在JSON中选择数据

Question

我想在Postgres中使用SQL来提取数据，但希望进入JSON结果并只提取我需要的数据。

如果我写（在瓦伦蒂娜工作室）：

Select "data" from "cars" 

第一行是这样的：

[{"Model": "Golf", "Make": "VW", "Engine": "2.9"}, 
{"Model": "M3", "Make": "BMW", "Engine": "3.0"}] 

我想什么是简单的：

Golf, M3或"Golf", "M3"

然后我也可以对012使用相同的方法或"Engine"

本质上我不希望我的结果在JSON中。

Answer 1

使用json_array_elements()：

with cars(data) as (
values 
    ('[ 
     {"Model": "Golf", "Make": "VW", "Engine": "2.9"}, 
     {"Model": "M3", "Make": "BMW", "Engine": "3.0"} 
    ]'::json) 
) 

select 
    elem->>'Model' as model, 
    elem->>'Make' as make, 
    elem->>'Engine' as engine 
from cars, 
lateral json_array_elements(data) elem 

model | make | engine 
-------+------+-------- 
Golf | VW | 2.9 
M3 | BMW | 3.0 
(2 rows)  

Answer 2

SELECT string_agg(x->>'Model', ',') 
FROM cars 
    CROSS JOIN LATERAL 
     jsonb_array_elements(data) x 
GROUP BY cars; 

┌────────────┐ 
│ string_agg │ 
├────────────┤ 
│ Golf,M3 │ 
└────────────┘ 
(1 row) 

Answer 3

我想我也得到了我想要的东西（和融化它，我正打算做的方式）与：从上面我

select obj->'Model' as model, obj->'Make' as make, 
obj->'Engine' as engine from data n, jsonb_array_elements(n.DATA) as obj 

还可以使用：

select 
    elem->>'Model' as model, 
    elem->>'Make' as make , 
    elem->>'Engine' as engine 
from cars, 
lateral jsonb_array_elements(data) elem; 

感谢您的帮助！