Sql 识别关系不一致的记录

您可以运行所有这些SQL并查看结果

• 跳到结果和随后的问题，了解问题的实质。

我有一张俱乐部的桌子（一个团体或组织中的俱乐部，如“游泳俱乐部”或“编织俱乐部”）

我有一桌成员

DECLARE @member TABLE (
   Id INT
   ,Name NVARCHAR(255)
   );
INSERT INTO @member VALUES
   (1, 'John Jones')
   ,(2, 'Sally Smith')
   ,(3, 'Rod Roosevelt')
   ,(4, 'Bobby Burns')
   ,(5, 'Megan Moore');

会员可以属于许多俱乐部，因此有一个会员表将俱乐部与会员联系起来（并说明会费价格）

大多数会员只需缴纳相关会费。然而，有些成员是由其他成员赞助的。因此，这些赞助会员的会费将由另一名会员（赞助商）支付。因此，我们有一个赞助表。赞助表将赞助商（支付会费）与特定俱乐部的赞助商（由赞助商支付会费）联系起来。因为赞助是特定于俱乐部的，所以赞助记录连接两个会员记录，而不是两个会员记录

DECLARE @sponsorship TABLE (
   Id INT
   ,Sponsee_Membership INT --FK to Sponsee's @membership record
   ,Sponsor_Membership INT --FK to Sponsor's @membership record
   );
INSERT INTO @sponsorship VALUES
   (1,5,1)
   ,(2,8,4)
   ,(3,9,3)
   ,(4,10,4);

为了全面了解我们的俱乐部/会员资格/赞助商，我们有：

SELECT
    mship.Id AS 'Mship'
    ,mem.Name AS 'Member'
    ,c.Name AS 'Club'
    ,mship.Dues
    ,spons_mem.Name AS 'Sponsor'
FROM
    @membership AS mship
    JOIN @member AS mem
        ON mship.Member = mem.Id
    JOIN @club AS c
        ON mship.Club = c.Id
    LEFT JOIN @sponsorship AS spons
        ON spons.Sponsee_Membership = mship.Id
    LEFT JOIN @membership AS spons_mship
        ON spons_mship.Id = spons.Sponsor_Membership
    LEFT JOIN @member AS spons_mem
        ON spons_mem.Id = spons_mship.Member;

这给了我们这些结果：

Mship   Member        Club         Dues  Sponsor
  1   John Jones     Swim Club      10    NULL
  2   John Jones     Knitting Club   5    NULL
  3   Sally Smith    Swim Club      10    NULL
  4   Sally Smith    Bridge Club    20    NULL
  5   Rod Roosevelt  Swim Club      10    John Jones
  6   Rod Roosevelt  Knitting Club   5    NULL
  7   Bobby Burns    Knitting Club   5    NULL
  8   Bobby Burns    Bridge Club    20    Sally Smith
  9   Megan Moore    Swim Club      10    Sally Smith
 10   Megan Moore    Bridge Club    20    Sally Smith

赞助应涵盖所有共享会员。

• 也就是说，如果Sally赞助Bobby，任何时候他们都在同一个俱乐部，Sally将被认定为Bobby的赞助商
• 我们可以在

  Mship=7

  和

  Mship=8

  行中看到这一点
• Bobby和Sally都在桥牌俱乐部，因此Sally被认定为Bobby桥牌俱乐部会员的赞助商
• 萨莉不是针织俱乐部的成员，因此鲍比的针织俱乐部会员资格并不表明萨莉是赞助商

很抱歉设置太长这是我的实际问题：

• 我如何确定缺少赞助的地方？
  从这个例子中，我们有行

  Mship=5

  和

  Mship=6

• 约翰是罗德的赞助商
• 我们可以看到罗德游泳俱乐部会员的赞助
• 罗德和约翰也是针织俱乐部的成员
• 但是，罗德没有表示约翰是他的针织俱乐部会员的赞助商
• 这是不正确的，这就是我所追求的
• 我想查询所有这些缺少的赞助

我可以使用游标/WHILE循环来实现这一点，但我知道这样的解决方案通常不会采用正确的基于集合的方法。对此的正确查询是什么样的？

---

非常感谢。

这是一个SQL查询，可能会响应您的要求

逻辑是使用子查询，根据映射

member.id

，而不是

memberships.id

，在发起人和发起人之间生成映射；为此，我们使用聚合。然后，外部查询搜索赞助商和赞助商都参与的俱乐部，但在赞助表中没有声明其关系

查询将为每个有问题的成员返回一条记录，其中包含sponsee和赞助商名称

SELECT mship1.Id, m1.Name Member, m2.Name Sponsor, c.Name Club, mship1.Dues
FROM 
    @membership mship1
    INNER JOIN @club c ON c.Id = mship1.Club
    INNER JOIN (
        SELECT ms1.Member Sponsee_Member , MAX(ms2.Member) Sponsor_Member
        FROM @sponsorship ss
        INNER JOIN @membership ms1 ON ms1.Id = Sponsee_Membership
        INNER JOIN @membership ms2 ON ms2.Id = Sponsor_Membership
        GROUP BY ms1.Member
    ) rels ON rels.Sponsee_Member = mship1.Member
    INNER JOIN @membership mship2 ON mship2.Member = rels.Sponsor_Member AND mship2.Club = mship1.Club
    INNER JOIN @member m1 ON m1.Id = mship1.Member
    INNER JOIN @member m2 ON m2.Id = mship2.Member
    LEFT JOIN @sponsorship sship  ON sship.Sponsor_Membership = mship2.Id
WHERE sship.Id IS NULL
;

在中，返回：

Id   | Member          | Sponsor      | Club           | Dues
-----|-----------------|--------------|----------------|-----
6    | Rod Roosevelt   | John Jones   | Knitting Club  | 5

---

创建这个查询让我想到，您可以优化数据库设计。目前的模式将难以保持一致性：你的问题本身就证明了这一点。将来，如果一个赞助商在一个新的俱乐部注册，而他的一个赞助商已经参加了，会发生什么？您需要再次检测缺少的

sponshorship

关系，并以某种方式创建它

实际上，赞助者和赞助者之间有1-1的关系，因为你说赞助者应该跨越所有共享的成员资格。看起来你不会允许一个赞助商有几个赞助商，即使是在不同的俱乐部

---

我建议您删除

sponshorship

表，并将赞助商的自外键直接存储在member表中。从这里开始，很容易检查两个成员有哪些共同的俱乐部，并使用SQL查询正确分配会费

我发现我可以在一个查询中收集到这一点，方法是收集所有与CTE的赞助关系，找到所有应该有赞助的会员资格，然后使用EXCEPT删除所有现有赞助资格。我剩下的赞助应该存在，但不存在

WITH sponsored_relationships AS (
    SELECT DISTINCT
        sponsee_member.Id AS Sponsee
        ,sponsor_member.Id AS Sponsor
    FROM
        @sponsorship AS s
        JOIN @membership AS sponsee_mship
            ON s.Sponsee_Membership = sponsee_mship.Id
        JOIN @member AS sponsee_member
            ON sponsee_mship.Member = sponsee_member.Id
        JOIN @membership AS sponsor_mship
            ON s.Sponsor_Membership = sponsor_mship.Id
        JOIN @member AS sponsor_member
            ON sponsor_mship.Member = sponsor_member.Id
    )
SELECT
    see_mem.Name AS Sponsee
    ,sor_mem.Name AS Sponsor
    ,c.Name AS Club
FROM
    sponsored_relationships AS sr
    JOIN @member AS see_mem
        ON sr.Sponsee = see_mem.Id
    JOIN @membership AS see_mship
        ON see_mship.Member = see_mem.Id
    JOIN @member AS sor_mem
        ON sr.Sponsor = sor_mem.Id
    JOIN @membership AS sor_mship
        ON sor_mship.Member = sor_mem.Id
    JOIN @club AS c
        ON (see_mship.Club = c.Id
            AND sor_mship.Club = c.Id
            )
EXCEPT
SELECT
    see_mem.Name AS Sponsee
    ,sor_mem.Name AS Sponsor
    ,c.Name AS Club
FROM
    sponsored_relationships AS sr
    JOIN @member AS see_mem
        ON sr.Sponsee = see_mem.Id
    JOIN @membership AS see_mship
        ON see_mship.Member = see_mem.Id
    JOIN @member AS sor_mem
        ON sr.Sponsor = sor_mem.Id
    JOIN @membership AS sor_mship
        ON sor_mship.Member = sor_mem.Id
    JOIN @club AS c
        ON (see_mship.Club = c.Id
            AND sor_mship.Club = c.Id
            )
    JOIN @sponsorship AS sship
        ON (sship.Sponsee_Membership = see_mship.Id
            AND sship.Sponsor_Membership = sor_mship.Id
            );

---

问题在于您的数据模型。您是否能够更改它，或者您是否必须使用此模型编写SQL work around？听起来赞助不应该是会员之间的1对1关系，而是俱乐部会员与赞助商之间的关系。您的表

赞助

需要一个额外的列（它链接到的俱乐部）。作为进一步的示例，让我们以3名会员和3个俱乐部为例。珍妮、简和约翰，还有国际象棋、网球和读书俱乐部。约翰属于这三个人，但珍妮只属于书本和网球，简只属于象棋和书本。珍妮和简都是约翰的赞助商，这对国际象棋和网球俱乐部来说很有意义，但是读书俱乐部呢。你有“双重泡泡”吗？：）是的，问题的根源在于数据模型。不幸的是，这是无法修改的。这种次优的数据结构产生了一致的数据清理量。此问题与定期运行的许多必需的清理查询中的一个有关。@JClark Thx以获取澄清。很高兴你找到了答案。希望我整理您的问题/添加rextester和要点能让其他人更容易提供帮助！谢谢，这个很好用。你对数据模型的评论是完全准确的。很遗憾，我无权修改它。@JClark欢迎！如果你认为我的回答正确地回答了你的问题，请考虑…谢谢

Id   | Member          | Sponsor      | Club           | Dues
-----|-----------------|--------------|----------------|-----
6    | Rod Roosevelt   | John Jones   | Knitting Club  | 5

WITH sponsored_relationships AS (
    SELECT DISTINCT
        sponsee_member.Id AS Sponsee
        ,sponsor_member.Id AS Sponsor
    FROM
        @sponsorship AS s
        JOIN @membership AS sponsee_mship
            ON s.Sponsee_Membership = sponsee_mship.Id
        JOIN @member AS sponsee_member
            ON sponsee_mship.Member = sponsee_member.Id
        JOIN @membership AS sponsor_mship
            ON s.Sponsor_Membership = sponsor_mship.Id
        JOIN @member AS sponsor_member
            ON sponsor_mship.Member = sponsor_member.Id
    )
SELECT
    see_mem.Name AS Sponsee
    ,sor_mem.Name AS Sponsor
    ,c.Name AS Club
FROM
    sponsored_relationships AS sr
    JOIN @member AS see_mem
        ON sr.Sponsee = see_mem.Id
    JOIN @membership AS see_mship
        ON see_mship.Member = see_mem.Id
    JOIN @member AS sor_mem
        ON sr.Sponsor = sor_mem.Id
    JOIN @membership AS sor_mship
        ON sor_mship.Member = sor_mem.Id
    JOIN @club AS c
        ON (see_mship.Club = c.Id
            AND sor_mship.Club = c.Id
            )
EXCEPT
SELECT
    see_mem.Name AS Sponsee
    ,sor_mem.Name AS Sponsor
    ,c.Name AS Club
FROM
    sponsored_relationships AS sr
    JOIN @member AS see_mem
        ON sr.Sponsee = see_mem.Id
    JOIN @membership AS see_mship
        ON see_mship.Member = see_mem.Id
    JOIN @member AS sor_mem
        ON sr.Sponsor = sor_mem.Id
    JOIN @membership AS sor_mship
        ON sor_mship.Member = sor_mem.Id
    JOIN @club AS c
        ON (see_mship.Club = c.Id
            AND sor_mship.Club = c.Id
            )
    JOIN @sponsorship AS sship
        ON (sship.Sponsee_Membership = see_mship.Id
            AND sship.Sponsor_Membership = sor_mship.Id
            );