2

考虑 SQL Server 2008 数据库中的下表和数据：

CREATE TABLE sponsorships 
(
    sponsorshipID INT NOT NULL PRIMARY KEY IDENTITY,
    sponsorshipLocationID INT NOT NULL,
    sponsorshipArtworkID INT NOT NULL
);

INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (1, 1);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (1, 2);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (2, 1);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (2, 2);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (3, 3);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (4, 3);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (5, 4);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (6, 1);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (7, 1);
INSERT INTO sponsorships (sponsorshipLocationID, sponsorshipArtworkID) 
VALUES (7, 3);

SELECT *
FROM sponsorships s
ORDER BY s.sponsorshipLocationID, s.sponsorshipArtworkID

我怎样才能产生以下输出？

CREATE TABLE sponGroups 
(
    rank INT,
    sponsorshipID INT,
    sponsorshipLocationID INT,
    sponsorshipArtworkID INT
);

INSERT INTO sponGroups VALUES (1, 1, 1, 1);
INSERT INTO sponGroups VALUES (1, 2, 1, 2);
INSERT INTO sponGroups VALUES (1, 3, 2, 1);
INSERT INTO sponGroups VALUES (1, 4, 2, 2);
INSERT INTO sponGroups VALUES (2, 5, 3, 3);
INSERT INTO sponGroups VALUES (2, 6, 4, 3);
INSERT INTO sponGroups VALUES (3, 7, 5, 4);
INSERT INTO sponGroups VALUES (4, 8, 6, 1);
INSERT INTO sponGroups VALUES (5, 9, 7, 1);
INSERT INTO sponGroups VALUES (5, 10, 7, 3);

SELECT *
FROM sponGroups sg
ORDER BY sg.rank, sg.sponsorshipID, sg.sponsorshipLocationID, sg.sponsorshipArtworkID

小提琴可用，在这里。

解释

艺术品展示在不同的地方。有些地方安装了双面艺术品（例如窗户），有些地方安装了单面艺术品（例如墙壁）。例如，位置 1 的艺术作品是双面的——它具有 SponsorshipArtworkID 1 和 2——而位置 5 的艺术作品是单面的（sponsorshipArtworkID 4）。

出于打印和安装目的，我需要一个查询来生成每件艺术品，无论是单面还是两面，以及与该作品相关的所有位置。（参考上面链接的小提琴中所需的输出。）例如，我需要告诉打印机：

  • 打印双面艺术品（1,2）并将其安装在位置（1,2）；
  • 打印单面图稿 (3) 并将其安装在位置 (3,4)；
  • 打印单面图稿（4）并安装在位置（5）；
  • 打印单面图稿（1）并安装在位置（6）；
  • 打印双面图稿 (1,3) 并将其安装在位置 (7)。

请注意，有时会重复使用艺术作品，因此在单面和双面位置都使用了赞助艺术作品 ID 1。

我尝试使用 DENSE_RANK()、递归 CTE 和按组除法来解决此问题，但目前还无法解决。在此先感谢您的帮助。

4

1 回答 1

1

它可能看起来有点难看，但想法很简单。

首先 group bysponsorshipLocationID并使用关联列表构建一个逗号分隔的字符串sponsorshipArtworkID。

然后根据这个以逗号分隔的 Artwork ID 字符串计算密集排名。

在下面的查询中，我使用FOR XML连接字符串。这不是唯一的方法。有很好的快速 CLR 函数可以做到这一点。

我建议逐步运行下面的查询，逐个 CTE 并检查中间结果以了解它是如何工作的。

样本数据

DECLARE @sponsorships TABLE
(
    sponsorshipID INT NOT NULL PRIMARY KEY IDENTITY,
    sponsorshipLocationID INT NOT NULL,
    sponsorshipArtworkID INT NOT NULL
);

INSERT INTO @sponsorships (sponsorshipLocationID, sponsorshipArtworkID) VALUES 
(1, 1),
(1, 2),
(2, 1),
(2, 2),
(3, 3),
(4, 3),
(5, 4),
(6, 1),
(7, 1),
(7, 3);

询问

WITH
CTE_Locations
AS
(
    SELECT
        sponsorshipLocationID
    FROM
        @sponsorships AS S
    GROUP BY
        sponsorshipLocationID
)
,CTE_Artworks
AS
(
    SELECT
        CTE_Locations.sponsorshipLocationID
        ,CA_Data.Artwork_Value
    FROM
        CTE_Locations
        CROSS APPLY
        (
            SELECT CAST(S.sponsorshipArtworkID AS varchar(10)) + ','
            FROM
                @sponsorships AS S
            WHERE
                S.sponsorshipLocationID = CTE_Locations.sponsorshipLocationID
            ORDER BY
                S.sponsorshipArtworkID
            FOR XML PATH(''), TYPE
        ) AS CA_XML(XML_Value)
        CROSS APPLY
        (
            SELECT CA_XML.XML_Value.value('.', 'NVARCHAR(MAX)')
        ) AS CA_Data(Artwork_Value)
)
,CTE_Rank
AS
(
    SELECT
        sponsorshipLocationID
        ,Artwork_Value
        ,DENSE_RANK() OVER (ORDER BY Artwork_Value) AS r
    FROM CTE_Artworks
)
SELECT
    CTE_Rank.r
    ,S.sponsorshipID
    ,CTE_Rank.sponsorshipLocationID
    ,S.sponsorshipArtworkID
FROM
    CTE_Rank
    INNER JOIN @sponsorships AS S 
        ON S.sponsorshipLocationID = CTE_Rank.sponsorshipLocationID
ORDER BY
    S.sponsorshipID
;

结果

+---+---------------+-----------------------+----------------------+
| r | sponsorshipID | sponsorshipLocationID | sponsorshipArtworkID |
+---+---------------+-----------------------+----------------------+
| 2 |             1 |                     1 |                    1 |
| 2 |             2 |                     1 |                    2 |
| 2 |             3 |                     2 |                    1 |
| 2 |             4 |                     2 |                    2 |
| 4 |             5 |                     3 |                    3 |
| 4 |             6 |                     4 |                    3 |
| 5 |             7 |                     5 |                    4 |
| 1 |             8 |                     6 |                    1 |
| 3 |             9 |                     7 |                    1 |
| 3 |            10 |                     7 |                    3 |
+---+---------------+-----------------------+----------------------+

rank 的实际值与您的预期结果并不完全相同，但它们对行进行了正确的分组。

于 2017-03-03T01:06:08.993 回答