python - 如何清除用户输入？
问题描述
所以我用数字制作了垄断交易系统。比如我要加东西就输入数字1，要删除东西就输入数字2等等。我的问题是，如果退出while循环，基本上就是代码“break”，先前的 break 输入也会激活主菜单的 if 命令。如果您对我想说的话感到困惑，我不知道是否允许我在此网站上发布链接，但链接是： https ://repl.it/@BrentTersol/Monopoly-Official
if Action == 2 :
while True:
replit.clear()
print("What would you like to give?\n1: Add\n2: Remove\n3: Clear\n4: Offer Trade\n5: Cancel\n6: Switch")
Action = int(input(">>> "))
if Action == 1:
while True:
replit.clear()
print("What would you like to add?\n1: Money\n2: Property\n3: Jail Free Card\n4: Back")
Action = int(input(">>> "))
if Action == 1:
if Turn == 1:
while True:
replit.clear()
print("How much money do you want to give for the trade?\nMoney:",str(Player1Money))
Action = int(input(">>> "))
if Action >= 0 and Action <= (Player1Money):
TMoney1 = (Action)
print("You added $"+str(TMoney1),"to the trade")
time.sleep(2)
break
else:
print("You do not have enough money")
break
if Turn == 2:
while True:
replit.clear()
print("How much money do you want to give for the trade?\nMoney:",str(Player2Money))
Action = int(input(">>> "))
if Action >= 0 and Action <= (Player2Money):
TMoney2 = (Action)
print("You added $"+str(TMoney2),"to the trade")
time.sleep(2)
break
if Action == "back":
break
else:
print("You do not have enough money")
break
if Action == 2:
while True:
replit.clear()
if Turn == 1:
print(Inventory1)
if Turn == 2:
print(Inventory2)
print("What property do you want to give?")
Action = int(input(">>> "))
if Turn == 1:
if (Action) in (Inventory1):
TProperty1.append((Action))
print("Added",(Action),"to the trade")
time.sleep(2)
break
else:
print("Item not found in your properties")
time.sleep(2)
break
if Turn == 2:
if (Action) in (Inventory2):
TProperty2 = (Action)
print("Added",(Action),"to the trade")
time.sleep(2)
break
else:
print("Item not found in your properties")
time.sleep(2)
break
if Action == 3:
if Turn == 1:
if JailCard1 == 1:
TCard1 = 1
print("Added Jail Free Card to the trade.")
time.sleep(2)
else:
print("You do not own a Jail Free Card")
time.sleep(2)
if Turn == 2:
if JailCard2 == 1:
TCard1 = 1
print("Added Jail Free Card to the trade.")
time.sleep(2)
else:
print("You do not own a Jail Free Card")
time.sleep(2)
if Action == 4:
break
if Action == 2:
while True:
replit.clear()
print("What would you like to remove?\n1: Money\n2: Property\n3: Jail Free Card\n4: Back")
Action = int(input(">>> "))
if Action == 1:
while True:
replit.clear()
if Turn == 1:
if TMoney1 == 0:
print("There wasn't any money to remove")
time.sleep(2)
else:
TMoney1 = 0
print("Removed Cash from offer")
time.sleep(2)
break
if Action == 2:
while True:
replit.clear()
print(TProperty1)
print("What property would you like to remove")
Action = input(">>> ")
Action = Action.lower()
if Turn == 1:
if Action == "back":
break
if (Action) in (TProperty1):
TProperty1.remove((Action))
print("Removed",(TProperty1),"from trade")
time.sleep(2)
break
else:
print("That item did not exist")
time.sleep(2)
if Turn == 2:
if (Action) in (TProperty2):
TProperty2.remove((Action))
print("Removed",(TProperty2),"from trade")
time.sleep(2)
else:
print("That item did not exist")
time.sleep(2)
if Action == 3:
if Turn == 1:
if JailCard1 == 1:
print("Removed Jail Free Card from trade")
TCard1 = 0
break
else:
print("Card does not exist in trade")
if Turn == 2:
if JailCard2 == 1:
print("Removed Jail Free Card from trade")
TCard2 = 0
break
else:
print("Card does not exist in trade")
if Action == 4:
break
if Action == 3:
TMoney1 = 0
TMoney2 = 0
TProperty1.clear()
TProperty2.clear()
TCard1 = 0
TCard2 = 0
if Action == 4:
if Turn == 1:
while True:
print("This is what",(Name1),"offers:\n--------------------")
time.sleep(2)
print("You get:\nMoney:",(TMoney1),"\nProperty:",(TProperty1),"\nGet out of Jail Free Card:",(TCard1),"\n")
time.sleep(2)
print("You give",(Name1)+":\nMoney:",(TMoney2),"\nProperty:",(TProperty2),"\nGet out of Jail Free Card:",(TCard2),"\n")
time.sleep(2)
print("Do you accept this Offer? (Y/N):")
Action = input(">>> ")
Action = Action.lower()
if Action == "y":
print("This is beyond what I can do at this point. Very sorry you took a long time not knowing that it wouldnt work. But this will soon be fixed as soon as I know how to do this")
time.sleep(5)
else:
print("Trade has been rejected")
time.sleep(2)
break
if Action == 5:
if Turn == 1:
Turn = 2
else:
Turn = 1
if Action == 6:
print("This is beyond what I can do. Please wait until I learn how to do this. Thank you.")
解决方案
您应该实现一个While True循环，这将重置每个循环的用户输入（只要您在循环内覆盖它）！我还建议您将代码重新创建为函数，而不是几个嵌套的逻辑语句。每个循环获取用户输入的代码示例（根据您的问题清除用户输入）：
while True:
big_prompt = """What would you like to do?
1) say hi
2) say bye
3) pay taxes
4) ???
5) profit
6) Exit D:"""
action = input(big_prompt) # reset user input each loop
if action not in ["1", "2", "3", "4", "5", "6"]:
print("Sorry I didn't understand that")
continue # get a new user input
# at this point we know user input fits our criteria
if action == "1":
print("Hi!")
elif action == "2":
print("Bye!")
# ... you should be putting function calls in these for more complex tasks
elif action == "6":
break # this only exits one loop, I think you are trying to use it to
带样本输出：
What would you like to do?
1) say hi
2) say bye
3) pay taxes
4) ???
5) profit
6) Exit D:
>>>1
Hi!
What would you like to do?
1) say hi
2) say bye
3) pay taxes
4) ???
5) profit
6) Exit D:
>>>6
Process finished with exit code 0
通过打开交互式终端并运行来查看Python 的禅宗也是一个好主意import this：
>>>import this
The Zen of Python, by Tim Peters
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!
特别值得注意的是，我们发现了一些适用于这里的东西：
平面优于嵌套。
可读性很重要。
维护单个 while 循环对这两个方面都有很大帮助。在您的代码中的某一时刻，您会被困在3 个 while True:循环中，然后尝试用一个来打破多个循环break，但总是会被困在至少 2 个级别的深度......
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