Ruby OOP tic tac toe哈希/条件/itiration

试图用Ruby创建一个经典的tic-tac-toe-OOP，但是我的

game\u results（）

方法遇到了问题

我意识到这并不完全，需要更多的功能，但现在我只是想用用户输入的对象填充我的棋盘，然后输出填充的棋盘和赢家

当我在棋盘上看到我们有一个赢家时，我调用

game\u results（）

方法，它会给我正确的赢家，但每当平局时，我都会收到一个错误

有人对我在操作人员身上犯的错误有什么想法或解决方案吗？顺便说一句，我知道这一切是相当混乱，但我是一个初学者

#require "pry"

class Game
 attr_reader :turn

 def initialize
  @turn = 1
  @board = { a1: " ", a2: " ", a3: " ", b1: " ", b2: " ", b3: " ", c1: " ", c2: " ", c3: " " }
 end

# def start_game
#   x = Game.new
#   x.player_symbol
#   x.player_turn
#   x.check_game
# end

 def intro
  puts "ULTIMATE TIC TAC TOE .. in ruby!\n"
  display_board
 end

 def player_symbol
  puts "Player 1, choose your marker. X or O?\n"
  i = gets.chomp.upcase 
  if i == "X"
    @p1 = "X"
    @p2 = "O"
  elsif i == "O"
    @p1 = "O"
    @p2 = "X"
  else
    puts "That is not a valid player!"
  end
end

def player_turn
  if @turn == 1
    puts "Player 1 turn."
    @turn = 2
    player_move(@p1)
  else
    puts "Player 2 turn."
    @turn = 1
    player_move(@p2)
  end
end

def display_board
  # Displays board grid using hash values.
  puts "   1   2   3 "
  puts "A  #{@board[:a1]} | #{@board[:a2]} | #{@board[:a3]} "
  puts "  ---+---+---"
  puts "B  #{@board[:b1]} | #{@board[:b2]} | #{@board[:b3]} "
  puts "  ---+---+---"
  puts "C  #{@board[:c1]} | #{@board[:c2]} | #{@board[:c3]} "
end


def player_move(n)
  # Player picks square to claim.
  # Player symbol populates hash.
  puts "pick move"
  x = gets.chomp.downcase.to_sym
  @board[x] = "#{n}" # Directs to player X/O
  display_board

end
def game_results
  # Determines winner, loser, or tie game
 if ((@board[:a1] == @p1) && (@board[:a2] == @p1) && (@board[:a3] == @p1)) ||
    ((@board[:b1] == @p1) && (@board[:b2] == @p1) && (@board[:b3] == @p1)) ||
    ((@board[:c1] == @p1) && (@board[:c2] == @p1) && (@board[:c3] == @p1)) ||
    ((@board[:a1] == @p1) && (@board[:b1] == @p1) && (@board[:c1] == @p1)) ||
    ((@board[:a2] == @p1) && (@board[:b2] == @p1) && (@board[:c2] == @p1)) ||
    ((@board[:a3] == @p1) && (@board[:b3] == @p1) && (@board[:c3] == @p1)) ||
    ((@board[:a1] == @p1) && (@board[:b2] == @p1) && (@board[:c3] == @p1)) ||
    ((@board[:a3] == @p1) && (@board[:b2] == @p1) && (@board[:c1] == @p1))
   puts "Player #{@p1} is the winner!"
 elsif ((@board[:a1] == @p2) && (@board[:a2] == @p2) && (@board[:a3] == @p2)) ||
    ((@board[:b1] == @p2) && (@board[:b2] == @p2) && (@board[:b3] == @p2)) ||
    ((@board[:c1] == @p2) && (@board[:c2] == @p2) && (@board[:c3] == @p2)) ||
    ((@board[:a1] == @p2) && (@board[:b1] == @p2) && (@board[:b1] == @p2)) ||
    ((@board[:a2] == @p2) && (@board[:b2] == @p2) && (@board[:c2] == @p2)) ||
    ((@board[:a3] == @p2) && (@board[:b3] == @p2) && (@board[:c3] == @p2)) ||
    ((@board[:a1] == @p2) && (@board[:b2] == @p2) && (@board[:c3] == @p2)) ||
    ((@board[:a3] == @p2) && (@board[:b2] == @p2) && (@board[:c1] == @p2))
   puts "Player #{@p2} is the winner!"
 elsif ((@board[:a1] == ("X" || "O")) &&
    (@board[:a2] == ("X" || "O")) &&
    (@board[:a3] == ("X" || "O")) &&
    (@board[:b1] == ("X" || "O")) &&
    (@board[:b2] == ("X" || "O")) &&
    (@board[:b3] == ("X" || "O")) &&
    (@board[:c1] == ("X" || "O")) &&
    (@board[:c2] == ("X" || "O")) &&
    (@board[:c3] == ("X" || "O")))
    # This should represent that empty elements in the hash are now filled not making a winning pair
   puts "Tie Game"
 else
   player_turn
  end
 end


end

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@纳特瓦伦感谢你的否决票和对这篇文章的精彩编辑。贡献的方式。兄弟，我没有投反对票！为了更好的可读性，我进行了编辑，只是为了让你更快地得到答案。哦，抱歉@nathvarun，我想既然你编辑了你，你就投了反对票，因为他们的观点只有5个。希望我能得到一些反馈。没问题，伙计。因为投反对票的人没有留下评论，所以我投了赞成票。我不是一个喜欢红宝石的人，所以我帮不了你。cheers@bigREDcode我敢肯定你被否决了，因为你发布了一大堆乱七八糟的代码，并要求我们仔细研究，以找出错误。退房通常，做一个MCVE，这样你就可以问一个非常好的so问题，这会帮助你自己发现错误，而你根本不需要问！这节省了每个人的时间！