Rank consecutive null values - sql-server

Question

I want to rank consecutive null value for my records. Every record will be rank as 1. For the null value that only appear once, the rank will also be 1. But for the null values that appear in a consecutive way, the rank will be 1 for the first record and 2 for the second record and so on. Here's my code.
CREATE TABLE #my_table
(
id BIGINT IDENTITY PRIMARY KEY
,fruit varchar(100)
);
INSERT INTO #my_table
SELECT 'apple'
UNION ALL SELECT 'apple'
UNION ALL SELECT NULL
UNION ALL SELECT 'pineapple'
UNION ALL SELECT 'banana'
UNION ALL SELECT NULL
UNION ALL SELECT NULL
UNION ALL SELECT 'orange'
select * from #my_table
Intended result
+----+-----------+------+
| id | fruit | rank |
+----+-----------+------+
| 1 | apple | 1 |
| 2 | apple | 1 |
| 3 | NULL | 1 |
| 4 | pineapple | 1 |
| 5 | banana | 1 |
| 6 | NULL | 1 |
| 7 | NULL | 2 |
| 8 | orange | 1 |
+----+-----------+------+
How should I query it?
Please help!

Answer 1

You can use difference of ROW_NUMBER to get the grouping of consecutive NULL values:
WITH Cte AS(
SELECT *,
g = ROW_NUMBER() OVER(ORDER BY id)
- ROW_NUMBER() OVER(PARTITION BY fruit ORDER BY id)
FROM #my_table
)
SELECT
id,
fruit,
CASE
WHEN fruit IS NULL THEN ROW_NUMBER() OVER(PARTITION BY fruit, g ORDER BY id)
ELSE 1
END AS rank
FROM Cte
ORDER BY id;
ONLINE DEMO

Answer 2

CREATE TABLE #my_table
(
id BIGINT IDENTITY PRIMARY KEY
,fruit varchar(100)
);
INSERT INTO #my_table
SELECT 'apple'
UNION ALL SELECT 'apple'
UNION ALL SELECT NULL
UNION ALL SELECT 'pineapple'
UNION ALL SELECT 'banana'
UNION ALL SELECT NULL
UNION ALL SELECT NULL
UNION ALL SELECT 'orange'
;
WITH REC_CTE (id,fruit,ranks)
AS (
-- Anchor definition
SELECT id,
fruit,
1 as ranks
FROM #my_table
WHERE fruit is not null
-- Recursive definition
UNION ALL
SELECT son.id,
son.fruit,
case when son.fruit is null AND father.fruit is null then
father.ranks + 1
else
1
end as ranks
FROM #my_table son INNER JOIN
REC_CTE father
on son.id = father.id +1
WHERE son.fruit is null
--AND father.fruit is null
)
SELECT * from REC_CTE order by id
DROP TABLE #my_table

Answer 3

Following solution doesn't use recursion (limited to 32767 level = ~ rows depending on solution) and also it uses only two agregate/ranking functions (SUM and DENSE_RANK):
;WITH Base
AS (
SELECT *, IIF(fruit IS NULL, SUM(IIF(fruit IS NOT NULL, 1, 0)) OVER(ORDER BY id), NULL) AS group_num
FROM #my_table t
)
SELECT *, IIF(fruit IS NULL, DENSE_RANK() OVER(PARTITION BY group_num ORDER BY id), 1) rnk
FROM Base b
ORDER BY id
Results:
id fruit group_num rnk
--- --------- --------- ---
100 apple NULL 1
125 apple NULL 1
150 NULL 2 1
175 pineapple NULL 1
200 banana NULL 1
225 NULL 4 1
250 NULL 4 2
275 orange NULL 1
300 NULL 5 1
325 NULL 5 2
350 NULL 5 3