我有一个这样的数据库结构:
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
);
我想在选择时得到如下结果:
{
"name": "<person.name>",
"age": "<person.age>"
"hometown": {
"name": "<tometown.name>",
"population": "<tometown.population>"
}
}
我已经在使用 psycopg2.extras.DictCursor
,所以我认为我需要使用 SQL 的 SELECT AS
。
这是我尝试过但没有结果的一个例子,我做了很多类似的小调整,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
有什么方法可以做到这一点,还是我应该单独选择所有列并在 Python 内部构建字典?
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.
最佳答案
像什么?
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)
关于python - Select into subdict 从多个表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42504724/