從兩個或多個變量的Mysql內部連接？

Question

我有一個表，如：叫做書

+------+-------------------+---------------+ 
| id | book_title  | author  | 
+------+-------------------+---------------+ 
| 1 | learning mysql | 1234 12  | 
+------+-------------------+---------------+ 
| 2 | learning php  | 125 50  | 
+------+-------------------+---------------+ 

，我想一個VIEW從書本和作家：

+------+-------------------+ 
| id | author   | 
+------+-------------------+ 
| 12 | JOHN    | 
+------+-------------------+ 
| 50 | PAUL    | 
+------+-------------------+ 
| 125 | CHRISTOPHER  | 
+------+-------------------+ 
| 1234 | PATRICK   | 
+------+-------------------+ 

，這樣我可以有我應該是這樣的表中查看低於，但應顯示而不是來自tabel AUTHOR的作者id作者姓名。

+------+-------------------+-----------------------------------+ 
| id | book_title  | author       | 
+------+-------------------+-----------------------------------+ 
| 1 | learning mysql | PATRICK       | 
+------+-------------------+-----------------------------------+ 
| 1 | learning mysql | CHRISTOPHER      | 
+------+-------------------+-----------------------------------+ 
| 2 | learning php  | JOHN        | 
+------+-------------------+-----------------------------------+ 
| 2 | learning php  | PAUL        | 
+------+-------------------+-----------------------------------+ 

Answer 1

您的數據庫的設計是錯誤的。 books.author應該是INT幷包含author.id的外鍵。表author應包含name VARCHAR(255)（或引用author_name的兩列，我仍然不確定您是否有兩位作者，或者您將姓名分爲兩個條目）。

所以正確的設計應該是：

BOOKS (
    id INT, 
    book_title VARCHAR(255), 
    author INT, -- only if each book has just one author 
    PRIMARY KEY (id) 
) 

AUTHOR (
    id INT, 
    name VARCHAR(255), 
    first_name_id INT, -- If you want to split names into more columns 
    PRIMARY KEY (id) 
) 

-- If you need more authors for one book 
-- you maybe should keep original (primary) author id 
BOOK_AUTHOR (
    book_id INT, 
    author_id INT, 
    PRIMARY KEY (book_id, author_id) 
); 

比你可以選擇與數據：

SELECT BOOKS.id, BOOKS.book_title, AUTHOR.name AS author 
FROM BOOKS 
-- Study difference between left and inner joins 
INNER JOIN AUTHOR on AUTHOR.id = BOOKS.author 

如果你需要有更多的作家的一本書：

SELECT BOOKS.id, BOOKS.book_title, AUTHOR.name AS author 
FROM BOOKS 
LEFT JOIN BOOK_AUTHOR on BOOK_AUTHOR.book_id = BOOKS.id 
LEFT JOIN AUTHOR on AUTHOR.id = BOOK_AUTHOR.author_id 

您可能需要輸出作者Wolfgang Goethe; Oscar Wilde，比您可以使用GROUP_CONCAT：

SELECT BOOKS.id, BOOKS.book_title, 
     GROUP_CONCAT(AUTHOR.name SEPARATOR '; ') AS author 
FROM BOOKS 
LEFT JOIN BOOK_AUTHOR on BOOK_AUTHOR.book_id = BOOKS.id 
LEFT JOIN AUTHOR on AUTHOR.id = BOOK_AUTHOR.author_id 
GROUP BY BOOKS.id 

Answer 2

正如註釋中的knittl註釋，您應該明確地修復您的數據庫設計。這就是說，這裏是使用現有的架構中的「蠻力和無知」的解決方案：

CREATE VIEW book_author (id, book_title, author) 
AS SELECT book.id, book.book_title, author.author 
FROM book 
    JOIN author ON FIND_IN_SET(author.id, REPLACE(book.author, ' ', ',')) > 0 

但是這將是一個非常緩慢的，醜陋的解決方案，即使它的工作原理。一個更好的解決辦法是在你的book表擺脫author列，而是添加一個鏈接表是這樣的：

CREATE TABLE book_author (
    book INTEGER NOT NULL, 
    author INTEGER NOT NULL, 
    PRIMARY KEY (book, author), 
    UNIQUE KEY (author, book) /* for "all books by author X" queries */ 
) 

（不像Vyktor，I don't feel that adding extra ID columns to simple link tables is a good idea有已在該表中一個非常好的自然主鍵。 —它並不需要一個代理鍵）

然後你就可以更輕鬆地創建視圖和高效。

CREATE VIEW book_author (id, book_title, author) 
AS SELECT book.id, book.book_title, author.author 
FROM book 
    JOIN book_author ON book_author.book = book.id 
    JOIN author ON book_author.author = author.id