Question

我有一个查询返回2列：

Product   | Price
------------------
Product01 | 10.00
Product02 | 10.00
Product03 | 10.00
Product04 | 10.00
Product05 | 10.00
Product06 | 10.00
Product07 | 10.00
Product08 | 10.00
Product09 | 10.00
Product10 | 10.00

我想做的就是以这种形式返回相同的数据：

Product1   | Price1 | Product2   | Price2 | Product3   | Price3
---------------------------------------------------------------
Product01  | 10.00  | Product05  | 10.00  | Product08  | 10.00
Product02  | 10.00  | Product06  | 10.00  | Product09  | 10.00
Product03  | 10.00  | Product07  | 10.00  | Product10  | 10.00
Product04  | 10.00  | NULL       | NULL   | NULL       | NULL

到目前为止，这是我的代码：

DECLARE @COLUMNS INT
SET @COLUMNS = 3

;
WITH CTE AS (
SELECT
 ROW_NUMBER() OVER ( PARTITION BY [COLUMN] ORDER BY ROW ASC ) AS ID_IN_GROUP
,*
FROM (
    SELECT
        ROW_NUMBER() OVER ( ORDER BY Name ASC ) AS ROW
       ,NTILE(@COLUMNS) OVER ( ORDER BY Name ASC ) AS [COLUMN]
       ,*
    FROM
      Products
) X
)
SELECT A.Name as Product1, A.Price as Price1, B.Name as Product2, B.Price as Price2, C.Name as Product3, C.Price as Price3 FROM
(SELECT Name,Price,ID_IN_GROUP FROM CTE WHERE [COLUMN]=1) A
CROSS APPLY
(SELECT Name,Price, ID_IN_GROUP FROM CTE WHERE [COLUMN]=2) B
CROSS APPLY
(SELECT Name,Price, ID_IN_GROUP FROM CTE WHERE [COLUMN]=3) C
WHERE 
B.ID_IN_GROUP=A.ID_IN_GROUP
AND C.ID_IN_GROUP=A.ID_IN_GROUP

http://sqlfiddle.com/#!18/bf3b4/30

它返回几乎想要的结果：

Product1   | Price1 | Product2   | Price2 | Product3   | Price3
---------------------------------------------------------------
Product01  | 10.00  | Product05  | 10.00  | Product08  | 10.00
Product02  | 10.00  | Product06  | 10.00  | Product09  | 10.00
Product03  | 10.00  | Product07  | 10.00  | Product10  | 10.00

但缺少最后一行-Product04。如何解决？

编辑：我已通过将CROSS APPLY更改为LEFT JOIN来解决此问题，

DECLARE @COLUMNS INT 
SET 
  @COLUMNS = 3;
WITH CTE AS (
  SELECT 
    ROW_NUMBER() OVER (
      PARTITION BY [COLUMN] 
      ORDER BY 
        ROW ASC
    ) AS ID_IN_GROUP, 
    * 
  FROM 
    (
      SELECT 
        ROW_NUMBER() OVER (
          ORDER BY 
            Name ASC
        ) AS ROW, 
        NTILE(@COLUMNS) OVER (
          ORDER BY 
            Name ASC
        ) AS [COLUMN], 
        * 
      FROM 
        Products
    ) X
) 
SELECT 
  A.Name as Product1, 
  A.Price as Price1, 
  B.Name as Product2, 
  B.Price as Price2, 
  C.Name as Product3, 
  C.Price as Price3 
FROM 
  (
    SELECT 
      Name, 
      Price, 
      ID_IN_GROUP 
    FROM 
      CTE 
    WHERE 
      [COLUMN] = 1
  ) A 
  LEFT JOIN (
    SELECT 
      Name, 
      Price, 
      ID_IN_GROUP 
    FROM 
      CTE 
    WHERE 
      [COLUMN] = 2
  ) B ON B.ID_IN_GROUP = A.ID_IN_GROUP 
  LEFT JOIN (
    SELECT 
      Name, 
      Price, 
      ID_IN_GROUP 
    FROM 
      CTE 
    WHERE 
      [COLUMN] = 3
  ) C ON C.ID_IN_GROUP = A.ID_IN_GROUP

http://sqlfiddle.com/#!18/bf3b4/61

这有可能使它更具动态性吗？因此，当我将@COLUMNS更改为4时，它将返回4组吗？

Answer 1

您要的东西比我想的要难。我还没有弄清楚逻辑，但这很接近：

SELECT MAX(CASE WHEN floor((seqnum - 1) / ceiling(cnt / 3.0)) = 0 THEN Name END) as Product1,
       MAX(CASE WHEN floor((seqnum - 1) / ceiling(cnt / 3.0)) = 0 THEN Price END) as Price1,
       MAX(CASE WHEN floor((seqnum - 1) / ceiling(cnt / 3.0)) = 1 THEN Name END) as Product2,
       MAX(CASE WHEN floor((seqnum - 1) / ceiling(cnt / 3.0)) = 1 THEN Price END) as Price2,
       MAX(CASE WHEN floor((seqnum - 1) / ceiling(cnt / 3.0)) = 2 THEN Name END) as Product3,
       MAX(CASE WHEN floor((seqnum - 1) / ceiling(cnt / 3.0)) = 2 THEN Price END) as Price3
FROM (SELECT p.*,
             ROW_NUMBER() OVER (ORDER BY Name) as seqnum,
             COUNT(*) OVER () as cnt
      FROM Products p
     ) p
GROUP BY (seqnum - 1) % ceiling(cnt / 3.0);

区别在于，这会“贪婪地”填充列，因此列是4-4-2而不是4-3-3。这种方法的一个优点是，如果添加第11行，则各列保持不变。

The SQL Fiddle.

将返回的行分成n组，并并排返回它们作为新列-动态列数

1 个答案: