Server 2008 -查找艺术品的位置

Question

请考虑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

如何产生以下输出？

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

提供小提琴，这里。

解释

艺术品在不同的地方展出。有些地点安装了双面艺术品(如窗户)，有些地方则安装了单面艺术品(如墙壁)。例如，位置1的艺术品是双面的-它有赞助artwork 1和2-在位置5 (sponsorshipArtworkID 4)有单面的艺术品。

为了打印和安装的目的，我需要一个查询，以产生每件艺术品，无论是一面或两个，以及所有的位置与该作品。(请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：

• 打印双面艺术作品(1,2)，并安装在位置(1,2)；
• 打印单面艺术品(3)，并安装在位置(3,4)；
• 打印单面艺术品(4)并安装在位置(5)；
• 打印单面艺术品(1)并安装在位置(6)；
• 打印双面艺术作品(1,3)，并安装在位置(7)。

请注意，art有时会被重用，因此sponsorshipArtworkID 1在单面和双面位置都被使用.

我尝试使用DENSE_RANK()，一种递归的CTE，并按设置分治来解决这个问题，但到目前为止还未能解决。提前谢谢你的帮助。

Answer 1

它可能看起来有点难看，但想法很简单。

首先由sponsorshipLocationID组成组，并使用关联的sponsorshipArtworkID列表构建一个逗号分隔的字符串。

然后根据这个以逗号分隔的艺术品ID字符串计算密集等级。

在下面的查询中，我使用FOR XML连接字符串。这不是唯一的办法。有好的，快速的CLR函数写的，做它。

我建议一步一步地运行下面的查询，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);

查询

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

秩的实际值与预期结果不完全相同，但它们正确地对行进行了分组。