带有2个左联接的MySQL select显示来自第二个左联接的所有记录

Question

我正在尝试获取已阅读的书籍（和藏书-我链接到书籍ID）。我使用2个左联接。

请参见下面的示例。

我正在获得以下结果：

我想获取此列表，但仅获取在a_books_subcollection中读取的集合。在此结果中，它显示了所有集合的读取时间，并且输出仅为集合3。

a_books

a_books_subcollection

a_read

内容

CREATE TABLE `a_books_subcollection` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `label` varchar(35) DEFAULT NULL,
  `book_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `a_books_subcollection` (`id`, `label`, `book_id`)
VALUES
    (1,'Collection 1',3),
    (2,'Collection 2',3),
    (3,'Collection 3',3);


CREATE TABLE `a_read` (
  `label` varchar(35) DEFAULT NULL,
  `book_id` int(11) DEFAULT NULL,
  `readtime` date DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `a_read` (`label`, `book_id`, `readtime`)
VALUES
    ('Story 1',1,'2020-09-09'),
    ('Collection 3',3,'2020-09-09');


CREATE TABLE `a_books` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `label` varchar(35) DEFAULT NULL,
  `is_collection` int(11) DEFAULT 0,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `a_books` (`id`, `label`, `is_collection`)
VALUES
    (1,'story 1',NULL),
    (2,'story 2',NULL),
    (3,'story 3',1);

我的查询

SELECT a_books.* ,a_read.readtime,a_read.label
FROM a_books

LEFT JOIN a_books_subcollection
ON a_books_subcollection.book_id=a_books.id

LEFT JOIN a_read
ON a_books.id=a_read.book_id

ORDER BY a_books.id

Answer 1

SELECT b.id,
       b.label book_label,
       b.is_collection,
       r.readtime,
       COALESCE(bs.label, b.label) label
FROM a_books b
LEFT JOIN a_books_subcollection bs ON bs.book_id=b.id
LEFT JOIN a_read r ON r.book_id = b.id AND r.label IN (b.label, bs.label)
ORDER BY id, label

fiddle

Answer 2

SELECT a.*,c.readtime ,IF(a.is_collection=1,b.label,c.label) 
FROM a_books a
LEFT JOIN a_books_subcollection b ON b.book_id=a.id
LEFT JOIN a_read c on c.book_id=a.id and (c.label=b.label or c.label=a.label)
ORDER BY a.id,b.label

也许这可以解决您的问题，但是您想要从表label或a_books_subcollection中获取的字段a_read的结果确实是模棱两可的。我添加条件来获取标签，如果来自表is_collection的数据a_books是1从表a_books_subcollection获取标签，否则从表a_read获取标签