2 
DECLARE @ProductFeature TABLE (ProductID int, FeatureID int)

INSERT INTO @ProductFeature
  SELECT 1,100
    UNION ALL
  SELECT 1,101
    UNION ALL
  SELECT 1,102
    UNION ALL
  SELECT 2,103
    UNION ALL
  SELECT 2,104
    UNION ALL
  SELECT 3,100
    UNION ALL
  SELECT 3,101
    UNION ALL
  SELECT 3,102
    UNION ALL
  SELECT 4,102
    UNION ALL
  SELECT 4,101
    UNION ALL
  SELECT 5,110
    UNION ALL
  SELECT 5,100
    UNION ALL
  SELECT 5,101

我的要求是，如果我通过ProductID = 1了，那么我必须选择Product具有类似于ProductID = 1.

由于ProductID = 1具有 3 个特征（100,101,102），因此只有ProductID = 3具有相同计数的特征和具有ProductID = 1

预期结果

ProductID   FeatureID
3              100
3              101
3              102
4

4 回答 4

2

带有EXCEPT操作的选项

DECLARE @ProductID int = 1

SELECT ProductID, FeatureID
FROM ProductFeature p1
WHERE p1.ProductID != @ProductID AND 
      NOT EXISTS (
                  SELECT p2.FeatureID         
                  FROM ProductFeature p2
                  WHERE p2.ProductID = @ProductID                                                
                  EXCEPT
                  SELECT p3.FeatureID
                  FROM ProductFeature p3
                  WHERE p3.ProductID = p1.ProductID               
                  )
于 2013-01-28T12:33:00.183 回答
1

它有点效率，但它有效

select pr.ProductID , pr.FeatureID
from @ProductFeature pr
where pr.ProductID in (
    select prd.ProductID
    from @ProductFeature pr
    join @ProductFeature prd
    on pr.ProductID != prd.ProductID
    and pr.FeatureID = prd.FeatureID
    where pr.ProductID = @ProductId
    group by prd.ProductID
    having count(prd.ProductID) = (select count(distinct pr.FeatureID) from @ProductFeature pr where pr.ProductID = @ProductId)
    )
于 2013-01-28T07:17:18.443 回答
1

您首先必须确定共享至少一项功能的产品。然后从这些产品中，找到具有完全相同数量功能的产品。

这应该可以解决问题：

DECLARE @productID int = 1

SELECT
  [p3].[ProductID],
  [p3].[FeatureID]
FROM
(
  SELECT 
    [p1].[ProductID]
  FROM [ProductFeature] [p1]
  INNER JOIN [ProductFeature] [p2] ON [p1].[FeatureID] = [p2].[FeatureID]
  WHERE [p1].[ProductID] <> [p2].[ProductID] AND [p2].[ProductID] = @productID
) AS [sub]
INNER JOIN [ProductFeature] [p3] ON [sub].[ProductID] = [p3].[ProductID]
GROUP BY
  [p3].[ProductID],
  [p3].[FeatureID]
HAVING COUNT(*) = (SELECT COUNT(*)
                   FROM [ProductFeature]
                   WHERE [ProductID] = @productID)
ORDER BY 
  [p3].[ProductID] ASC,
  [p3].[FeatureID] ASC

在这里寻找Fiddle。

于 2013-01-28T07:19:46.787 回答
1

通常我使用 cte -bit clearer（不知道是否更慢/更快）。

    DECLARE @fromProductID int = 1;

    with cteGroup (ProductID) AS
     (
        select ProductID
        from @ProductFeature
        where FeatureID in (select FeatureID
                        from @ProductFeature 
                        where ProductID = @fromProductID)
           and ProductID <> @fromProductID
        group by ProductID
        having COUNT(FeatureID)= (select COUNT(FeatureID) as NoOfRecords
                                    from @ProductFeature 
                                  where ProductID = @fromProductID) 
    ) 
     select a.ProductID,b.FeatureID
     from cteGroup a
     inner join @ProductFeature b
     on a.ProductID = b.ProductID
于 2013-01-28T07:52:24.523 回答