我需要从完全结构化的 JSON 查询中获取结果。 我可以在 postgres 中看到一些可能有用的内置函数。
作为示例,我创建了一个结构如下:
-- Table: person
-- DROP TABLE person;
CREATE TABLE person
(
id integer NOT NULL,
name character varying(30),
CONSTRAINT person_pk PRIMARY KEY (id)
)
WITH (
OIDS=FALSE
);
ALTER TABLE person
OWNER TO postgres;
-- Table: car
-- DROP TABLE car;
CREATE TABLE car
(
id integer NOT NULL,
type character varying(30),
personid integer,
CONSTRAINT car_pk PRIMARY KEY (id)
)
WITH (
OIDS=FALSE
);
ALTER TABLE car
OWNER TO postgres;
-- Table: wheel
-- DROP TABLE wheel;
CREATE TABLE wheel
(
id integer NOT NULL,
whichone character varying(30),
serialnumber integer,
carid integer,
CONSTRAINT "Wheel_PK" PRIMARY KEY (id)
)
WITH (
OIDS=FALSE
);
ALTER TABLE wheel
OWNER TO postgres;
还有一些数据:
INSERT INTO person(id, name)
VALUES (1, 'Johny'),
(2, 'Freddy');
INSERT INTO car(id, type, personid)
VALUES (1, 'Toyota', 1),
(2, 'Fiat', 1),
(3, 'Opel', 2);
INSERT INTO wheel(id, whichone, serialnumber, carid)
VALUES (1, 'front', '11', 1),
(2, 'back', '12', 1),
(3, 'front', '21', 2),
(4, 'back', '22', 2),
(5, 'front', '3', 3);
因此,我想要一个包含人员列表的 JSON 对象,每个人都有汽车列表和每辆汽车的车轮列表。
我试过类似的东西,但这不是我想要的:
select json_build_object(
'Persons', json_build_object(
'person_name', person.name,
'cars', json_build_object(
'carid', car.id,
'type', car.type,
'comment', 'nice car', -- this is constant
'wheels', json_build_object(
'which', wheel.whichone,
'serial number', wheel.serialnumber
)
))
)
from
person
left join car on car.personid = person.id
left join wheel on wheel.carid = car.id
我想我缺少一些 group by 和 json_agg 但我不确定该怎么做。
我想得到这样的结果:
{ "persons": [
{
"person_name": "Johny",
"cars": [
{
"carid": 1,
"type": "Toyota",
"comment": "nice car",
"wheels": [{
"which": "Front",
"serial number": 11
},
{
"which": "Back",
"serial number": 12
}]
},
{
"carid": 2,
"type": "Fiat",
"comment": "nice car",
"wheels": [{
"which": "Front",
"serial number": 21
},{
"which": "Back",
"serial number": 22
}]
}
]
},
{
"person_name": "Freddy",
"cars": [
{
"carid": 3,
"type": "Opel",
"comment": "nice car",
"wheels": [{
"which": "Front",
"serial number": 33
}]
}]
}]
}
http://www.jsoneditoronline.org/?id=7792a0a2bf11be724c29bb86c4b14577
最佳答案
您应该构建分层查询以获得分层结构作为结果。
你想在一个 json 对象中有很多人,所以使用 json_agg()
在 json 数组中收集人员。
类似地,一个人可以拥有多辆汽车,您应该将属于一个人的汽车放在一个 json 数组中。这同样适用于汽车和车轮。
select
json_build_object(
'persons', json_agg(
json_build_object(
'person_name', p.name,
'cars', cars
)
)
) persons
from person p
left join (
select
personid,
json_agg(
json_build_object(
'carid', c.id,
'type', c.type,
'comment', 'nice car', -- this is constant
'wheels', wheels
)
) cars
from
car c
left join (
select
carid,
json_agg(
json_build_object(
'which', w.whichone,
'serial number', w.serialnumber
)
) wheels
from wheel w
group by 1
) w on c.id = w.carid
group by personid
) c on p.id = c.personid;
(格式化的)结果:
{
"persons": [
{
"person_name": "Johny",
"cars": [
{
"carid": 1,
"type": "Toyota",
"comment": "nice car",
"wheels": [
{
"which": "front",
"serial number": 11
},
{
"which": "back",
"serial number": 12
}
]
},
{
"carid": 2,
"type": "Fiat",
"comment": "nice car",
"wheels": [
{
"which": "front",
"serial number": 21
},
{
"which": "back",
"serial number": 22
}
]
}
]
},
{
"person_name": "Freddy",
"cars": [
{
"carid": 3,
"type": "Opel",
"comment": "nice car",
"wheels": [
{
"which": "front",
"serial number": 3
}
]
}
]
}
]
}
如果您不熟悉嵌套派生表,您可以使用公用表表达式。 此变体说明应该从最嵌套的对象开始向最高级别构建查询:
with wheels as (
select
carid,
json_agg(
json_build_object(
'which', w.whichone,
'serial number', w.serialnumber
)
) wheels
from wheel w
group by 1
),
cars as (
select
personid,
json_agg(
json_build_object(
'carid', c.id,
'type', c.type,
'comment', 'nice car', -- this is constant
'wheels', wheels
)
) cars
from car c
left join wheels w on c.id = w.carid
group by c.personid
)
select
json_build_object(
'persons', json_agg(
json_build_object(
'person_name', p.name,
'cars', cars
)
)
) persons
from person p
left join cars c on p.id = c.personid;
关于sql - 从 sql 查询 postgres 9.4 创建嵌套的 json,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42222968/