我正在编写一个程序，与两名玩家一起玩tic tac toe。我已经完成了基本代码（尽管效率很低），但我一直收到错误，说明没有定义player2。我已经尝试了许多方法来修复错误但是想知道你们是否有任何想法。它被第一个player1 == player2条件所吸引。这是代码：
def main():
board1 = [" "," "," "]
board2 = [" "," "," "]
board3 = [" "," "," "]
game(board1,board2,board3)
def play1():
global player1
player1 = int(input("Player 1, where would you like to move? "))
return player1
def play2():
global player2
player2 = int(input("Player 2, where would you like to move? "))
return player2
def game(brd1,brd2,brd3):
isvalid = False
while(not(isvalid)):
play1()
try:
if player1 == player2:
print("You can't both go to the same spot!")
except NameError:
if player1 == 0:
brd1.pop(0)
brd1.insert(0,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 1:
brd1.pop(1)
brd1.insert(1,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 2:
brd1.pop(2)
brd1.insert(2,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 3:
brd2.pop(0)
brd2.insert(0,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 4:
brd2.pop(1)
brd2.insert(1,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 5:
brd2.pop(2)
brd2.insert(2,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 6:
brd3.pop(0)
brd3.insert(0,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 7:
brd3.pop(1)
brd3.insert(1,"x")
print(brd1)
print(brd2)
print(brd3)
elif player1 == 8:
brd3.pop(2)
brd3.insert(2,"x")
print(brd1)
print(brd2)
print(brd3)
play2()
if player2 == player1:
print("You can't both go to the same spot!")
elif player2 == 0:
brd1.pop(0)
brd1.insert(0,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 1:
brd1.pop(1)
brd1.insert(1,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 2:
brd1.pop(2)
brd1.insert(2,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 3:
brd2.pop(0)
brd2.insert(0,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 4:
brd2.pop(1)
brd2.insert(1,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 5:
brd2.pop(2)
brd2.insert(2,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 6:
brd3.pop(0)
brd3.insert(0,"o")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 7:
brd3.pop(1)
brd3.insert(1,"x")
print(brd1)
print(brd2)
print(brd3)
elif player2 == 8:
brd3.pop(2)
brd3.insert(2,"o")
print(brd1)
print(brd2)
print(brd3)
if __name__ == '__main__':
main()
答案 0 :(得分：0)
您的变量 - player1和player2 - 在本地范围（play1和play2）中定义，您尝试在该范围之外访问它们。在函数开头，键入global player1（或player2）作为其自己的行，以在全局范围内定义它。另外，在所有功能的基础上添加player1, player2 = None, None以启动。至于为什么在错误中指定player2，那是因为它是条件中的第一个变量，并且Python从左到右进行评估。
答案 1 :(得分：0)
由于您的错误表明未定义播放器2，请尝试添加try except条件以检查播放器2是否已定义。如果它达到NameError，它可以继续游戏（打印你的棋盘），否则，通过你的if条件。
while(not(isvalid)):
play1()
try:
# Try going through all your conditions
if player 1 == player 2:
...
except NameError:
# But if your player 2 is not defined, go through them all except the player 1 == player 2 clause
if player 1 == 0:
.....
print(brd1)
....
答案 2 :(得分：0)
您只能在循环中调用play1()。您还需要play2()。但是，由于player1和player2在函数中是本地定义的，因此无法解决您的问题。要解决此问题，请在global player1中添加play1()，在global player2中添加play2()。然后代码看起来像这样
def play1():
global player1
player1 = int(input("Player 1, where would you like to move? "))
def play2():
global player2
player2 = int(input("Player 2, where would you like to move? "))
循环：
while(not(isvalid)):
play1()
play2()
if player2 == player1:
另一种解决方法是返回它们，如下所示：
def play1():
player1 = int(input("Player 1, where would you like to move? "))
return player1
def play2():
player2 = int(input("Player 2, where would you like to move? "))
return player2
然后在你的循环中：
while(not(isvalid)):
player1 = play1()
player2 = play2()
if player2 == player1:
这样可以避免全局变种。
编辑：为了在每次转弯后打印电路板，请在文件开头初始化player2 = None：
player2 = None
def main():
...
您仍然需要play1()和play2()
def play1():
global player1
player1 = int(input("Player 1, where would you like to move? "))
def play2():
global player2
player2 = int(input("Player 2, where would you like to move? "))
那时不需要适应循环。
答案 3 :(得分：0)
从你的代码可以清楚地看出，player1和player2是在第二行游戏方法中初始化的，但它们没有定义。您可能需要定义一些值。
添加此行
player1 = player2 = None
在查看代码之后，我建议尝试在开始时将变量定义到全局范围。
global player1, player2
player1 = player2 = None