SQL在Redshift中拆分日期范围

Question

我有一张日期范围的表格

----------------------------------------------------------------
|     id    | date_start               | date_end              |
----------------------------------------------------------------
|     1     | 2017-02-03 08:00:00.000 | 2017-02-03 17:00:00.000|
|     2     | 2017-02-04 15:00:00.000 | 2017-02-05 10:00:00.000|
|     3     | 2017-02-06 14:00:00.000 | 2017-02-07 23:00:00.000|
----------------------------------------------------------------

正如您所看到的，某些范围可以涵盖超过1天的时间段（例如＃2，＃3），我需要将这些记录分开几天才能获得如下结果：

----------------------------------------------------------------
|     id    | date_start               | date_end              |
----------------------------------------------------------------
|     1     | 2017-02-03 08:00:00.000 | 2017-02-03 17:00:00.000|
|     2     | 2017-02-04 15:00:00.000 | 2017-02-04 23:59:59.999|
|     2     | 2017-02-05 00:00:00.000 | 2017-02-05 10:00:00.000|
|     3     | 2017-02-06 14:00:00.000 | 2017-02-06 23:59:59.999|
|     3     | 2017-02-07 00:00:00.000 | 2017-02-07 23:00:00.000|
----------------------------------------------------------------

如何在Redshift上使用SQL？

Answer 1

1. 我创建了一个数字列表来创建1天范围。在这种情况下，我创造了100天
2. 我只为调试保留了很多列
3. 创建1天范围new_start和new_end
4. 基本情况是第一个间隔，如果两个日期都在同一天，那么你不需要改变任何东西
5. 现在在第一个间隔我选择原始date_start相同的最后一个间隔我使用date_end
6. 其余的我使用new_start和new_end - 1 second
7. 只有一天范围与原始范围重叠

SQL DEMO

WITH days as  (
    SELECT a.n
    from generate_series(1, 100) as a(n)
), ranges as (
    SELECT *, (d.n::text || ' DAY')::interval as i,
           t1.date_start::date + ((d.n - 1)::text || ' DAY')::interval as new_start,
           t1.date_start::date + (d.n::text || ' DAY')::interval as new_end,

           CASE WHEN t1.date_start::date = t1.date_end::date AND d.n = 1
                THEN t1.date_start    
                WHEN t1.date_start::date < t1.date_end::date 
                THEN t1.date_start
                ELSE NULL
           END as date_start1,
           CASE WHEN t1.date_start::date = t1.date_end::date AND d.n = 1
                THEN t1.date_end
           END  date_end1
    FROM Table1 t1
    CROSS JOIN days d
)
SELECT *, CASE WHEN date_start < new_end AND date_end > new_start
               THEN 'overlap'
          END as overlap,
          CASE WHEN date_end1 IS NOT NULL 
               THEN date_start1
               WHEN date_start < new_end AND date_end > new_start
               THEN CASE WHEN date_start > new_start
                         THEN date_start
                         ELSE new_start
                    END
          END as final_start,

          CASE WHEN date_end1 IS NOT NULL 
               THEN date_end1
               WHEN date_start < new_end AND date_end > new_start
               THEN CASE WHEN date_end < new_end
                         THEN date_end
                         ELSE new_end - '1 second'::interval
                    END
          END as final_end                      
FROM ranges
WHERE date_start < new_end AND date_end > new_start
ORDER BY "id", new_start

OUTPUT

Answer 2

最后，我已经这样做了。连续工作时间最长为2天（即2017-12-02开始，2017-12-04结束 - 不会在此数据集中进行; 2017-12-02 - 2017-12-03还可以） 。

 -- Select 1-st day's interval for two-days sessions:
     SELECT sessions.date_start
            ,DATE_TRUNC('day',sessions.date_end) as date_end
       FROM sessions
      WHERE DATEDIFF(day,sessions.date_start,sessions.date_end) = 1


     UNION ALL
     -- Select 2-nd day's interval for two-days sessions:
     SELECT DATE_TRUNC('day',sessions.date_end) as date_start
            ,sessions.date_end as date_end
       FROM sessions
      WHERE DATEDIFF(day, sessions.date_start, sessions.date_end) = 1

    UNION ALL
    -- Select one-day sessions:
     SELECT sessions.date_start as date_start
            ,sessions.date_end as date_end
       FROM sessions
      WHERE DATEDIFF(day, sessions.date_start, sessions.date_end) = 0