5

I'm trying to solve the below problem.

I feel like it is possible, but I can't seem to get it.

Here's the scenario:

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

If I am looking for something having the same attributes as Asset-A, then it should identify Asset-B since Asset-B has all the same attributes as Asset-A (it should discard heavy, since Asset-A didn't specify anything different or the similar). Also, if I wanted the attributes for only Asset-A AND Asset-B that were common, how would I get that?

Seems simple, but I can't nail it...

The actual table I am using, is almost precisely Table2, simply an association of an AssetId, and an AttributeId so: PK: Id
int: AssetId
int: AttributeId

I only included the idea of the asset table to simplify the question.

4

9 に答える 9

4 
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

、またはSQL Server 2005(確認するサンプル データを使用して):

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
于 2009-07-21T15:30:13.370 に答える
0

おそらく、LINQでこれを実行してから、次のように逆方向に作業できると思いました。

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()

これをLinqPadを使用してSQLに戻すと、かなりのSQLクエリになると思いました。しかし、そうではありませんでした:)。DevProductsは、ProductandAttribute列を持つ私のテストテーブルです。とにかくLINQクエリを投稿すると思いましたが、LINQで遊んでいる人には役立つかもしれません。

上記のLINQクエリを最適化できる場合は、お知らせください（SQLが改善される可能性があります;））

于 2009-07-22T08:52:17.407 に答える
0 
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");
于 2009-07-21T15:56:58.247 に答える
0

あなたの質問の最初の部分、つまり属性に基づいて資産を特定することを完全には理解していません。

列名についていくつかの仮定を立てると、次のクエリは Asset-A と Asset-B の間の共通の属性を生成します。

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
于 2009-07-21T15:32:31.883 に答える
0

「A」が持つすべての属性を持つすべてのアセットを検索します (ただし、追加の属性を持つ場合もあります)。

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
      )
    )

つまり、「このアセットの属性でもなく、A の属性がないアセットを見つける」ということです。

「A」とまったく同じ属性を持つすべての資産を検索します。

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
      )
   )

つまり、「このアセットの属性でもなく、このアセットの属性でもなく、A の属性でもない A の属性がないアセットを見つけます。」

于 2009-07-21T17:38:09.120 に答える
0 
 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'
于 2009-07-21T15:40:42.163 に答える
0

このソリューションは規定どおりに機能します。ご意見ありがとうございます。

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]
于 2009-07-21T18:47:25.190 に答える
0

次のDDLを使用しています

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

2問目は簡単

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

最初の質問はそれほど難しくありません

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')

編集：

簡潔にするために、2 番目のスニペットAND Asset != 'A'のWHERE節は省略しました

于 2009-07-22T21:16:00.663 に答える
0

asset-a とすべて同じ属性を持つすべてのアセットを検索します。

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)
于 2009-07-21T19:29:11.610 に答える