获取至少一个关联对象的记录 [英] get record with at least one associated object

问题描述

我在 mongoid 中具有以下架构:

用户有很多任务-has_many:tasks

User has many tasks - has_many :tasks

任务属于用户-属于:user

Task belongs to user - belongs_to :user

我如何只让10位第一位用户完成至少一项任务?

How can I get only 10 first users with at least one task?

类似这样的东西:

User.where(:tasks.ne => [] ).limit(10)

推荐答案

您的问题是Mongoid的

Your problem is that Mongoid's has_many doesn't leave anything in the parent document so there are no queries on the parent document that will do anything useful for you. However, the belongs_to :user in your Task will add a :user_id field to the tasks collection. That leaves you with horrific things like this:

user_ids = Task.all.distinct(:user_id)
users    = User.where(:id => user_ids).limit(10)

当然，如果您使用的是 embeds_many :tasks 而不是has_many :tasks，然后您可以根据需要查询users集合内的:tasks.太太，这可能会破坏其他东西.

Of course, if you had embeds_many :tasks instead of has_many :tasks then you could query the :tasks inside the users collection as you want to. OTOH, this would probably break other things.

如果您需要将任务分开放置(即未嵌入)，则可以在User中设置一个计数器来跟踪任务的数量，然后您可以说:

If you need to keep the tasks separate (i.e. not embedded) then you could set up a counter in User to keep track of the number of tasks and then you could say things like:

User.where(:num_tasks.gt => 0).limit(10)

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