T- SQL оконные функции - подсчет пробелов

Question

Я хочу посчитать пробелы (столбцы, заполненные NULL) в окне v, но я не знаю, как.

IF OBJECT_ID('tempdb..#X') IS NOT NULL DROP TABLE #X;

CREATE TABLE #X
(
    ID INT IDENTITY(1,1) PRIMARY KEY,
    v INT
);

INSERT INTO #X
    SELECT 121 UNION ALL SELECT NULL UNION ALL SELECT NULL 
    UNION ALL SELECT 312 UNION ALL SELECT NULL UNION ALL SELECT NULL UNION ALL SELECT NULL
    UNION ALL SELECT 123 UNION ALL SELECT NULL UNION ALL SELECT NULL 
    UNION ALL SELECT 415 UNION ALL SELECT 416 UNION ALL SELECT NULL UNION ALL SELECT NULL UNION ALL SELECT NULL
    UNION ALL SELECT 200;


SELECT 
    ID, v, s, n, m, x, c
FROM
    (SELECT 
         ID, v, 
         MAX(v) OVER (PARTITION BY c) s,
         ROW_NUMBER() OVER (PARTITION BY c ORDER BY ID DESC) n,
         ROW_NUMBER() OVER (PARTITION BY c ORDER BY ID) - 1 m,
         COUNT(CASE WHEN v IS NULL THEN 1 END) OVER (PARTITION BY c) x,
         c
     FROM
         (SELECT 
              ID, v,
              -- c =  COUNT(CASE WHEN v IS NULL THEN 1 END) OVER (ORDER BY ID)
              c = COUNT(v) OVER (ORDER BY ID)
          FROM 
              #X) a
    ) a
ORDER BY 
    ID;

Расчет c близко, но он не работает на двух заполненных столбцах подряд.

Может кто-нибудь дать подсказку, пожалуйста?

---

Спасибо, ребята, теперь это работает для меня.

Answer 1

select *, case when v is null then count(null_grp_start) over (order by id) end null_grp_id
from   (select *,case when v is null and lag(v,1,1) over (order by id) is not null then 1 end as null_grp_start
        from   t
        ) t

-

+----+--------+----------+-------------+
| ID |   v    | null_grp | null_grp_id |
+----+--------+----------+-------------+
|  1 | 121    | (null)   | (null)      |
|  2 | (null) | 1        | 1           |
|  3 | (null) | (null)   | 1           |
|  4 | 312    | (null)   | (null)      |
|  5 | (null) | 1        | 2           |
|  6 | (null) | (null)   | 2           |
|  7 | (null) | (null)   | 2           |
|  8 | 123    | (null)   | (null)      |
|  9 | (null) | 1        | 3           |
| 10 | (null) | (null)   | 3           |
| 11 | 415    | (null)   | (null)      |
| 12 | 416    | (null)   | (null)      |
| 13 | (null) | 1        | 4           |
| 14 | (null) | (null)   | 4           |
| 15 | (null) | (null)   | 4           |
| 16 | 200    | (null)   | (null)      |
+----+--------+----------+-------------+

Answer 2

Если вы хотите перечислить периоды NULL-ness, вы можете использовать кумулятивный подсчет, чтобы получить ненулевые значения, а затем перечислить их:

select t.*,
       (case when v is null then dense_rank() over (partition by v order by null_grp)
        end) as newcolumn
from (select t.*,
             count(v) over (order by id) as null_grp
      from t
     ) t;

Здесь is дб <> скрипка.

Answer 3

declare @t table
(
    id int identity primary key clustered,
    val int
);


insert into @t(val)
values (null), (2), (3), (null), (5), (null), (null), (null), (9), (10), (null), (null), (null), (null), (15);


select *, case when first_value(val) over(order by id) is null then 1 else 0 end + case when val is null then sum(addme) over(order by id) end as null_group_ordinal
from
(
select *, case when lag(val) over(order by id) is null and val is not null then 1 else 0 end as addme
from @t
) as src;