Question

我对SQL查询有点麻烦，并且认为我会征求人群的智慧，看看我缺少什么。我很确定下面有效，但看起来真的很差，我想知道是否有更聪明的方法（理想情况下使用连接而不是子选择）来做到这一点。

问题

假设我有一些表格：

Prize
  - PrizeId

RulePrize_Map
  - PrizeId
  - RuleId

Rule
  - RuleId

Conditional
  - ConditionalId
  - RuleId
  - InputId
  - ExpectedValue (bit)

Input
  - InputId

当至少有一个Prize为真时赢得Rule。当Rule的所有为真时，Conditionals为真。当Conditional在InputId表中存在或不存在时，Input为“真”，由ExpectedValue字段指定。对于Count(InputId in Input table) ExpectedValue，这可能被视为等同于：Conditional's = InputId。

一些例子：

Conditional (InputId = 11, ExpectedValue = 1) -> True if InputId 11 in Input Table
Conditional (InputId = 12, ExpectedValue = 0) -> True if Inputid 12 NOT in Input Table

我的目标

我想让所有Prizes至少有一个Rule为“true”。我愿意接受：“所有Rules都是真的”。

我的尝试

select p.PrizeId from Prize p INNER JOIN RulePrize_Map rpm ON rpm.PrizeId = p.PrizeId
WHERE p.PrizeId IN
(select r.PrizeId from Rule r
where
  (select count(*) from Conditional c1 where c1.RuleId = r.RuleId)
  = 
  (select count(*) from Conditional c2
   where c2.RuleId = r.RuleId AND
   (select count(*) from Input i where i.InputId = c2.InputId) = c2.ExpectedValue
  )
)
GROUP BY p.prizeId

Answer 1

问题有所改变，所以我重新回答了问题......

SELECT
  PrizeId
FROM
  (
  SELECT
    PrizeRule_Map.PrizeId,
    PrizeRule_Map.RuleId
  FROM
    PrizeRule_Map
  INNER JOIN
    Rule
      ON Rule.RuleId = PrizeRule_Map.RuleId
  INNER JOIN
    Conditional
      ON Conditional.RuleId = Rule.RuleID
  LEFT JOIN
    Input
      ON Input.InputId = Conditional.InputID
  GROUP BY
    PrizeRule_Map.PrizeId,
    PrizeRule_Map.RuleId
  HAVING
    COUNT(*) = SUM(CASE Conditional.ExpectedValue
                     WHEN 1 THEN CASE WHEN Input.InputId IS NULL THEN 0 ELSE 1 END
                     WHEN 0 THEN CASE WHEN Input.InputId IS NULL THEN 1 ELSE 0 END
                   END
                   )
  )
    AS map
GROUP BY
  PrizeId

Answer 2

获取ALL条件为真的所有RuleId：

SELECT r.RuleID
FROM Rule r
  JOIN Conditional c
    ON c.RuleId = r.RuleId
  LEFT JOIN Input i
    ON i.InputId = c.InputId
GROUP BY r.RuleID
HAVING COUNT( CASE WHEN (c.ExpectedValue=1) AND (i.InputId IS NOT NULL)
                     OR (c.ExpectedValue=0) AND (i.InputId IS NULL)
                   THEN 1
                   ELSE NULL 
              END ) 
       = COUNT( * )

另一种方式 - 可能更慢，但测试速度并没有坏处。它不使用CASE，而是两个EXCEPT的差异（JOIN），其中只有一个使用GROUP BY：

    SELECT r.RuleID
    FROM Rule r
      JOIN Conditional c
        ON c.RuleId = r.RuleId
      LEFT JOIN Input i
        ON i.InputId = c.InputId
    WHERE c.ExpectedValue = 1
    GROUP BY r.RuleID
    HAVING COUNT( i.InputId ) = COUNT( * )
  EXCEPT
    SELECT r.RuleID
    FROM Rule r
      JOIN Conditional c
        ON c.RuleId = r.RuleId
      JOIN Input i
        ON i.InputId = c.InputId
    WHERE c.ExpectedValue = 0

Answer 3

尝试以下方法：

SELECT Rule.RuleId, Rule.RuleName
FROM Rule
INNER JOIN Conditional ON Rule.RuleId = Conditional.RuleId
Where Conditional.ExpectedValue == true

至少有一个孩子拥有自己所有孩子的所有行都会通过一个条件

3 个答案: