SQL - 检查前一行，直到找到请求的值

Question

I've got a dataset like below:

ID   ReportingDate  Status
123  05/05/2020     GREEN
123  12/05/2020     NONE
123  19/05/2020     NONE
123  26/05/2020     AMBER
123  02/06/2020     RED
123  09/06/2020     NONE
123  16/06/2020     GREEN
123  23/06/2020     NONE
123  30/06/2020     AMBER

I want to ignore the NONE statuses and take the previous value, which may be previous row, but sometimes 2 or 3 rows before. Basically the final output should like the column "FINAL"

ID   ReportingDate  Status   FINAL
123  05/05/2020     GREEN    GREEN 
123  12/05/2020     NONE     GREEN
123  19/05/2020     NONE     GREEN
123  26/05/2020     AMBER    AMBER
123  02/06/2020     RED      RED
123  09/06/2020     NONE     RED
123  16/06/2020     GREEN    GREEN
123  23/06/2020     NONE     GREEN
123  30/06/2020     AMBER    AMBER

I've tried to use LAG() or LEAD() functions, but it doesn't work as requested.

UPPER
(
   CASE 
      WHEN psh.Status = 'NONE'
      THEN LAG(psh.Status,1,psh.Status) OVER
           (
              PARTITION BY psh.ID 
              ORDER BY rp.ReportingDate
           ) 
      ELSE psh.Status 
   END
) AS OverallStatusHistory

Could you advise me, if there is a way how to achieve it, please?

Many thanks!

Answer 1

You can assign groups based on the count of non- 'NONE' values. Then spread the value:

select t.*,
       max(nullif(status, 'NONE')) over (partition by id, grp) as imputed_status
from (select t.*,
             sum(case when status = 'NONE' then 0 else 1 end) over (partition by id order by reportingdate) as grp
      from t
     ) t;

Here is a db<>fiddle.

The SQL standard actually supports the IGNORE NULLS option on LAG() (and various other window functions). This would make it possible to solve this without a subquery. Unfortunately, SQL Server does not (yet???) support that functionality.