我的数据库中有5个表，表示继承的EAV模型：
CREATE TABLE AttributeNames
("ID" int, "Name" varchar(8))
;
INSERT INTO AttributeNames
("ID", "Name")
VALUES
(1, 'Color'),
(2, 'FuelType'),
(3, 'Doors'),
(4, 'Price')
;
CREATE TABLE MasterCars
("ID" int, "Name" varchar(10))
;
INSERT INTO MasterCars
("ID", "Name")
VALUES
(5, 'BMW'),
(6, 'Audi'),
(7, 'Ford')
;
CREATE TABLE MasterCarAttributes
("ID" int, "AttributeNameId" int, "Value" varchar(10), "MasterCarId" int)
;
INSERT INTO MasterCarAttributes
("ID", "AttributeNameId", "Value", "MasterCarId")
VALUES
(100, 1, 'Red', 5),
(101, 2, 'Gas', 5),
(102, 3, '4', 5),
(102, 4, '$100K', 5),
(103, 1, 'Blue', 6),
(104, 2, 'Diesel', 6),
(105, 3, '3', 6),
(106, 4, '$80k', 6),
(107, 1, 'Green', 7),
(108, 2, 'Diesel', 7),
(109, 3, '5', 7),
(110, 4, '$60k', 7)
;
CREATE TABLE LocalCars
("ID" int, "MasterCarId" int)
;
INSERT INTO LocalCars
("ID", "MasterCarId")
VALUES
(8, '5'),
(9, '6'),
(10, NULL)
;
CREATE TABLE LocalCarAttributes
("ID" int, "AttributeNameId" int, "Value" varchar(6), "LocalCarId" int)
;
INSERT INTO LocalCarAttributes
("ID", "AttributeNameId", "Value", "LocalCarId")
VALUES
(43, 1, 'Yellow', 8),
(44, 3, '6', 9),
(45, 1, 'Red', 10),
(46, 2, 'Gas', 10),
(47, 3, '2', 10),
(48, 4, '$60k', 10)
;
我可以按如下方式检索所有主车属性：
SELECT MC.ID, MCA.AttributeNameId, MCA.Value
FROM MasterCars MC
left join MasterCarAttributes MCA on MC.ID = MCA.MasterCarId
order by MC.ID;
同样，我可以检索所有本地汽车属性，如下所示：
SELECT LC.ID, LCA.AttributeNameId, LCA.Value
FROM LocalCars LC
left join LocalCarAttributes LCA on LC.ID = LCA.LocalCarId
order by LC.ID;
如果LocalCars.MasterCarId不为NULL，则该本地汽车可以继承该主汽车的属性。具有相同AttributeNameId的本地car属性将覆盖具有相同AttributeNameId的任何主属性。
因此，鉴于上述数据，我有3辆本地车，每辆车都有4个属性（颜色，燃料类型，车门，价格）。以粗体显示的继承属性值：
本地车辆识别码= 1（黄色，气体， 4 ， $ 100K ）
本地车辆识别码= 2（蓝色，柴油，6， $ 80k ）
本地车辆识别码= 3（红色，气体，2，$ 60k）
我正在尝试找到加入上述两个查询所需的必要连接，以提供一组完整的本地汽车属性，一些继承：
LocalCarId AttributeNameId Value
------------------------------------------
1 1 Yellow
1 2 Gas
1 3 4
1 4 $100K
2 1 Blue
2 2 Diesel
2 3 6
2 4 $80K
3 1 Red
3 2 Gas
3 3 2
3 4 $60K
或甚至可能：
LocalCarId AttributeNameId LocalValue MasterValue
-------------------------------------------------------------
1 1 Yellow Red
1 2 NULL Gas
1 3 NULL 4
1 4 NULL $100K
2 1 NULL Blue
2 2 NULL Diesel
2 3 6 3
2 4 NULL $80K
3 1 Red NULL
3 2 Gas NULL
3 3 2 NULL
3 4 $60K NULL
答案 0 :(得分：0)
<强> SQL Fiddle Demo
SELECT LC."ID" as LocalCarID,
COALESCE(LCA."AttributeNameId", MCA."AttributeNameId") as "AttributeNameId",
COALESCE(LCA."Value", MCA."Value") as "Value"
FROM LocalCars LC
LEFT JOIN MasterCars MC
ON LC."MasterCarId" = MC."ID"
LEFT JOIN MasterCarAttributes MCA
ON MC."ID" = MCA."MasterCarId"
LEFT JOIN LocalCarAttributes LCA
ON ( MCA."AttributeNameId" = LCA."AttributeNameId"
OR MCA."AttributeNameId" IS NULL)
-- This is the important part
-- Try to join with a MasterAtribute otherwise use the Car Atribute.
AND LC."ID" = LCA."ID"
<强>输出
| LocalCarID | AttributeNameId | Value |
|------------|-----------------|--------|
| 1 | 1 | Blue |
| 1 | 2 | Gas |
| 2 | 1 | Green |
| 2 | 2 | Diesel |
答案 1 :(得分：0)
可以通过对所有本地汽车属性和主汽车属性执行联合来解决问题。每条记录都标有[IsMasterAttribute]标志。然后，下一步是使用ROW_NUMBER（）窗口函数对每个重复属性进行排名。最后一步是仅选择排名为1的属性。
;WITH CTE_CombinedAttributes
AS
(
SELECT 1 AS IsMasterAttribute
,LC.ID
,MC.ID AS MasterCarId
,MCA.AttributeNameId
,MCA.Value
FROM MasterCars MC
LEFT OUTER JOIN MasterCarAttributes MCA on MC.ID = MCA.MasterCarId
INNER JOIN LocalCars LC ON LC.MasterCarId = MC.ID
UNION ALL
SELECT 0 AS IsMasterAttribute
,LC.ID
,LC.MasterCarId
,LCA.AttributeNameId
,LCA.Value
FROM LocalCars LC
LEFT OUTER JOIN LocalCarAttributes LCA on LC.ID = LCA.LocalCarId
)
,
CTE_RankedAttributes
AS
(
SELECT [IsMasterAttribute]
,[ID]
,[AttributeNameId]
,[Value]
,ROW_NUMBER() OVER (PARTITION BY [ID], [AttributeNameId] ORDER BY [IsMasterAttribute]) AS [AttributeRank]
FROM CTE_CombinedAttributes
)
SELECT [IsMasterAttribute]
,[ID]
,[AttributeNameId]
,[Value]
FROM CTE_RankedAttributes
WHERE [AttributeRank] = 1
ORDER BY [ID]
通过对最终结果执行简单的旋转，也可以进行第二次输出：
;WITH CTE_CombinedAttributes
AS
(
SELECT 1 AS IsMasterAttribute
,LC.ID
,MC.ID AS MasterCarId
,MCA.AttributeNameId
,MCA.Value
FROM MasterCars MC
LEFT OUTER JOIN MasterCarAttributes MCA on MC.ID = MCA.MasterCarId
INNER JOIN LocalCars LC ON LC.MasterCarId = MC.ID
UNION ALL
SELECT 0 AS IsMasterAttribute
,LC.ID
,LC.MasterCarId
,LCA.AttributeNameId
,LCA.Value
FROM LocalCars LC
LEFT OUTER JOIN LocalCarAttributes LCA on LC.ID = LCA.LocalCarId
)
,
CTE_RankedAttributes
AS
(
SELECT [IsMasterAttribute]
,[ID]
,[AttributeNameId]
,[Value]
,ROW_NUMBER() OVER (PARTITION BY [ID], [AttributeNameId] ORDER BY [IsMasterAttribute]) AS [AttributeRank]
FROM CTE_CombinedAttributes
)
SELECT [ID]
,[AttributeNameId]
,MAX(
CASE [IsMasterAttribute]
WHEN 0 THEN [Value]
END
) AS LocalValue
,MAX(
CASE [IsMasterAttribute]
WHEN 1 THEN [Value]
END
) AS MasterValue
FROM CTE_RankedAttributes
GROUP BY [ID], [AttributeNameId]
ORDER BY [ID]