如何在 postgresql 的列中返回 ARRAY/JSON 结构？

Question

您好，我有一组表格如下，用一个最小的示例来解释我的问题（我想知道如何检索该结构，而不是更改数据库结构）：

fruit: 
id | name | form| texture_id
-------------------------------
1 | Apple | round | 1
2 | Banana | long | 1

fruit_varieties: 
id | name | fruit_id | color
-------------------------------
1 | golden   | 1| green
2 | fuji     | 1| red
3 | canarias | 2| yellow

fruit_texture
id | name
-------------------------------
1 | soft
2 | hard

variety_countries
id | name | fruit_variety_id | average_temperature
-------------------------------
1 | france | 1 | 21
2 | spain | 1  | 24
3 | italy | 2  | 23

我想得到这个结构如下： 对于给定的fruit.name=Apple：

{
    "fruit_name" = "Apple",
    "form" = "round",
    "texture" = "soft",
    "fruit_properties" = [{
            "variety_name" = "Golden",
            "color" = "green",
            "countries" = [{
                    "country" = "france",
                    "avg_temperature" = "21",
                }, {
                    "country" = "spain",
                    "avg_temperature" = "24",
                }
            ]
        }, {
            "variety_name" = "fuji",
            "color" = "red",
            "countries" = [{
                    "country" = "italy",
                    "avg_temperature" = "23",
                }
            ]
        }
    ]
}

所以我从这样的事情开始

SELECT 
fruit.name AS fruit_name,
fruit.form AS form,
fruit_texture.name AS texture,
(
    # I don't know how to handle this
) AS fruit_properties
FROM fruit
JOIN fruit_varieties
ON fruit.id = fruit_varieties.fruit_id
WHERE fruit.name = 'Apple'

现在我不知道如何在列中返回该数组，或创建包含整个响应的 JSON。我已经尝试使用一些 JSON PATH 函数已经有几个小时了

有人可以用这个简单的例子给我一个提示吗？

Answer 1

您的输出结构不是标准的 JSON 格式。键和值之间应该是: 而不是=。考虑到您想要标准 JSON 输出，请尝试下面提到的查询：

select row_to_json(d2) from (
select
name,
form,
texture,
json_agg(json_build_object('variety_name',variety_name,'color',color,'countries',countries)) "fruit_properties"
from 
(
select 
t1.name "name",
t1.form "form",
t3.name "texture",
t2.name "variety_name",
t2.color "color",
json_agg(json_build_object( 'country',t4.name,'temp',t4.average_temperature)) "countries"
from 
fruit t1 inner join fruit_varieties t2 on t1.id=t2.fruit_id
inner join fruit_texture t3 on t1.texture_id=t3.id
inner join variety_countries  t4 on t4.fruit_variety_id=t2.id
group by 1,2,3,4,5
) d1
group by 1,2,3
) d2
where d2.name='Apple'

DEMO

如果您不使用 where 子句，上述查询将为每个水果返回一个带有 JSON 值的行。

如果您确实想要问题中提到的输出，请在上述查询中将 row_to_json(d2) 替换为 replace(row_to_json(d2)::text,':', ' = ')。