我正在尝试通过While循环重新启动程序。编写完整个程序后,我将其告知“ Continue”并再次启动while循环,但它始终给我带来语法错误。有人可以解释一下我在做什么错吗?请忽略代码的质量,着重指出我的错误。谢谢!
while True:
input1 = int(input("Player one please enter your desired choice through a number\n 1 for Rock\n 2 for Scissors\n 3 for Paper\n Enter here: "))
input2 = int(input("Player two please enter your desired choice through a number\n 1 for Rock\n 2 for Scissors\n 3 for Paper\n Enter here: "))
if input1 == input2:
print("It's a tie")
elif input1 == 1 and input2 == 2 or input1 == 1 and input2 == 3:
print("Player one WINS!")
elif input1 == 2 and input2 == 1:
print("Player two WINS!")
elif input1 == 3 and input2 == 1:
print("Player one WINS!")
elif input1 == 2 and input2 == 3:
print("Player one WINS")
elif input1 == 3 and input2 == 2:
print("Player two WINS!")
else:
pass
x = int(input("Do you want to play another game?: Y/N: "))
if x == 'y':
continue
else:
break
预期结果: 程序再次运行
实际结果: SyntaxError:“继续”无法正确循环
答案 0 :(得分:0)
您的x是一个int,必须将其设置为String才能比较字符串。 (使用str())
Int:
x = int(input("Do you want to play another game?: Y/N: "))
字符串:
x = str(input("Do you want to play another game?: Y/N: "))
工作版本:
try:
while True:
input1 = int(input("Player one please enter your desired choice through a number\n 1 for Rock\n 2 for Scissors\n 3 for Paper\n Enter here: "))
input2 = int(input("Player two please enter your desired choice through a number\n 1 for Rock\n 2 for Scissors\n 3 for Paper\n Enter here: "))
if input1 == input2:
print("It's a tie")
elif input1 == 1 and input2 == 2 or input1 == 1 and input2 == 3:
print("Player one WINS!")
elif input1 == 2 and input2 == 1:
print("Player two WINS!")
elif input1 == 3 and input2 == 1:
print("Player one WINS!")
elif input1 == 2 and input2 == 3:
print("Player one WINS")
elif input1 == 3 and input2 == 2:
print("Player two WINS!")
else:
pass
# x = int(input("Do you want to play another game?: Y/N: "))
x = str(input("Do you want to play another game?: Y/N: "))
if x == 'y':
continue
else:
break
except ValueError:
raise ValueError('no negative numbers')
except TypeError:
raise TypeError('wrong type')