我有这样的数据库结构:
CREATE TABLE person (
id SERIAL PRIMARY KEY,
name TEXT NOT NULL,
age INTEGER NOT NULL,
hometown_id INTEGER REFERENCES town(id)
);
CREATE TABLE town (
id SERIAL PRIMARY KEY,
name TEXT NOT NULL,
population INTEGER NOT NULL
);
我想在选择时获得以下结果:
{
"name": "<person.name>",
"age": "<person.age>"
"hometown": {
"name": "<tometown.name>",
"population": "<tometown.population>"
}
}
我已经在使用psycopg2.extras.DictCursor
了,所以我想我需要使用SQL的SELECT AS
。
这是我尝试过没有resullt的例子,我做了很多类似的微调,所有这些都引发了不同的错误:
SELECT
person(name, age),
town(name, population) as town,
FROM person
JOIN town ON town.id = person.hometown_id
有没有办法做到这一点,或者我应该单独选择所有列并在Python中构建dict?
Postgres版本信息:
psql (9.4.6, server 9.5.2)
WARNING: psql major version 9.4, server major version 9.5.
Some psql features might not work.
smth喜欢?..
t=# with t as (
select to_json(town),* from town
)
select json_build_object('name',p.name,'age',age,'hometown',to_json) "NameItAsYou Wish"
from person p
join t on t.id=p.hometown_id
;
NameItAsYou Wish
--------------------------------------------------------------------------------
{"name" : "a", "age" : 23, "hometown" : {"id":1,"name":"tn","population":100}}
(1 row)