Question

我想为此查询返回的每个string_agg元素强制添加新列（即Fiction，Mystery改为一列为“ Fiction”，下一列为“ Mystery”），并且2.）我需要能够将标签栏最多扩展到五个标签：

SELECT books.isbn_13 as "ISBN", title as "Title", 
  author as "Author", 
  string_agg(tag_name, ', ') as "Tags"
FROM books 
LEFT JOIN book_tags on books.isbn_13 = book_tags.isbn_13 
GROUP BY books.isbn_13;

现在一切看起来都不错，除了我希望每个Tag都用一列而不是逗号分隔的值。这是我当前的结果：

    ISBN      |            Title            |      Author       |       Tags
1111111111111 | The Adventures of Steve     | Russell Barron    | Fiction, Mystery
2222222222222 | It's all a mystery to me    | Mystery Man       | Mystery
3333333333333 | Biography of a Programmer   | Solo Artist       | Biography
4444444444444 | Steve and Russel go to Mars | Russell Groupon   |
6666666666666 | Newest Book you Must Have   | Newbie Onthescene |

所需的结果（将标签分成多个的列）：

    ISBN      |            Title            |      Author       |       Tag1    | Tag2    | Tag3              | Tag4    
1111111111111 | The Adventures of Steve     | Russell Barron    | Fiction       | Mystery | Male Protagonists | Fantasy| 
2222222222222 | It's all a mystery to me    | Mystery Man       | Mystery
3333333333333 | Biography of a Programmer   | Solo Artist       | Biography
4444444444444 | Steve and Russel go to Mars | Russell Groupon   |
6666666666666 | Newest Book you Must Have   | Newbie Onthescene |

SCHEMA for books table (parent):
CREATE TABLE public.books
(
    isbn_13 character varying(13) COLLATE pg_catalog."default" NOT NULL,
    title character varying(100) COLLATE pg_catalog."default",
    author character varying(80) COLLATE pg_catalog."default",
    publish_date date,
    price numeric(6,2),
    content bytea,
    CONSTRAINT books_pkey PRIMARY KEY (isbn_13)
)

SCHEMA book_tags table:
CREATE TABLE public.book_tags
(
isbn_13 character varying(13) COLLATE pg_catalog."default" NOT NULL,
tag_name character varying(30) COLLATE pg_catalog."default" NOT NULL,
CONSTRAINT book_tags_pkey PRIMARY KEY (isbn_13, tag_name),
CONSTRAINT book_tags_isbn_13_fkey FOREIGN KEY (isbn_13)
    REFERENCES public.books (isbn_13) MATCH SIMPLE
    ON UPDATE NO ACTION
    ON DELETE CASCADE
)

我已经对分组依据进行了研究，但交叉表/数据透视表却没有运气。这似乎应该是一件简单的事情，但是我是一个初学者，还没有找到答案。预先感谢您的指导。

Answer 1

With CTE as (    
    SELECT books.isbn_13 as "ISBN", 
           title as "Title", 
           author as "Author", 
           tag_name as "Tag",
           row_number() over (partition by books.isbn_13) as rn
    FROM books 
    LEFT JOIN book_tags 
      on books.isbn_13 = book_tags.isbn_13
)
SELECT "ISBN", "Title", "Author", 
       MAX( CASE WHEN rn = 1 THEN Tag END) as Tag1,
       MAX( CASE WHEN rn = 2 THEN Tag END) as Tag2,
       MAX( CASE WHEN rn = 3 THEN Tag END) as Tag3,
       MAX( CASE WHEN rn = 4 THEN Tag END) as Tag4,
       MAX( CASE WHEN rn = 5 THEN Tag END) as Tag5
FROM CTE
GROUP BY  "ISBN", "Title", "Author";

PostgreSQL为每个array_agg元素创建一列，而不是用逗号分隔

1 个答案: