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Rails::逆和关联扩展

inverse associations ruby-on-rails-3 ruby ruby-on-rails

9

Chowlett · 技术社区 · 15 年前

我有以下设置

class Player < ActiveRecord::Base
  has_many :cards, :inverse_of => :player do
    def in_hand
      find_all_by_location('hand')
    end
  end
end

class Card < ActiveRecord::Base
  belongs_to :player, :inverse_of => :cards
end

p = Player.find(:first)
c = p.cards[0]
p.score # => 2
c.player.score # => 2
p.score += 1
c.player.score # => 3
c.player.score += 2
p.score # => 5

但以下行为并不相同:

p = Player.find(:first)
c = p.cards.in_hand[0]
p.score # => 2
c.player.score # => 2
p.score += 1
c.player.score # => 2
c.player.score += 2
p.score # => 3

d = p.cards.in_hand[1]
d.player.score # => 2

我怎样才能 :inverse_of

2 回复 | 直到 15 年前

1

4

RajeshT 14 年前

它不起作用,因为“in-hand”方法有一个返回到数据库的查询。

由于选项的逆,工作代码知道如何使用已经在内存中的对象。

http://api.rubyonrails.org/classes/ActiveRecord/Associations/ClassMethods.html

2

7

Chowlett 15 年前

我已经找到了一个解决方法,如果(我是)你愿意放弃Arel授予的SQL优化,只需要用Ruby就可以了。

class Player < ActiveRecord::Base
  has_many :cards, :inverse_of => :player do
    def in_hand
      select {|c| c.location == 'hand'}
    end
  end
end

class Card < ActiveRecord::Base
  belongs_to :player, :inverse_of => :cards
end

:inverse_of :

p = Player.find(:first)
c = p.cards[0]
p.score # => 2
c.player.score # => 2
p.score += 1
c.player.score # => 3
c.player.score += 2
p.score # => 5

d = p.cards.in_hand[0]
d.player.score # => 5
d.player.score += 3
c.player.score # => 8