将连续的日期有效期合并在一起

Question

我有一系列记录，其中包含一些具有时间有效性的信息（产品类型）。

如果分组信息（产品类型）保持不变，我想将相邻的有效期合并在一起。我不能GROUP BY与MIN和一起使用简单的方法MAX，因为某些产品类型（A在示例中为）可以“消失”和“返回”。

使用Oracle 11g。

输入数据 ：

| PRODUCT |                       START_DATE |                         END_DATE |
|---------|----------------------------------|----------------------------------|
|       A |      July, 01 2013 00:00:00+0000 |      July, 31 2013 00:00:00+0000 |
|       A |    August, 01 2013 00:00:00+0000 |    August, 31 2013 00:00:00+0000 |
|       A | September, 01 2013 00:00:00+0000 | September, 30 2013 00:00:00+0000 |
|       B |   October, 01 2013 00:00:00+0000 |   October, 31 2013 00:00:00+0000 |
|       B |  November, 01 2013 00:00:00+0000 |  November, 30 2013 00:00:00+0000 |
|       A |  December, 01 2013 00:00:00+0000 |  December, 31 2013 00:00:00+0000 |
|       A |   January, 01 2014 00:00:00+0000 |   January, 31 2014 00:00:00+0000 |
|       A |  February, 01 2014 00:00:00+0000 |  February, 28 2014 00:00:00+0000 |
|       A |     March, 01 2014 00:00:00+0000 |     March, 31 2014 00:00:00+0000 |

预期结果 ：

| PRODUCT |                      START_DATE |                         END_DATE |
|---------|---------------------------------|----------------------------------|
|       A |     July, 01 2013 00:00:00+0000 | September, 30 2013 00:00:00+0000 |
|       B |  October, 01 2013 00:00:00+0000 |  November, 30 2013 00:00:00+0000 |
|       A | December, 01 2013 00:00:00+0000 |     March, 31 2014 00:00:00+0000 |

请参阅完整的SQL Fiddle。

Answer 1

这是一个孤岛问题。有多种方法可以解决此问题。这使用lead和lag分析功能：

select distinct product,
  case when start_date is null then lag(start_date)
    over (partition by product order by rn) else start_date end as start_date,
  case when end_date is null then lead(end_date)
    over (partition by product order by rn) else end_date end as end_date
from (
  select product, start_date, end_date, rn
  from (
    select t.product,
      case when lag(end_date)
          over (partition by product order by start_date) is null
        or lag(end_date)
          over (partition by product order by start_date) != start_date - 1
        then start_date end as start_date,
      case when lead(start_date)
          over (partition by product order by start_date) is null
        or lead(start_date)
          over (partition by product order by start_date) != end_date + 1
        then end_date end as end_date,
      row_number() over (partition by product order by start_date) as rn
    from t
  )
  where start_date is not null or end_date is not null
)
order by start_date, product;

PRODUCT START_DATE END_DATE
------- ---------- ---------
A       01-JUL-13  30-SEP-13 
B       01-OCT-13  30-NOV-13 
A       01-DEC-13  31-MAR-14

SQL小提琴

最里面的查询查看产品的前后记录，并且仅在记录不连续时才保留开始和/或结束时间：

select t.product,
  case when lag(end_date)
      over (partition by product order by start_date) is null
    or lag(end_date)
      over (partition by product order by start_date) != start_date - 1
    then start_date end as start_date,
  case when lead(start_date)
      over (partition by product order by start_date) is null
    or lead(start_date)
      over (partition by product order by start_date) != end_date + 1
    then end_date end as end_date
from t;

PRODUCT START_DATE END_DATE
------- ---------- ---------
A       01-JUL-13            
A                            
A                  30-SEP-13 
A       01-DEC-13            
A                            
A                            
A                  31-MAR-14 
B       01-OCT-13            
B                  30-NOV-13

select的下一个级别将删除那些处于中期的日期，其中两个日期都被内部查询所遮盖，从而得到：

PRODUCT START_DATE END_DATE
------- ---------- ---------
A       01-JUL-13            
A                  30-SEP-13 
A       01-DEC-13            
A                  31-MAR-14 
B       01-OCT-13            
B                  30-NOV-13

然后，外部查询会折叠那些相邻的对；我使用了创建重复项，然后使用消除重复项的简单distinct方法，但是您可以通过其他方式进行操作，例如将两个值都放入一对行中，并将两个值都保留为另一个空值，然后用另一个值消除它们选择层，但我认为在这里完全可以。

如果您的实际用例有时间，而不仅仅是日期，那么您需要在内部查询中调整比较；而不是+/-
1，可能是1秒的间隔，或者如果您愿意，则是1/86400，但取决于值的精度。