[英]how to use an if else statement in another while loop
我是编码新手。 我想尝试编写一个简单的剪刀石头布游戏。 但是我不知道如何结束比赛。
在该程序的最后,如果用户输入错误,我想再次转到end变量。 我尝试使用注释行,但无法正常工作。
player1 = input("What is player 1's name ? ")
player2 = input("What is player 2's name ? ")
player1 = player1.title()
player2 = player2.title()
while True:
print(player1 + " What do you choose ? rock / paper / scissors : ")
a = input()
print(player2 + " What do you choose ? rock / paper / scissors : ")
b = input()
if a == "rock" and b == "scissors" :
print(player1, "won !!!")
elif a == "scissors" and b == "rock":
print(player2, "won !!!")
elif a == "paper" and b == "rock":
print(player1, "won !!!")
elif a == "rock" and b == "paper":
print(player2, "won !!!")
elif a == "scissors" and b == "paper":
print(player1, "won !!!")
elif a == "paper" and b == "scissors":
print(player2, "won !!!")
elif a == b:
print("Its a tie :-(")
elif a or b != "rock" or "paper" or "scissors":
print("Wrong input, Try again")
end = input("Do you want to play again ? yes/no ") == "yes"
if input == "yes":
continue
else:
print('''
GAME OVER''')
break
# elif input != "yes" or "no":
# print("Wrong input, Try again. yes or no ?")
如果输入为“ no”,我希望它结束游戏;如果输入为“ yes”,则重新启动游戏;如果输入不正确,我希望再次出现提示。
只需检查end的值
if end is True:
continue
else:
break
既然,通过将input()与“ yes”进行比较,将end的值设置为布尔值,它将表明用户是否要结束游戏? 另外,您没有初始化输入变量,并且最后的elif条件将始终为true,如注释中所述。
您的代码有一些需要解决的问题,还有一些可以简化的地方。 我对您的程序进行了一些更改,并添加了一些注释来解释这些更改。
player1 = input("What is player 1's name ? ").title() #This uses chaining to streamline code
player2 = input("What is player 2's name ? ").title() #Same as above
while True:
a = input(player1 + " What do you choose ? rock / paper / scissors : ") #no need to use a separate print statement
b = input(player2 + " What do you choose ? rock / paper / scissors : ")
valid_entries = ["rock", "paper", "scissors"] #To check for valid inputs
if (a not in valid_entries) or (b not in valid_entries):
print("Wrong input, try again")
continue
a_number = valid_entries.index(a) #Converting it to numbers for easier comparison
b_number = valid_entries.index(b)
if(a_number == b_number):
print("Its a tie :-(")
else:
a_wins = ((a_number > b_number or (b_number == 2 and a_number == 0)) and not (a_number == 2 and b_number == 0)) #uses some number comparisons to see who wins instead of multiple if/elif checks
if(a_wins):
print(player1, "won !!!")
else:
print(player2, "won !!!")
end = input("Do you want to play again ? yes/no ")
while (end !="yes") and (end != "no"):
print("invalid input, try again")
end = input("Do you want to play again ? yes/no ")
if end == "yes":
continue
else:
print("GAME OVER")
break
这些更改还通过使用另一个while循环进行检查,以查看重新启动游戏的输入是否有效
*请注意,我尚未测试这些更改,因此可能需要进行一些修改
好吧,您可以使用列表简化代码,然后简化if测试。 您可以检查选项的顺序并根据其做出决定。 您也可以使测试成为标准,以最大程度地减少if语句的数量。 这是我改善您的代码的建议。 希望对您有所帮助:
# get playe names
player1 = input("What is player 1's name ? ")
player2 = input("What is player 2's name ? ")
player1 = player1.title()
player2 = player2.title()
# init vars
options = ["rock", "paper", "scissors"]
players = [player1, player2]
# start game
while True:
a = input(player1 + " What do you choose ? rock / paper / scissors : ")
b = input(player2 + " What do you choose ? rock / paper / scissors : ")
# check if inputs are correct
while (a not in options or b not in options):
print("Wrong input, Try again")
a = input(player1 + " What do you choose ? rock / paper / scissors : ")
b = input(player2 + " What do you choose ? rock / paper / scissors : ")
# check who won
if abs(options.index(a) - options.index(b)) == 1:
print(players[1*int(options.index(a) > options.index(b))], "won !!!")
elif abs(options.index(b) - options.index(a)) > 1:
print(players[1*int(options.index(a) > options.index(b))], "won !!!")
elif a == b:
print("Its a tie :-(")
# continue or drop game
end = input("Do you want to play again ? yes/no ")
if end == "yes":
continue
else:
print('''
GAME OVER''')
break
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