在我脑海爆炸之前,我真的可以在这里使用一些帮助...... 鉴于以下数据结构:
SELECT * FROM (VALUES (1, 1, 1, 1), (2, 2, 2, 2)) AS t(day, apple, banana, orange);
day | apple | banana | orange
-----+-------+--------+--------
1 | 1 | 1 | 1
2 | 2 | 2 | 2
我想构建一个JSON
对象,如下所示:
{
"data": [
{
"day": 1,
"fruits": [
{
"key": "apple",
"value": 1
},
{
"key": "banana",
"value": 1
},
{
"key": "orange",
"value": 1
}
]
}
]
}
也许我离目标不远:
SELECT json_build_object(
'data', json_agg(
json_build_object(
'day', t.day,
'fruits', t)
)
) FROM (VALUES (1, 1, 1, 1), (2, 2, 2, 2)) AS t(day, apple, banana, orange);
结果:
{
"data": [
{
"day": 1,
"fruits": {
"day": 1,
"apple": 1,
"banana": 1,
"orange": 1
}
}
]
}
我知道有json_each
可以解决问题。但我正在努力将其应用于查询。
修改 :
这是我更新的查询,我猜,它非常接近。我放弃了用json_each
解决问题的想法。现在我只需要返回一个fruits
数组,而不是附加到fruits
对象:
SELECT json_build_object(
'data', json_agg(
json_build_object(
'day', t.day,
'fruits', json_build_object(
'key', 'apple',
'value', t.apple,
'key', 'banana',
'value', t.banana,
'key', 'orange',
'value', t.orange
)
)
)
) FROM (VALUES (1, 1, 1, 1), (2, 2, 2, 2)) AS t(day, apple, banana, orange);
我是否需要添加子查询以防止嵌套聚合函数?
答案 0 :(得分:3)
使用函数jsonb_each()
获取对(key, value),
,这样您无需知道列数及其名称即可获得正确的输出:
select jsonb_build_object('data', jsonb_agg(to_jsonb(s) order by day))
from (
select day, jsonb_agg(jsonb_build_object('key', key, 'value', value)) as fruits
from (
values (1, 1, 1, 1), (2, 2, 2, 2)
) as t(day, apple, banana, orange),
jsonb_each(to_jsonb(t)- 'day')
group by 1
) s;
上面的查询给出了这个对象:
{
"data": [
{
"day": 1,
"fruits": [
{
"key": "apple",
"value": 1
},
{
"key": "banana",
"value": 1
},
{
"key": "orange",
"value": 1
}
]
},
{
"day": 2,
"fruits": [
{
"key": "apple",
"value": 2
},
{
"key": "banana",
"value": 2
},
{
"key": "orange",
"value": 2
}
]
}
]
}