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使用has_many时,在给定两个实体的情况下找到连接模型的最简单方法是什么:通过

ruby-on-rails

Denis Hennessy · 技术社区 · 16 年前

has_many :through

class Group < ActiveRecord::Base
    has_many :memberships
    has_many :members, :through => :memberships
end 

class Member < ActiveRecord::Base
    has_many :memberships
    has_many :groups, :through => :memberships
end

create_table :memberships do |t|
  t.integer :group_id
  t.integer :member_id
  t.boolean :admin

  t.timestamps
end

membership = @member.memberships.find_by_group_id(@group.id)

find_by_group_id

2 回复 | 直到 16 年前

eggdrop 16 年前

避免使用辅助方法,您可以编写:

  @member.admin?(@group)

像这样的东西(未经测试!)。

  class Member < ActiveRecord::Base

    has_many :memberships
    has_many :groups, :through => :memberships
    named_scope :admin?, :through => :memberships, 
    lambda {|group|:conditions=> ["admin = 'true' and group_id = ?"], group.id }
  end

eggdrop 16 年前

你可以问:

  @member.is_admin?(@group)

  class Member < ActiveRecord::Base

    has_many :memberships
    has_many :groups, :through => :memberships

    def is_admin?(group)
      memberships.find_by_group_id(group.id).admin?
    end

  end

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