按对象键将 SQL JSONB 数组转换为列

Question

我的表格结构如下：

CREATE TABLE artists (artist TEXT UNIQUE, facts JSONB);

INSERT INTO artists (artist, facts) 
  VALUES ('adele', '[{"type": "full_name", "value": "Adele Laurie"}, {"type": "age", "value": "25"}]');

INSERT INTO artists (artist, facts) 
  VALUES ('taylor', '[{"type": "age", "value": "25"}, {"type": "last_album", "value": "1989"}]');

事实“类型”的数量是固定的，但并不是每个艺术家都会拥有每个事实。如何选择包含每种事实类型的列和缺失事实名称的空值的结果？

期望的输出：

| artist |  full_name    | age  | last_album |
|--------|---------------|------|------------|
| adele  | Adele Laurie  | 25   | null       |
| taylor | null          | 25   | 1989       |

Answer 1

你可以这样做：

select a.artist, 
     max(case when b.value->>'type' = 'full_name' 
         then b.value->>'value' 
         else b.value->>'full_name' end) as full_name,
     max(case when b.value->>'type' = 'age' 
         then b.value->>'value' 
         else b.value->>'age' end) as age,
     max(case when b.value->>'type' = 'last_album' 
         then b.value->>'value' 
         else b.value->>'last_album' end) as last_album
from artists a, 
     json_array_elements(a.facts) b
group by a.artist
order by a.artist

在这里查看：http://sqlfiddle.com/#!15/e376b/2

在小提琴中，我将字段创建为 JSON，因为没有可用的 JSONB 类型

如果您需要添加更多类型，只需像其他类型一样将其添加为案例条件。我想你可以从这里弄清楚:)

编辑

即使您更改了格式，此查询也应该可以解决您的问题。刚刚编辑了小提琴。在这里看到它：http://sqlfiddle.com/#!15/1c2b6/2

唯一的区别是你并不真的需要 else 的 case 语句。

这是没有 else 语句的查询

select a.artist, 
     max(case when b.value->>'type' = 'full_name' 
         then b.value->>'value' end) as full_name,
     max(case when b.value->>'type' = 'age' 
         then b.value->>'value' end) as age,
     max(case when b.value->>'type' = 'last_album' 
         then b.value->>'value' end) as last_album
from artists2 a, 
     json_array_elements(a.facts) b
group by a.artist
order by a.artist;

我在这里编辑了 SqlFiddle 链接。

Answer 2

我会使用crosstab() 函数。 @Eriwn 真的有good answers 使用这种方法。

例子：

SELECT * FROM crosstab(
    'WITH data AS (
      SELECT artist,fact
      FROM artists a, jsonb_array_elements(a.facts) fact
    )
    SELECT artist,type,value
    FROM data,jsonb_to_record(data.fact) AS x(type text,value text)',
    'WITH data AS (
      SELECT artist,fact
      FROM artists a, jsonb_array_elements(a.facts) fact
     )
    SELECT DISTINCT type
   FROM data,jsonb_to_record(data.fact) AS x(type text,value text)'
) AS ct (artist text,age text,full_name text,last_album TEXT);

结果：

 artist | age  |  full_name   | last_album 
--------+------+--------------+------------
 adele  |      | Adele Laurie | 25
 taylor | 1989 |              | 25
(2 rows)