2011-06-14 49 views
8

我有一个看起来像这三个表:查找具有在许多完全匹配记录,以一对多的关系

PROD

Prod_ID|Desc 
------------ 
P1|Foo1 
P2|Foo2 
P3|Foo3 
P4|Foo4 
... 

RAM

Ram_ID|Desc 
------------ 
R1|Bar1 
R2|Bar2 
R3|Bar3 
R4|Bar4 
... 

PROD_RAM

Prod_ID|Ram_ID 
------------ 
P1|R1 
P2|R2 
P3|R1 
P3|R2 
P3|R3 
P4|R3 
P5|R1 
P5|R2 
... 

之间PRODRAM有由PROD_RAM表中描述的许多一对多的关系。

给定一组Ram_ID(R1,R3)我想找到所有具有完全相同ONE在给定的RAM的ALLPROD

鉴于(R1,R3)应返回例如P1P4P5; P3不应退还,因为有R1R3,但也R2

什么来获得最快的查询都具有完全相同ONE给定RAM集的Ram_ID的ALLPROD

编辑:
PROD_RAM表可以包含关系比1-> 3更大,所以,对于计数 “硬编码” 检查= 1 OR = 2不是一个可行的解决方案。

+2

+1问清楚可理解的问题 – Stuti 2011-06-14 07:38:19

回答

2

你可以尝试的速度另一种解决办法是这样的

;WITH CANDIDATES AS (
    SELECT pr1.Prod_ID 
      , pr2.Ram_ID 
    FROM PROD_RAM pr1 
      INNER JOIN PROD_RAM pr2 ON pr2.Prod_ID = pr1.Prod_ID 
    WHERE pr1.Ram_ID IN ('R1', 'R3') 
) 
SELECT * 
FROM CANDIDATES 
WHERE CANDIDATES.Prod_ID NOT IN (
      SELECT Prod_ID 
      FROM CANDIDATES 
      WHERE Ram_ID NOT IN ('R1', 'R3') 
     )   

,或者如果你不喜欢重复设置条件

;WITH SUBSET (Ram_ID) AS (
    SELECT 'R1' 
    UNION ALL SELECT 'R3' 
) 
, CANDIDATES AS (
    SELECT pr1.Prod_ID 
      , pr2.Ram_ID 
    FROM PROD_RAM pr1 
      INNER JOIN PROD_RAM pr2 ON pr2.Prod_ID = pr1.Prod_ID 
      INNER JOIN SUBSET s ON s.Ram_ID = pr1.Ram_ID  
) 
, EXCLUDES AS (
    SELECT Prod_ID 
    FROM CANDIDATES 
      LEFT OUTER JOIN SUBSET s ON s.Ram_ID = CANDIDATES.Ram_ID 
    WHERE s.Ram_ID IS NULL 
) 
SELECT * 
FROM CANDIDATES 
     LEFT OUTER JOIN EXCLUDES ON EXCLUDES.Prod_ID = CANDIDATES.Prod_ID 
WHERE EXCLUDES.Prod_ID IS NULL   
+0

测试了你的第一个解决方案a nd它的工作:) – systempuntoout 2011-06-14 09:12:03

+0

@systempuntoout - 现在你让我想知道...和第二? – 2011-06-14 17:26:45

+0

还没有尝试过 – systempuntoout 2011-06-15 06:29:06

0
SELECT Prod_ID 
FROM 
    (SELECT Prod_ID 
     , COUNT(*) AS cntAll 
     , COUNT(CASE WHEN Ram_ID IN (1,3) 
         THEN 1 
         ELSE NULL 
        END 
       ) AS cntGood 
    FROM PROD_RAM 
    GROUP BY Prod_ID 
) AS grp 
WHERE cntAll = cntGood 
    AND (cntGood = 1 
    OR cntGood = 2    --- number of items in list (1,3) 
    ) 

完全不知道这是否是最快的方法。您必须尝试使用​​不同的方法来编写此查询(使用JOIN s和NOT EXISTS)并测试速度。要做到这一点

0

一种方式是类似以下内容:

SELECT PROD.Prod_ID FROM PROD WHERE 

(SELECT COUNT(*) FROM PROD_RAM WHERE PROD_RAM.Prod_ID = PROD.Prod_ID) > 0 AND 

(SELECT COUNT(*) FROM PROD_RAM WHERE PROD_RAM.Prod_ID = PROD.Prod_ID AND PROD.Ram_ID <> 
IFNULL((SELECT TOP 1 Ram_ID FROM PROD_RAM WHERE PROD_RAM.Prod_ID = PROD.Prod_ID),0)) = 0 
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