE t.rn_a = 1OR t.rn_b = 1OR t.rn_c = 1;其二：
SELECT a.date_id ,a.a ,b.date_id ,b.b ,c.date_id ,c.cFROM( 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) aLEFT 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) cON 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) t1cross 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) t2cross 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、工夫序列--补全数据
表名：t20
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表字段及内容：
date_id a b c2014 AB 12 bc2015 23 2016 d2017 BC 问题一：如何应用最新数据补全表格
输入后果如下所示：
date_id a b c2014 AB 12 bc2015 AB 23 bc2016 AB 23 d2017 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 cfrom( 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、工夫序列--取最新实现状态的前一个状态
表名：t21
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表字段及内容：
date_id a b2014 1 A2015 1 B2016 1 A2017 1 B2013 2 A2014 2 B2015 2 A2014 3 A2015 3 A2016 3 B2017 3 A上表中B为实现状态。
问题一：取最新实现状态的前一个状态
输入后果如下所示：
date_id a b2016 1 A2013 2 A2015 3 A参考答案:
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此处给出两种解法，其一：
select t21.date_id, t21.a, t21.bfrom ( 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_idand t1.a = t21.a;其二：
select next_date_id as date_id ,a ,next_b as bfrom( 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) twhere minb = nk and b = 'B';问题二：如何将实现状态的过程合并
输入后果如下所示:
a b_merge1 A、B、A、B2 A、B3 A、A、B参考答案：
select a ,collect_list(b) as bfrom( 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) twhere nk >= minbgroup by a;二十二、非等值连贯--范畴匹配
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相当于拉链过的变动维，但日期范畴可能是不全的。
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：范畴匹配
输入后果如下所示：
date_id p_id p_value 2017 C null 2018 B 2 2019 A null 2013 C 3*参考答案*：
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此处给出两种解法，其一：
select f.date_id, f.p_id, A.p_valuefrom 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)AON f.date_id = A.date_id;其二：
select date_id, p_id, flag as p_valuefrom ( 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) tmpwhere d_end = max_end;二十三、非等值连贯--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班的每个学生问题在t23_2班中找出问题最靠近的问题。
表t23_1：a中无反复值
a1245810表t23_2：b中无反复值
b2371113问题一：单向最近匹配
输入后果如下所示：
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留神：b的值可能会被抛弃
a b1 22 24 35 35 78 710 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
假如表A为事件流水表，客户当天有一条记录则视为当天沉闷。
表A：
time_id user_id2018-01-01 10:00:00 0012018-01-01 11:03:00 0022018-01-01 13:18:00 0012018-01-02 08:34:00 0042018-01-02 10:08:00 0022018-01-02 10:40:00 0032018-01-02 14:21:00 0022018-01-02 15:39:00 0042018-01-03 08:34:00 0052018-01-03 10:08:00 0032018-01-03 10:40:00 0012018-01-03 14:21:00 005假如客户沉闷十分，一天产生的事件记录均匀达千条。
问题一：累计去重
输入后果如下所示：
日期 当日沉闷人数 月累计沉闷人数_截至当日date_id user_cnt_act user_cnt_act_month2018-01-01 2 22018-01-02 3 42018-01-03 3 5参考答案：
SELECT tt1.date_id ,tt2.user_cnt_act ,tt1.user_cnt_act_monthFROM( -- ④ 依照t.date_id分组求出user_cnt_act_month，失去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，依照a.date_id,b.user_id分组，保障截止到当日的用户惟一，得出表t。 SELECT a.date_id ,b.user_id FROM ( -- ① 依照日期分组，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 依照date_id、user_id分组，保障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) tt1LEFT JOIN( -- ⑥ 依照date_id分组求出user_cnt_act，失去tt2 SELECT date_id ,COUNT(user_id) AS user_cnt_act FROM ( -- ⑤ 依照日期分组，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) tt2ON tt2.date_id = tt1.date_id参考：
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