Question

我想为记录排名连续的空值。每个记录将排名为1.对于仅出现一次的空值，等级也将为1.但对于以连续方式出现的空值，第一个记录的等级将为1，第二个记录的等级为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    |
+----+-----------+------+

我该如何查询？

请帮忙！

Answer 1

您可以使用ROW_NUMBER的差异来获取连续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（限制为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

排名连续的空值

3 个答案: