对连续的空值进行排序

Question

我想对我的记录的连续空值进行排名。每条记录都会排名为1。对于只出现一次的空值，排名也为1。但是对于连续出现的空值，第一条记录排名为1，第二条记录排名为2记录等。这是我的代码。

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

预期结果

+----+-----------+------+
| 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    |
+----+-----------+------+

如何查询？

请帮忙！

Answer 1

您可以使用 ROW_NUMBER 的差异来获得连续 NULL 值的分组：

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

以下解决方案不使用递归（限于 32767 级 = ~ 行，具体取决于解决方案）并且它仅使用两个 agregate/ranking 函数（SUM 和 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

结果：

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

对连续的空值进行排序

Rank consecutive null values

sql-server

null

gaps-and-islands