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Arrays 多对多表组合成数组_Arrays_Json_Postgresql_Aggregate Functions_Jsonb - Fatal编程技术网

Arrays 多对多表组合成数组

Arrays 多对多表组合成数组,arrays,json,postgresql,aggregate-functions,jsonb,Arrays,Json,Postgresql,Aggregate Functions,Jsonb,如何选择此格式的书籍: { title: 'Avengers', description:'Good book', tags: [{id:1, name: 'Drama'},{id:2, name: 'Horror'}] authors: [{id:1, name: 'Alex'},{id:2, name: 'Tanya'}] } 对于具有多对多关系的表: 或者最好是提出3个不同的池请求 书 作者 |id|name |1 |'Alex' |2 |'Tanya' 作

如何选择此格式的书籍:

{
   title: 'Avengers', 
   description:'Good book', 
   tags: [{id:1, name: 'Drama'},{id:2, name: 'Horror'}]
   authors: [{id:1, name: 'Alex'},{id:2, name: 'Tanya'}]
}
对于具有多对多关系的表:

或者最好是提出3个不同的池请求

作者

|id|name
|1 |'Alex'
|2 |'Tanya'
作家与书籍

|book_id|author_id
|1      |1
|1      |2
|2      |1
标签

标签和书籍

|book_id|tag_id
|1      |1
|1      |2
|2      |1
我以前在没有作者表的情况下使用过:

SELECT b.title, b.id, b.description,
 json_agg(json_build_object('id', t.id, 'name', t.name, 'description', 
 t.description, 'likes', tb.likes) ORDER BY tb.likes DESC) AS tags
 FROM books AS b
 INNER JOIN tags_books AS tb ON (b.id = tb.book_id)
 INNER JOIN tags AS t ON (tb.tag_id = t.id)
 WHERE b.id = $1
 GROUP BY b.id;`

但是,对于表autor的新内部联接,将有重复的值。

您应该将数据(
作者
标记
)聚合到单独的派生表中(在
FROM
子句中的子查询):


但是,您认为使用json_agg和json_build_对象进行联接会比单独请求更好吗?一般规则是,在联接之前在派生表中聚合数据比在联接表上聚合数据更快(或至少不慢),因为该操作是在较小的数据集上执行的。此外,单个查询(使用派生表)比两个单独的查询(除非在非常大的表上执行)更好。
|book_id|tag_id
|1      |1
|1      |2
|2      |1
SELECT b.title, b.id, b.description,
 json_agg(json_build_object('id', t.id, 'name', t.name, 'description', 
 t.description, 'likes', tb.likes) ORDER BY tb.likes DESC) AS tags
 FROM books AS b
 INNER JOIN tags_books AS tb ON (b.id = tb.book_id)
 INNER JOIN tags AS t ON (tb.tag_id = t.id)
 WHERE b.id = $1
 GROUP BY b.id;`
select 
    b.title, 
    b.description,
    t.tags, 
    a.authors
from books as b
join (
    select 
        tb.book_id,
        json_agg(json_build_object('id', t.id, 'name', t.name) order by tb.likes desc) as tags
    from tags_books as tb
    join tags as t on tb.tag_id = t.id
    group by tb.book_id
    ) t on b.id = t.book_id
join (
    select 
        ab.book_id,
        json_agg(json_build_object('id', a.id, 'name', a.name) order by ab.author_id) as authors
    from authors_books as ab
    join authors as a on ab.author_id = a.id
    group by ab.book_id
    ) a on b.id = a.book_id

  title   | description |                             tags                              |                           authors                           
----------+-------------+---------------------------------------------------------------+-------------------------------------------------------------
 Avengers | Good book   | [{"id" : 2, "name" : "Horror"}, {"id" : 1, "name" : "Drama"}] | [{"id" : 1, "name" : "Alex"}, {"id" : 2, "name" : "Tanya"}]
 Fear     | Scary       | [{"id" : 1, "name" : "Drama"}]                                | [{"id" : 1, "name" : "Alex"}]
(2 rows)