过去几天我一直在努力查询,我希望有人可以提供帮助。我有3张桌子:
CREATE TABLE user (
ID SERIAL PRIMARY KEY,
USERNAME TEXT NOT NULL UNIQUE
);
CREATE TABLE skill (
ID SERIAL PRIMARY KEY,
NAME TEXT NOT NULL
);
CREATE TABLE user_skill (
ID SERIAL PRIMARY KEY,
USER_ID INTEGER REFERENCES user(id),
SKILL_ID INTEGER REFERENCES skill(id)
);
我正在尝试运行一个查询,以便我可以返回一个具有JSON格式的单个对象,如下所示:
{
"id": 1,
"username": "test123",
"skills": [{
"name": "skillOne"
}, {
"name": "skillTwo"
}, {
"name": "skillThree"
}]
}
到目前为止我的样子是这样的:
SELECT
json_build_object(
'id', u.id,
'username', u.username,
'skills', jsonb_agg(skill)
)
FROM (
SELECT
jsonb_build_object(
'name', s.name
) skill
FROM user_skill us
JOIN skill s ON us.skill_id = s.id
) xombi_user u;
编辑:
我现在有了这个正确返回json数组,但现在需要构建一个json对象。
SELECT u.id, u.username, jsonb_agg(s)
FROM user u
LEFT JOIN user_skill us ON u.id = us.user_id
LEFT JOIN skill s ON us.skill_id = s.id
WHERE u.id = 60
GROUP BY u.id;
答案 0 :(得分:0)
知道了!希望它可以帮助某人:
SELECT
json_build_object(
'id', u.id,
'username', u.username,
'skills', jsonb_agg(s)
)
FROM user u
LEFT JOIN user_skill us ON u.id = us.user_id
LEFT JOIN skill s ON us.skill_id = s.id
WHERE u.id = 60
GROUP BY u.id;