从多个表创建视图而不复制输出

Question

我想从3个不同的表创建一个视图，而不是重复输出。我目前的代码似乎正确地组织了主题书籍。但是，它会将每个书名打印到同一个作者，一旦完成，它将移动到下一个作者。而不是15行，它最终是270。

CREATE VIEW book_summary(author_first_name, author_last_name, book_title, subject)
AS SELECT  DISTINCT first_name, last_name,title, subject
FROM authors, books  left join subjects
ON subjects.subject_id = books.subject_id;

我试过通过SOF找到答案。我从Thread遇到了DISTINCT和LEFT JOIN。我只是觉得我没有正确使用它

Output:
Andrew            | Brookins         | 2001: A Space Odyssey       | Science Fiction
 Andrew            | Brookins         | Bartholomew and the Oobleck | Children's Books
 Andrew            | Brookins         | Dune                        | Science Fiction
 Andrew            | Brookins         | Dynamic Anatomy             | Arts
 Andrew            | Brookins         | Franklin in the Dark        | Children's Books
 Andrew            | Brookins         | Goodnight Moon              | Children's Books
 Andrew            | Brookins         | Learning Python             | Computers
 Andrew            | Brookins         | Little Women                | Drama
 Andrew            | Brookins         | Perl Cookbook               | Computers
 Andrew            | Brookins         | Practical PostgreSQL        | Computers
 Andrew            | Brookins         | Programming Python          | Computers
 Andrew            | Brookins         | The Cat in the Hat          | Children's Books
 Andrew            | Brookins         | The Shining                 | Horror
 Andrew            | Brookins         | The Tell-Tale Heart         | Horror
 Andrew            | Brookins         | The Velveteen Rabbit        | Classics

（对于我在另一张桌子上的每位作者继续这种趋势）。

Answer 1

使用CROSS JOIN/comma syntax时会发生这种情况：

CREATE VIEW book_summary
   (author_first_name, author_last_name, book_title, subject)
AS 
SELECT DISTINCT first_name, last_name,title, subject
FROM authors, books  --here is the problem
left join subjects  
ON subjects.subject_id = books.subject_id;

应该是：

CREATE VIEW book_summary
   (author_first_name, author_last_name, book_title, subject)
AS 
SELECT first_name, last_name,title, subject
FROM authors a 
JOIN books b
  ON a.col_name = b.col_name
left join subjects s 
ON s.subject_id = b.subject_id;