我是个编程新手，我创建的石头剪刀布游戏并没有按预期运行。如果有人输入单词rock、paper或scissors，程序就会按预期运行。但是，当有人输入单词rock、paper或scissors以外的单词时，程序会说“我不明白，请再试一次”，并提示用户输入另一个输入，它确实做到了。但是程序并没有按预期继续工作，而是结束了。代码如下：
# The game of Rock Paper Scissors
import random
choices = ['rock', 'paper', 'scissors']
computer_choice = random.choice(choices)
selections = 'The computer chose ' + computer_choice + ' so'
computer_score = 0
user_score = 0
def choose_option():
user_choice = input('Please choose Rock, Paper or Scissors. (q to quit)\n>>> ')
while user_choice != 'q':
if user_choice in ['ROCK', 'Rock', 'rock', 'R', 'r']:
user_choice = 'rock'
elif user_choice in ['PAPER', 'Paper', 'paper', 'P', 'p']:
user_choice = 'paper'
elif user_choice in ['SCISSORS','Scissors', 'scissors', 'S', 's']:
user_choice = 'scissors'
else:
print("I don't understand, please try again.")
choose_option()
return user_choice
user_choice = choose_option()
while user_choice != 'q':
if user_choice == computer_choice:
print(selections + ' it\'s a tie')
elif (user_choice == 'rock' and computer_choice == 'scissors'):
user_score += 1
print(selections + ' you won! :)')
elif (user_choice == 'paper' and computer_choice == 'rock'):
user_score += 1
print(selections + ' you won! :)')
elif (user_choice == 'scissors' and computer_choice == 'paper'):
user_score += 1
print(selections + ' you won! :)')
elif (user_choice == 'rock' and computer_choice == 'paper'):
computer_score += 1
print(selections + ' you lost :(')
elif (user_choice == 'paper' and computer_choice == 'scissors'):
computer_score += 1
print(selections + ' you lost :(')
elif (user_choice == 'scissors' and computer_choice == 'rock'):
computer_score += 1
print(selections + ' you lost :(')
else:
break
print('You: ' + str(user_score) + " VS " + "Computer: " + str(computer_score))
computer_choice = random.choice(choices)
selections = 'The computer chose ' + computer_choice + ' so'
user_choice = choose_option()
2条答案
按热度按时间px9o7tmv1#
您的问题似乎是由于在函数中调用函数而引起的。这为我解决了问题：
这样，如果计算机不喜欢输入，它可以只要求另一个。此外，我会建议使用
if user_choice.lower() in []，只是比键入所有选项容易一点。希望这对你有帮助！
xa9qqrwz2#
代替
您可以使用
continue将导致程序跳过循环的其余部分（这里，跳过return），并继续while循环的下一次迭代（返回到user_choice = ...）。还请注意，我认为还有另一个bug，如果user_choice是"q"，则实际上不会返回结果。