hiveSQL面试题14__业务逻辑的分类与抽象--时效
2021/12/24 19:07:09
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目录
- 0- 描述
- 1- 问题一
0- 描述
描述: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
1- 问题一
描述: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;
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