Database 从多个表创建视图而不复制输出

我想从3个不同的表中创建一个视图，并且没有重复的输出。我目前的代码似乎正确地组织了主题的书名。但是，它会将每本书的标题打印给同一位作者，打印完成后会转到下一位作者。它不是15行，而是270行

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;

我试着通过软件查找答案。在这篇文章中，我遇到了DISTINCT和LEFT JOIN。我只是觉得我用的不对

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

---

（对我在另一个表中的每一位作者都是如此）。

当您使用

交叉连接/逗号语法时，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;

谢谢你的救命恩人：D