Question

我有这种关系：

class Action < ApplicationRecord
   has_many :actions_users

我试图进行如下查询：

select *
from actions left outer join actions_users
on actions_users.action_id = actions.id and actions_users.user_id = 1
where actions.user_id = 1

同时，根据我的经验，我尝试的所有结果，

select *
from actions left outer join actions_users
on actions_users.action_id = actions.id
where actions.user_id = 1 and actions_users.user_id = 1

加入条件代码和一般条件在where函数中。

我该如何解决？

Answer 1

您可以在联接查询中传递字符串，并为此使用rails表命名约定。

Action.joins("left outer join action_users on (action_users.id = actions.id and action_users.id = 1")).where('action_users.user_id = ? ', 1)

Answer 2

因为您有一般的where条件，所以可以使用includes。这将生成一个LEFT OUTER JOIN查询：

Action.includes(:actions_users).where(actions_users: { user_id: true })

或者，如果您使用的是Rails 5+，则Active Record提供了一种查找器方法left_outer_joins：

Action.left_outer_joins(:actions_users).where(actions_users: { user_id: true })

Answer 3

Action.left_outer_joins(:actions_users).where(user_id: 1)

select *
from actions left outer join actions_users
on actions_users.action_id = actions.id and actions_users.user_id = 1
where actions.user_id = 1

尽管您还没有要求，...

Action.left_outer_joins(:actions_users).where(actions_users: {status: 'active'})

select *
from actions left outer join actions_users
on actions_users.action_id = actions.id and actions_users.user_id = 1
where actions_users.status = 'active'

如何使外部连接脱离条件

3 个答案: