SQL Server 2008-查找艺术品的位置

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

如何生成以下输出

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

解释

艺术品陈列在不同的地方。一些位置安装了双面艺术品（如窗户），一些位置安装了单面艺术品（如墙壁）。例如，位置1处的艺术品是双面的——它有sponsorshipArtworkID 1和2——而位置5处的艺术品是单面的（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，无论是单面还是双面，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。）因此，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）
• 打印单面艺术品（3）并将其安装在位置（3,4）
• 打印单面艺术品（4）并将其安装在位置（5）
• 打印单面艺术品（1）并将其安装在位置（6）
• 打印双面艺术品（1,3）并将其安装在位置（7）处

请注意，有时会重复使用art，因此在单面和双面位置都会使用sponsorshipArtworkID 1

---

我曾尝试使用DENSE_RANK（）（一种递归CTE）和set-wise divide来解决这个问题，但到目前为止还没有实现。提前感谢您的帮助。

这可能看起来有点难看，但想法很简单

首先按

sponsorshipLocationID

分组，并用关联的

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字符串计算稠密等级

在下面的查询中，我使用

FOR XML

连接字符串。这不是唯一的方法。有很好的快速CLR函数可以实现这一点

我建议一步一步地运行下面的查询，逐个CTE并检查中间结果，以了解其工作原理

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中的值不完全相同，但它们对行进行了正确分组。

我已经阅读了说明，但仍然不知道

rank

列应该是什么represent@SqlZim安装双面艺术品的位置指定了两件艺术品。例如，位置7分配了两件艺术品，因此它是一个双面位置。@Lamak“rank”列用于说明我想要的分组；我解释底部的项目符号列表直接对应于“排名”列。基本上，在本例中，我需要按顺序将每个项目符号/列组作为5个单独的请求发送给打印机/安装程序。我知道它代表您想要的结果。拉马克说：“我只是不知道该怎么解释，但很难用语言来表达这个问题。”。如果你能具体告诉我解释的哪一部分很难理解，那么我将尝试澄清问题的这一部分。谢谢弗拉基米尔，这确实实现了目标。看起来这完全是可以实现的（没有连接），但是这比我提出的任何方法都好。非常感谢您的帮助。@avejidah，我在这里的回答中也有类似的方法和解决方案。我用

CHECKSUM\u AGG

给出了一个近似解。