Question

我想获取表A中每个相关ID的数据从表B到表A的价格。我的目标是根据表中ID的顺序使价格比表A中的价格低六倍。 B，如果没有更低的价格，我带来相同的价格值。

我已经编写了一个有效的查询，但是我想知道如何对其进行优化。

在我编写的查询中，我使用左联接六次以从表B中获取表A的每个ID的值。

那么还有什么其他方法可以写此查询，而不是将同一张表连接六次？

表A：

|Price| |ID||ID_1||ID_2||ID_3||ID_4||ID_5||ID_6|

|129.90| |1| |1|   |1|   |1|   |1|   |1|    |1|
|149.90| |2| |1|   |1|   |1|   |1|   |1|    |1|
|199.90| |3| |2|   |1|   |1|   |1|   |1|    |1|
|249.90| |4| |3|   |2|   |1|   |1|   |1|    |1|
|299.90| |5| |4|   |3|   |2|   |1|   |1|    |1|
|399.90| |6| |5|   |4|   |3|   |2|   |1|    |1|

表B：

| Price ||PriceID|  

|129.90|  |1|  
|149.90|  |2|  
|199.90|  |3|   
|249.90|  |4|   
|299.90|  |5|   
|399.90|  |6|

查询：

>      SELECT i.*, 
              f.Price AS  NewPrice1, 
              f1.Price AS NewPrice2, 
              f2.Price AS NewPrice3,
              f3.Price AS NewPrice4, 
              f4.Price AS NewPrice5, 
              f5.Price AS NewPrice6
>      FROM #TableA i
>       LEFT JOIN #TableB f ON  i.ID_1 = f.PriceID      
        LEFT JOIN #TableB f1 ON i.ID_2 = f1.PriceID     
        LEFT JOIN #TableB f2 ON i.ID_3 = f2.PriceID     
        LEFT JOIN #TableB f3 ON i.ID_4 = f3.PriceID     
        LEFT JOIN #TableB f4 ON i.ID_5 = f4.PriceID     
        LEFT JOIN #TableB f5 ON i.ID_6 = f5.PriceID

输出：

|Price| |ID||ID_1||ID_2||ID_3||Price_1||Price_2||Price_3||Price_4||Price_5||Price_6|

|129.90| |1| |1|   |1|   |1| |129.90|  |129.90| |129.90||129.90||129.90||129.90|
|149.90| |2| |1|   |1|   |1| |129.90|  |129.90| |129.90||129.90||129.90||129.90|
|199.90| |3| |2|   |1|   |1| |149.90|  |129.90| |129.90||129.90||129.90||129.90|
|249.90| |4| |3|   |2|   |1| |199.90|  |149.90| |129.90||129.90||129.90||129.90|
|299.90| |5| |4|   |3|   |2| |249.90|  |199.90| |149.90||129.90||129.90||129.90|
|399.90| |6| |5|   |4|   |3| |299.90|  |249.90| |199.90||129.90||129.90||129.90|

Answer 1

从另一个角度看：

DECLARE @A TABLE
(
    Price INT,
    ID_1 INT,
    ID_2 INT,
    ID_3 INT
)
INSERT INTO @A
(
    Price,
    ID_1,
    ID_2,
    ID_3
)
VALUES
(129.90, 1, 1, 1),
(199.90, 2, 1, 1),
(249.90, 3, 2, 1),
(299.90, 4, 4, 3),
(399.90, 6, 5, 4)


DECLARE @B TABLE
(
    Price INT,
    PriceID INT
)
INSERT INTO @B
(
    Price,
    PriceID
)
VALUES
(129.90, 1),  
(149.90, 2),  
(199.90, 3),   
(249.90, 4),  
(299.90, 5),   
(399.90, 6)

SELECT
    a.Price,
    SUM(b.NewPrice1),
    SUM(b.NewPrice2),
    SUM(b.NewPrice3)
FROM @A a
CROSS APPLY
(
    SELECT
        CASE WHEN PriceID = a.ID_1 THEN Price ELSE NULL END AS  NewPrice1,
        CASE WHEN PriceID = a.ID_2 THEN Price ELSE NULL END AS  NewPrice2,
        CASE WHEN PriceID = a.ID_3 THEN Price ELSE NULL END AS  NewPrice3
    FROM @B
) b
GROUP BY a.Price

Answer 2

请尝试以下查询。

SELECT
  i.*,
  (SELECT
    f.Price
  FROM #TableB f
  WHERE i.ID_1 = f.PriceID)
  AS NewPrice1,
  (SELECT
    f1.Price
  FROM #TableB f1
  WHERE i.ID_2 = f1.PriceID)
  AS NewPrice2,
  (SELECT
    f2.Price
  FROM #TableB f2
  WHERE i.ID_3 = f2.PriceID)
  AS NewPrice3,
  (SELECT
    f3.Price
  FROM #TableB f3
  WHERE i.ID_4 = f3.PriceID)
  AS NewPrice4,
  (SELECT
    f4.Price
  FROM #TableB f4
  WHERE i.ID_5 = f4.PriceID)
  AS NewPrice5,
  (SELECT
    f5.Price
  FROM #TableB f5
  WHERE i.ID_6 = f5.PriceID)
  AS NewPrice6
FROM #TableA i

如何优化查询以根据每个ID获得值？

2 个答案: