sql - 如何摆脱 Postgres 查询中继承的 SELECT？

Question

Postgres 9.4
我想，就数据库性能而言，这样的查询不是最佳方法:

SELECT t.name,
    t.description,
    t.rating,
    t.readme,
    t.id AS userid,
    t.notifications
   FROM ( SELECT "user".name,
            "user".description,
            "user".rating,
            "user".readme,
            "user".id,
            ( SELECT array_to_json(array_agg(row_to_json(notifications.*))) AS array_to_json
                   FROM ( SELECT notification.id,
                            notification.action_type,
                            notification.user_id,
                            notification.user_name,
                            notification.resource_id,
                            notification.resource_name,
                            notification.resource_type,
                            notification.rating,
                            notification.owner
                           FROM notification
                          WHERE (notification.owner = "user".id)
                          ORDER BY notification.created DESC) notifications) AS notifications
           FROM "user") t

notification 列包含 json 对象以及来自 notification 表的所有匹配行。
我应该如何重建此查询以同样的方式接收数据？我想，我应该以某种方式使用 JOIN 命令。
我有一个请求，它利用了多个继承的 SELECT。

感谢您的宝贵时间!

Answer 1

最外层查询仅将id 别名为userid。您可以将别名移至内部查询，并完全省略外部查询。

然后您可以创建一个函数来创建通知 JSON:

create or replace function get_user_notifications(user_id bigint)
returns json language sql as
$$
    select  array_to_json(array_agg(row_to_json(n)))
    from    (
            select  id
            ,       action_type
            ,       ... other columns from notification ...
            from    notification
                    -- Use function name to refer to parameter not column
            where   user_id = get_user_notifications.user_id 
            order by
                    created desc
            ) n
$$;

现在您可以将查询写成:

select  id as userid
,       ... other columns from "user" ...
,       get_user_notifications(id) as notifications
from    "user" u;

这看起来好多了，代价是必须维护 Postgres 函数。