PostgreSQL生成JSON字符串的方法汇总

-- 示例：用户表单行转 JSON
SELECT row_to_json(users) FROM users WHERE id = 1;

{"id":1,"name":"张三","age":25,"email":"zhangsan@test.com"}

-- 自定义键名，只返回需要的字段
SELECT 
  json_build_object(
    'userId', id,
    'userName', name,
    'userAge', age
  ) AS user_json
FROM users 
WHERE id = 1;

{"userId":1,"userName":"张三","userAge":25}

-- 所有用户转 JSON 数组
SELECT json_agg(users) FROM users;

-- 只筛选指定字段转数组
SELECT json_agg(
  json_build_object('id',id,'name',name)
) FROM users WHERE age > 20;

[{"id":1,"name":"张三"},{"id":2,"name":"李四"}]

SELECT array_to_json(ARRAY[1,2,3,4]);  -- [1,2,3,4]
SELECT array_to_json(ARRAY['苹果','香蕉']);  -- ["苹果","香蕉"]

SELECT json_build_object(
  'code', 200,
  'msg', 'success',
  'data', json_build_object(
    'user', row_to_json(users),
    'createTime', now()
  )
) FROM users WHERE id = 1;

{
  "code":200,
  "msg":"success",
  "data":{
    "user":{"id":1,"name":"张三","age":25},
    "createTime":"2025-05-20 10:00:00"
  }
}

SELECT json_build_object(
  'total', 2,
  'list', json_agg(users)
) FROM users;

-- 输出标准 JSON 字符串
SELECT row_to_json(users)::text FROM users WHERE id = 1;