如何找到按日期排序的两个连续行，包含特定值？

Question

I have a table with the following structure and data in it:

| ID    | Date          | Result    |
|----   |------------   |--------   |
| 1     | 30/04/2020    | +         |
| 1     | 01/05/2020    | -         |
| 1     | 05/05/2020    | -         |
| 2     | 03/05/2020    | -         |
| 2     | 04/05/2020    | +         |
| 2     | 05/05/2020    | -         |
| 2     | 06/05/2020    | -         |
| 3     | 01/05/2020    | -         |
| 3     | 02/05/2020    | -         |
| 3     | 03/05/2020    | -         |
| 3     | 04/05/2020    | -         |

I'm trying to write an SQL query (I'm using SQL Server) which returns the date of the first two consecutive negative results for a given ID. For example, for ID no. 1, the first two consecutive negative results are on 01/05 and 05/05. The first two consecutive results for ID No. 2 are on 05/05 and 06/05. The first two consecutive negative results for ID No. 3 are on on 01/05 and 02/05.

So the query should produce the following result:

| ID    | FirstNegativeDate     |
|----   |-------------------    |
| 1     | 01/05                 |
| 2     | 05/05                 |
| 3     | 01/05                 |

Please note that the dates aren't necessarily one day apart. Sometimes, two consecutive negative tests may be several days apart. But they should still be considered as "consecutive negative tests". In other words, two negative tests are not 'consecutive' only if there is a positive test result in between them.

How can this be done in SQL? I've done some reading and it looks like maybe the PARTITION BY statement is required but I'm not sure how it works.

Answer 1

This is a gaps-and-island problem, where you want the start of the first island of '-' s that contains at least two rows.

I would recommend lead() and aggregation:

select id, min(date) first_negative_date
from (
    select t.*, lead(result) over(partition by id order by date) lead_result
    from mytable t
) t
where result = '-' and lead_result = '-'
group by id

Answer 2

Use LEAD or LAG functions over ID partition ordered by your Date column. Then simple check where LEAD/LAG column is equal to Result. You'll need also to filter the top ones.

The image attached just shows what LEAD/LAG would return