首先,您需要进行一些更改。
-
class C 需要与D 关联
class C < ActiveRecord::Base
belongs_to :B
has_one :D
end
-
如果你想访问A的D,你也需要指定这个。
class A < ActiveRecord::Base
has_many :B
has_many :C, :through => :B
has_many :D, :through => :C
end
现在,访问A 的所有C 的:
-> a = A.where(:id => 1).includes(:C).first
A Load (0.2ms) SELECT "as".* FROM "as" WHERE "as"."id" = 1 LIMIT 1
B Load (0.1ms) SELECT "bs".* FROM "bs" WHERE "bs"."a_id" IN (1)
C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" IN (1, 2)
=> #<A id: 1, created_at: "2012-01-10 04:28:42", updated_at: "2012-01-10 04:28:42">
-> a.C
=> [#<C id: 1, b_id: 1, created_at: "2012-01-10 04:30:10", updated_at: "2012-01-10 04:30:10">, #<C id: 2, b_id: 1, created_at: "2012-01-10 04:30:11", updated_at: "2012-01-10 04:30:11">, #<C id: 3, b_id: 2, created_at: "2012-01-10 04:30:21", updated_at: "2012-01-10 04:30:21">, #<C id: 4, b_id: 2, created_at: "2012-01-10 04:30:21", updated_at: "2012-01-10 04:30:21">]
请注意,当您调用 a.C 时不会执行另一个查询。这是因为 ActiveRecord 知道您将希望通过include 调用访问找到的A 的C,并生成最少数量的查询。 D's 也是如此:
-> a = A.where(:id => 1).includes(:D).first
A Load (0.1ms) SELECT "as".* FROM "as" WHERE "as"."id" = 1 LIMIT 1
B Load (0.1ms) SELECT "bs".* FROM "bs" WHERE "bs"."a_id" IN (1)
C Load (0.1ms) SELECT "cs".* FROM "cs" WHERE "cs"."b_id" IN (1, 2)
D Load (0.1ms) SELECT "ds".* FROM "ds" WHERE "ds"."c_id" IN (1, 2, 3, 4)
假设您想要所有A 的D 但想要订购C:
A.where(:id => 1).includes(:C).order('cs.created_at DESC').includes(:D)
注意,您也可以将其设置为关联的默认值:
:order 选项规定了接收关联对象的顺序(在 SQL ORDER BY 子句使用的语法中)。
class Customer < ActiveRecord::Base
has_many :orders, :order => "date_confirmed DESC"
end