Question

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。

以下是我尝试过没有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

任何方法都可以这样做，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.

Answer 1

是什么样的？..

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)

从多个表中选择subict

1 个答案: