PostgreSQL实例以JSONB格式存储数据:
CREATE TABLE myschema.mytable (
id BIGSERIAL PRIMARY KEY,
data JSONB NOT NULL
)
数据数组可能包含以下对象:
{
"observations": {
"temperature": {
"type": "float",
"unit": "C",
"value": "23.1"
},
"pressure": {
"type": "float",
"unit": "mbar",
"value": "1011.3"
}
}
}
所选行应作为键值对返回,格式如下:
temperature,type,float,value,23.1,unit,C,pressure,type,float,value,1011.3,unit,mbar
以下查询至少返回每个对象,同时仍为JSON:
SELECT id, value FROM mytable JOIN jsonb_each_text(mytable.data->'observations') ON true;
1 | {"type": "float", "unit": "mbar", "value": 1140.5}
1 | {"type": "float", "unit": "C", "value": -0.9}
5 | {"type": "float", "unit": "mbar", "value": "1011.3"}
5 | {"type": "float", "unit": "C", "value": "23.1"}
但是结果是分开的并且不是文本格式。
如何返回data
中所有对象的键值对?
答案 0 :(得分:1)
这将拉平json结构,并有效地将值和顶级键名(例如温度和压力)连接起来,以达到预期的“深度”级别。看看这是否是您的初衷。
SELECT
id,
(
SELECT STRING_AGG(conc, ',')
FROM (
SELECT CONCAT_WS(',', key, STRING_AGG(value, ',')) AS conc
FROM (
SELECT key, (jsonb_each_text(value)).value
FROM jsonb_each(data->'observations')
) AS x
GROUP BY key
) AS csv
) AS csv
FROM mytable
结果:
| id | csv |
| --- | --------------------------------------------------- |
| 1 | pressure,float,mbar,1011.3,temperature,float,C,23.1 |
| 2 | pressure,bigint,unk,455,temperature,int,F,45 |