Question

我正在尝试解决以下问题。

我觉得这是可能的，但我似乎无法得到它。

以下是该方案：

Table 1 (Assets)
1 Asset-A
2 Asset-B
3 Asset-C
4 Asset-D

Table 2 (Attributes)
1 Asset-A Red
2 Asset-A Hard
3 Asset-B Red
4 Asset-B Hard
5 Asset-B Heavy
6 Asset-C Blue
7 Asset-C Hard

如果我正在寻找具有与Asset-A相同属性的东西，那么它应该识别Asset-B，因为Asset-B具有与Asset-A相同的所有属性（它应该丢弃重，因为Asset-A没有' t指定任何不同或类似的东西）。另外，如果我只想要Asset-A和Asset-B的属性，那么我该如何获得呢？

看似简单，但我无法指出......

我使用的实际表几乎就是Table2，只是AssetId和AttributeId的关联，所以： PK：Id
int：AssetId
int：AttributeId

我只包含了资产表的概念来简化问题。

Answer 1

SELECT  ato.id, ato.value
FROM    (
        SELECT  id
        FROM    assets a
        WHERE   NOT EXISTS
                (
                SELECT  NULL
                FROM    attributes ata
                LEFT  JOIN
                        attributes ato
                ON      ato.id = ata.id
                        AND ato.value = ata.value
                WHERE   ata.id = 1
                        AND ato.id IS NULL
                )
        ) ao
JOIN    attributes ato
ON      ato.id = ao.id
JOIN    attributes ata
ON      ata.id = 1
        AND ata.value = ato.value

，或SQL Server 2005（包含要检查的样本数据）：

WITH    assets AS 
        (
        SELECT 1 AS id, 'A' AS name
        UNION ALL
        SELECT 2 AS id, 'B' AS name
        UNION ALL
        SELECT 3 AS id, 'C' AS name
        UNION ALL
        SELECT 4 AS id, 'D' AS name
        ),
        attributes AS
        (
        SELECT 1 AS id, 'Red' AS value
        UNION ALL
        SELECT 1 AS id, 'Hard' AS value
        UNION ALL
        SELECT 2 AS id, 'Red' AS value
        UNION ALL
        SELECT 2 AS id, 'Hard' AS value
        UNION ALL
        SELECT 2 AS id, 'Heavy' AS value
        UNION ALL
        SELECT 3 AS id, 'Blue' AS value
        UNION ALL
        SELECT 3 AS id, 'Hard' AS value
        )
SELECT  ato.id, ato.value
FROM    (
        SELECT  id
        FROM    assets a
        WHERE   a.id <> 1
                AND NOT EXISTS
                (
                SELECT  ata.value
                FROM    attributes ata
                WHERE   ata.id = 1
                EXCEPT
                SELECT  ato.value
                FROM    attributes ato
                WHERE   ato.id = a.id
                )
        ) ao
JOIN    attributes ato
ON      ato.id = ao.id
JOIN    attributes ata
ON      ata.id = 1
        AND ata.value = ato.value

Answer 2

我不完全理解您问题的第一部分，根据属性识别资产。

对列名进行一些假设，以下查询将产生Asset-A和Asset-B之间的公共属性：

SELECT [Table 2].Name
FROM [Table 2]
JOIN [Table 1] a ON a.ID = [Table 2].AssetID AND a.Name = 'Asset-A'
JOIN [Table 1] b ON b.ID = [Table 2].AssetID AND b.Name = 'Asset-B'
GROUP BY [Table 2].Name

Answer 3

 Select * From Assets A
    Where Exists 
      (Select * From Assets
       Where AssetId <> A.AssetID
          And (Select Count(*)
               From Attributes At1 Join Attributes At2
                  On At1.AssetId <> At2.AssetId
                      And At1.attribute <> At2.Attribute
               Where At1.AssetId = A.AssetId Asset) = 0 )
    And AssetId = 'Asset-A'

Answer 4

select at2.asset, count(*)
from       attribute at1
inner join attribute at2 on at1.value = at2.value
where at1.asset =  "Asset-A"
and   at2.asset != "Asset-A"
group by at2.asset
having count(*) = (select count(*) from attribute where asset = "Asset-A");

Answer 5

查找具有“A”所有属性的所有资产（但也可能具有其他属性）：

SELECT Other.ID
FROM Assets Other
WHERE
  Other.AssetID <> 'Asset-A'   -- do not return Asset A as a match to itself
  AND NOT EXISTS (SELECT NULL FROM Attributes AttA WHERE
    AttA.AssetID='Asset-A' 
    AND NOT EXISTS (SELECT NULL FROM Attributes AttOther WHERE
      AttOther.AssetID=Other.ID AND AttOther.AttributeID = AttA.AttributeID
      )
    )

即，“找到任何没有A属性且不属于此资产属性的资产”。

查找与“A”具有完全相同属性的所有资产：

SELECT Other.ID
FROM Assets Other
WHERE
  Other.AssetID <> 'Asset-A'   -- do not return Asset A as a match to itself
  AND NOT EXISTS (SELECT NULL FROM Attributes AttA WHERE
    AttA.AssetID='Asset-A' 
    AND NOT EXISTS (SELECT NULL FROM Attributes AttOther WHERE
      AttOther.AssetID=Other.ID 
      AND AttOther.AttributeID = AttA.AttributeID
      )
    ) 
  AND NOT EXISTS (SELECT NULL FROM Attributes AttaOther WHERE
    AttaOther.AssetID=Other.ID 
    AND NOT EXISTS (SELECT NULL FROM Attributes AttaA WHERE
      AttaA.AssetID='Asset-A' 
      AND AttaA.AttributeID = AttaOther.AttributeID
      )
   )

即，“找到任何没有A属性且不属于此资产的属性的资产，以及此资产的属性不属于A的属性。”

Answer 6

此解决方案按照规定运行，感谢输入。

WITH Atts AS 
(
    SELECT
    DISTINCT
        at1.[Attribute]
    FROM
        Attribute at1
    WHERE
        at1.[Asset] = 'Asset-A'
)

SELECT 
    DISTINCT
    Asset,
    (
        SELECT 
            COUNT(ta2.[Attribute]) 
        FROM 
            Attribute ta2 
        INNER JOIN
            Atts b 
            ON
                b.[Attribute] = ta2.[attribute]
        WHERE 
            ta2.[Asset] = ta.Asset
    ) 
    AS [Count]
FROM 
    Atts a
INNER JOIN
    Attribute ta
    ON
    a.[Attribute] = ta.[Attribute]

Answer 7

查找与asset-a具有所有相同属性的所有资产：

select att2.Asset from attribute att1
inner join attribute att2 on att2.Attribute = att1.Attribute and att1.Asset <> att2.Asset
where att1.Asset = 'Asset-A'
group by att2.Asset, att1.Asset
having COUNT(*) = (select COUNT(*) from attribute where Asset=att1.Asset)

Answer 8

我想也许我可以用LINQ做到这一点然后用我的方式向下工作：

var result = from productsNotA in DevProducts
             where  productsNotA.Product != "A" && 
            (
                from productsA in DevProducts
                where productsA.Product == "A"
                select productsA.Attribute
            ).Except
            (
                from productOther in DevProducts
                where productOther.Product == productsNotA.Product
                select productOther.Attribute
            ).Single() == null
            select new {productsNotA.Product};

result.Distinct()

我认为使用LinqPad将其转换回SQL会产生一个漂亮的SQL查询。但它没有:)。 DevProducts是我的testtable，其中包含Product和Attribute列。我以为我会发布LINQ查询，可能对正在使用LINQ的人有用。

如果你可以优化上面的LINQ查询，请告诉我（它可能会产生更好的SQL;）

Answer 9

我正在使用以下DDL

CREATE TABLE Attributes (
    Asset      VARCHAR(100)
    , Name     VARCHAR(100)
    , UNIQUE(Asset, Name)
    )

第二个问题很简单

SELECT   Name
FROM     Attributes
WHERE    Name IN (SELECT Name FROM Attributes WHERE Asset = 'A')
         AND Asset = 'B'

第一个问题并不困难

SELECT   Asset
FROM     Attributes
WHERE    Name IN (SELECT Name FROM Attributes WHERE Asset = 'A')
GROUP BY Asset
HAVING   COUNT(*) = (SELECT COUNT(*) FROM FROM Attributes WHERE Asset = 'A')

编辑：

为了简洁，我将AND Asset != 'A'从第二个代码段的WHERE子句中删除了

SQL Elaborate加入查询

9 个答案: