Question

class Manager < ApplicationRecord
  has_many :bands
end

class Band < ApplicationRecord
  belongs_to :manager
  has_many :member_bands
  has_many :members, through: :member_bands
end

class Member < ApplicationRecord
  has_many :member_bands
  has_many :bands, through: :member_bands
end

class MemberBand < ApplicationRecord
  belongs_to :member
  belongs_to :band
end

我最终要做的是让所有会员为特定经理。

我觉得我需要做一些连接，但是无法确定这一点。

我已经尝试了manager.bands.joins(:members)，但我只获得了多个bands而没有成员。

我现在正在与多个循环一起攻击这个：

manager.bands.each do |band|
  band.members.each do |member|
    ##
  end
end

那么，我怎样才能获得与给定members相关联的所有manager（通过bands连接的人）？

Answer 1

根据您的关联，您应该只需添加

irb(main):003:0> Manager.first.members
  Manager Load (0.2ms)  SELECT  "managers".* FROM "managers" ORDER BY "managers"."id" ASC LIMIT $1  [["LIMIT", 1]]
  Member Load (0.6ms)  SELECT  "members".* FROM "members" INNER JOIN "member_bands" ON "members"."id" = "member_bands"."member_id" INNER JOIN "bands" ON "member_bands"."band_id" = "bands"."id" WHERE "bands"."manager_id" = $1 LIMIT $2  [["manager_id", 1], ["LIMIT", 11]]
=> #<ActiveRecord::Associations::CollectionProxy [#<Member id: 1>, #<Member id: 2>, #<Member id: 3>, #<Member id: 5>, #<Member id: 6>, #<Member id: 7>]>

给你的经理。

{{1}}

如何使用has_many获取嵌套记录：通过关联？

1 个答案: