Question

我是Rails的新手，所以我为问题的标题道歉，我不知道怎么说。随意改变它。我正在建立一个扑克游戏来学习rails，我有以下的关联......

class Game < ActiveRecord::Base
    has_many :players, :dependent => :destroy
    has_many :community_cards, :class_name => "Card", :dependent => :destroy, :conditions => { :is_community_card => true }
    has_many :used_cards, :class_name => "Card", :dependent => :destroy, :conditions => { :is_community_card => false }
    attr_accessible :pot, :name, :status

class Player < ActiveRecord::Base
    belongs_to :game
    has_many :cards, :dependent => :destroy
    attr_accessible :chip_count, :position, :fb_id
end

class Card < ActiveRecord::Base
    belongs_to :player
    belongs_to :game
    attr_accessible :face, :suit, :is_community_card
end

当我尝试向所有玩家发放随机牌时，所有牌都以一名玩家结束......

def deal_players_hole_cards
    players.all.each do |p|
        if(p.cards.count < 2)
            first_card = deal_card()
            second_card = deal_card()
            p.cards << first_card
            p.cards << second_card
        end
    end
end

这是交易卡方法......

def deal_card
    card_was_found = false
    while(!card_was_found) do
        card_was_found = true
        random_suit = (0..3).to_a.sample
        random_face = (1..13).to_a.sample

        used_cards.all.each do |used_card|
            if(random_suit == used_card.suit and random_face == used_card.face)
                card_was_found = false
            end
        end
    end

    new_card = Card.create(:suit => random_suit, :face => random_face, :is_community_card => false)
    used_cards << new_card 
end

有两个玩家，每个玩家应该有两张牌，相反，一个玩家拥有全部四张牌......

ruby-1.9.2-p290 :001 > Game.last.players.last.cards.count
  Game Load (0.1ms)  SELECT "games".* FROM "games" ORDER BY "games"."id" DESC LIMIT 1
  Player Load (0.1ms)  SELECT "players".* FROM "players" WHERE "players"."game_id" = 2 ORDER BY "players"."id" DESC LIMIT 1
   (0.2ms)  SELECT COUNT(*) FROM "cards" WHERE "cards"."player_id" = 6
 => 4 
ruby-1.9.2-p290 :002 > Game.last.players.first.cards.count
  Game Load (0.4ms)  SELECT "games".* FROM "games" ORDER BY "games"."id" DESC LIMIT 1
  Player Load (0.3ms)  SELECT "players".* FROM "players" WHERE "players"."game_id" = 2 LIMIT 1
   (0.2ms)  SELECT COUNT(*) FROM "cards" WHERE "cards"."player_id" = 5
 => 0

非常感谢您的所有智慧！

Answer 1

我最后通过将播放器对象传递给交易卡方法并使用构建方法将新创建的卡附加到播放器来解决此问题...

####################################################################
    # Deals each player their two hole cards
    ####################################################################
    def deal_players_hole_cards
        players.all.each do |p|
            if(p.cards.count < 2)
                deal_card(p)
                deal_card(p)
            end
        end
    end

    ####################################################################
    # returns a random, unused card
    ####################################################################
    def deal_card(p)
        card_was_found = false
        while(!card_was_found) do
            card_was_found = true
            random_suit = (0..3).to_a.sample
            random_face = (1..13).to_a.sample

            used_cards.all.each do |used_card|
                if(random_suit == used_card.suit and random_face == used_card.face)
                    card_was_found = false
                end
            end
        end

        new_card = p.cards.build(:suit => random_suit, :face => random_face, :is_community_card => false)
        used_cards << new_card 
    end

结果...

ruby-1.9.2-p290 :005 > Game.last.players.first.cards.count
  Game Load (0.3ms)  SELECT "games".* FROM "games" ORDER BY "games"."id" DESC LIMIT 1
  Player Load (0.4ms)  SELECT "players".* FROM "players" WHERE "players"."game_id" = 2 LIMIT 1
   (0.3ms)  SELECT COUNT(*) FROM "cards" WHERE "cards"."player_id" = 5
 => 2 
ruby-1.9.2-p290 :006 > Game.last.players.last.cards.count
  Game Load (0.4ms)  SELECT "games".* FROM "games" ORDER BY "games"."id" DESC LIMIT 1
  Player Load (0.4ms)  SELECT "players".* FROM "players" WHERE "players"."game_id" = 2 ORDER BY "players"."id" DESC LIMIT 1
   (0.2ms)  SELECT COUNT(*) FROM "cards" WHERE "cards"."player_id" = 6
 => 2

我仍然很好奇为什么我的旧代码不起作用。

Rails 3.1：为什么所有发行的牌最终都附加到一个玩家？

1 个答案: