Mysql 嵌套hasManyThrough性能_Mysql_Laravel_Eloquent

Mysql 嵌套hasManyThrough性能

mysql laravel

Mysql 嵌套hasManyThrough性能,mysql,laravel,eloquent,Mysql,Laravel,Eloquent,我的数据库布局遇到问题。基本上，布局以以下方式呈现： - Community - - Each community has officers (Model `User`) - - Each community has participants (Model `User`) - Chat rooms (different section) - - each chat room has participants (Model `User`) - - each chat has officers,

我的数据库布局遇到问题。基本上，布局以以下方式呈现：

- Community
- - Each community has officers (Model `User`)
- - Each community has participants (Model `User`)

- Chat rooms (different section)
- - each chat room has participants (Model `User`)
- - each chat has officers, which are generated based on the participant's community's officers. This means: It's not hard-coded into a new table, instead, it should fetch the officers of the community, to which the participant (non-officer) belongs to.

我的奋斗目标是实现最后一部分，那就是军官可以进入。我现在的查询完全是一团糟，尤其是当前服务器的计划。考虑到代码很长，我将直截了当地说：

- Fetch chat rooms
- - fetch participants of each chat room
- - - fetch communities of each participant 
- - - - fetch officers of each community 
- - - - add these officers to a main array in the model
- - - - remove any duplicated officers

现在，您可以看到，这是一种非常松散的获得结果的方法，而且它需要大量的性能。官员和非官员之间唯一相关的模式是社区

最好的方法是什么？

您如何看待创建新的桌面聊天室社区，您将在其中存储社区的聊天室。该表将包含community\u id user\u id和chat\u room\u id列。每次用户加入/离开聊天室时，在此表中添加/删除记录。

每当用户加入新社区或离开新社区时，您也必须修改此新表，但使用此方法，您可以通过简单的查询从chat_room_community中选择*，其中community_id=？

获取所有可供官员使用的聊天室，我确实想到了这一点，但随后需要更新两个表。添加/删除官员时的社区官员表和新的聊天室社区表。对于雄辩的标准来说，这不是多余的吗？了解JOIN。我确实想过，但我认为使用雄辩的模型是最佳实践？我是否应该在关系内部进行原始查询？