在同一个 table 中加入两个外键引用同一个主键

Question

我正在尝试加入多个 tables(BooksReview, Followers, Books) 与 Users table 其中Books 引用了 userId 来自 Users，BooksReview 引用了 userId，userId 评论了一本书并引用了 bookId 来自 Books 这是被审查的书 ID，最后但并非最不重要的是 Followers table（我认为这里是问题所在）有两个来自相同主键的引用它有 follower 来自 userId in Users table 和 followed 也来自 userId in Users table.

问题：我创建了一个 MySQL 查询来获取基于特定用户的特定图书的书评数量，并获取该用户的图书数量和关注者数量 he/she 有，但是当我将 Followers 连接部分添加到我的查询时，它显示所有值的结果为 0，预期值为 4 本书、4 条评论和 1 个关注者。

我试图更改查询中的连接类型，但结果相同，并搜索了在同一 table 中连接两个外键以获得相同的主键，但我没有找到任何有用的东西.

```
 CREATE TABLE IF NOT EXISTS `Authors`.`Users` (
`userId` VARCHAR(100) NOT NULL,
`username` VARCHAR(25) NOT NULL,
`password` VARCHAR(16) NOT NULL,
`email` VARCHAR(254) NOT NULL,
`birthday` DATE NULL,
`aboutMe` TEXT(300) NOT NULL,
`facebookAccount` VARCHAR(25) NULL,
`twitterAccount` VARCHAR(25) NULL,
`linkedinAccount` VARCHAR(25) NULL,
`profileImage` VARCHAR(200) NULL,
PRIMARY KEY (`userId`),
UNIQUE INDEX `username_UNIQUE` (`username` ASC),
UNIQUE INDEX `email_UNIQUE` (`email` ASC))
ENGINE = InnoDB;
```

 CREATE TABLE IF NOT EXISTS `Authors`.`Books` (
`bookId` VARCHAR(100) NOT NULL,
`bookCategory` VARCHAR(25) NOT NULL,
`title` VARCHAR(25) NOT NULL,
`bookCover` VARCHAR(45) NOT NULL,
`bookDescription` VARCHAR(200) NOT NULL,
 `userId` VARCHAR(100) NOT NULL,
`price` DECIMAL(2,2) NOT NULL,
`introduction` VARCHAR(300) NOT NULL,
 PRIMARY KEY (`bookId`),
 INDEX `userId_idx` (`userId` ASC),
 CONSTRAINT `userId`
 FOREIGN KEY (`userId`)
 REFERENCES `Authors`.`Users` (`userId`)
 ON DELETE NO ACTION
 ON UPDATE NO ACTION)
 ENGINE = InnoDB;    
 ```
 ```
  CREATE TABLE IF NOT EXISTS `Authors`.`BooksReview` (
`bookId` VARCHAR(100) NOT NULL,
`rateMessage` VARCHAR(100) NULL,
`rateNumber` DECIMAL(1,1) NULL,
`userId` VARCHAR(100) NOT NULL,
 INDEX `userId_idx` (`userId` ASC),
 CONSTRAINT `bookId`
 FOREIGN KEY (`bookId`)
 REFERENCES `Authors`.`Books` (`bookId`)
 ON DELETE CASCADE
 ON UPDATE CASCADE,
 CONSTRAINT `userId`
 FOREIGN KEY (`userId`)
 REFERENCES `Authors`.`Users` (`userId`)
 ON DELETE CASCADE
 ON UPDATE CASCADE)
 ENGINE = InnoDB;
 ```
CREATE TABLE IF NOT EXISTS `Authors`.`Followers` (
`follower` VARCHAR(100) NOT NULL,
`followed` VARCHAR(100) NOT NULL,
 INDEX `follower_idx` (`follower` ASC),
 INDEX `followed_idx` (`followed` ASC),
 CONSTRAINT `follower`
 FOREIGN KEY (`follower`)
 REFERENCES `Authors`.`Users` (`userId`)
 ON DELETE CASCADE
 ON UPDATE CASCADE,
 CONSTRAINT `followed`
 FOREIGN KEY (`followed`)
 REFERENCES `Authors`.`Users` (`userId`)
 ON DELETE CASCADE
 ON UPDATE CASCADE)
ENGINE = InnoDB;

THIS IS THE QUERY
SELECT count(br.bookId) AS reviewsCount, count(b.bookId) AS booksCount, count(f.follower) AS followersCount
FROM Users AS u
LEFT JOIN Books AS b ON b.userId = u.userId
JOIN Followers AS f ON b.userId = f.followed AND f.follower = u.userId
INNER JOIN BooksReview AS br ON br.bookId = b.bookId 
                    AND b.bookId IN (SELECT bookId 
                                     FROM Books 
                                     WHERE userId = 'dbb21849-ccce-4af1-aa0f-6653919bf956');

我预计结果应该是 1 个关注者、4 本书和 4 条评论但实际结果全部为0。

DML:

Users ->

       userId: dbb21849-ccce-4af1-aa0f-6653919bf956

       username: mostafabbbaron

等...

Books ->

       userId: dbb21849-ccce-4af1-aa0f-6653919bf956

       bookId: 5f39c1ae-5e99-4b3a-8ee0-97a80c1ba9b1

等...

Followers ->

       follower: dbb21849-ccce-4af1-aa0f-6653919bf956

       folllowed: b39c8e0c-4124-4339-8c30-e1fc8db5f2d4

等...

BooksReviews ->

       userId: dbb21849-ccce-4af1-aa0f-6653919bf956

       bookId: aa44a455-dc28-476f-b4b9-47563a717f03

等...

Answer 1

无论哪种方式，您的查询都是错误的。但是结果为零的原因可能是这些条件：b.userId = u.userId 和 b.userId = f.followed AND f.follower = u.userId.

如果 b.userId = u.userId 和 b.userId = f.followed 那么 f.followed = u.userId

如果 f.followed = u.userId 和 f.follower = u.userId 那么 f.followed = f.follower

这意味着用户必须遵循 himself/herself，我怀疑是这样。

我将按以下方式编写查询：

SELECT
count(DISTINCT b.bookId) AS booksCount,
count(br.bookId) AS reviewsCount,
(SELECT COUNT(*) FROM Followers AS f WHERE f.followed = u.userId) AS followersCount
FROM Users AS u
LEFT JOIN Books       AS b  ON b.userId  = u.userId
LEFT JOIN BooksReview AS br ON br.bookId = b.bookId 
WHERE u.userId = 'dbb21849-ccce-4af1-aa0f-6653919bf956'

注意：虽然 Users LEFT JOIN Books LEFT JOIN BooksReview 很好，因为您有 "relation chain" Users <- Books <- BooksReview。但是你不应该只是加入 Followers table，因为它与 Books 或 BooksReview 并没有真正的关系，也不适合那个链。这就是为什么我在 SELECT 子句中使用子查询来计算关注者的原因。

在同一个 table 中加入两个外键引用同一个主键

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