Question

我正在尝试实现通用通知系统..我有这样的数据结构;

notification_base:id,type,object
notificiation_sub:id,user_id,not_base_id,lastNotifyTime
notification_action:id,user_id,not_base_id,action,creationDate

所以基本情况是，用户通过发布状态或上传照片等来创建通知基础项目。（在这种情况下，照片指的是notification_base表中的类型字段，object_id是post_id或photo_id取决于类型）然后用户订阅此notification_base项目..（用户3订阅notification_base 5和最后通知时间x）

之后，另一个用户触摸此notification_base项。（例如评论状态或照片）此操作记录在notification_action表中（用户5在12/02/2011上执行'动作'）..

我想要的是从用户订阅中获取notification_base项目，如果上一次通知时间小于通知操作，则通过notification_action加入它们。我可以用这个sql继承它;

对于用户ID 3;

select * from notification_action
                inner join notification_base on notification_action.not_base_id = notification_base.id
                inner join notification_sub on notification_action.not_base_id = notification_sub.not_base_id
                where notification_sub.user_id = 3 and notification_sub.lastShowDate < notification_action.creationDate ;

结果几乎是我想要的，例如

用户x对你的对象做了'动作'，在时间t有'type'和object_id

但我也希望加入object_id取决于类型..所以我实际上可以学习触摸哪个对象..但是你可以看到类型是动态的，如果type = post对象id指的是post表上的post_id，如果type =照片对象id引用照片桌上的photo_id等。

我尝试这样做但得到一些语法错误;

SELECT *
FROM notification_action
INNER JOIN notification_base
    ON notification_action.not_base_id = notification_base.id
INNER JOIN notification_sub
    ON notification_action.not_base_id = notification_sub.not_base_id CASE notification_base.type
            WHEN 'photo'
                THEN (
                        INNER JOIN photo
                            ON photo.id = notification_base.object_id
                        )
            ELSE (
                    INNER JOIN post
                        ON post.id = notification_base.object_id
                    )
            END
WHERE notification_sub.user_id = 3
    AND notification_sub.lastShowDate < notification_action.creationDate;

我知道这不正确，就像伪代码

Answer 1

您无法有条件地跨表创建联接。最好的方法是使用LEFT JOIN并在CASE语句中使用SELECT。您必须指定要显示的公共列，

SELECT *,
        (CASE notification_base.type
            WHEN 'photo'
            THEN photo.columnName1
            ELSE post.ColumnName1
        END) as ColumnName1,
        (CASE notification_base.type
            WHEN 'photo'
            THEN photo.columnName2
            ELSE post.ColumnName2
        END) as ColumnName2
FROM notification_action
    INNER JOIN notification_base
        ON notification_action.not_base_id = notification_base.id
    INNER JOIN notification_sub
        ON notification_action.not_base_id = notification_sub.not_base_id
    INNER JOIN photo
        ON photo.id = notification_base.object_id
    INNER JOIN post
        ON post.id = notification_base.object_id
WHERE notification_sub.user_id = 3
    AND notification_sub.lastShowDate < notification_action.creationDate;

mysql条件连接取决于列

1 个答案: