Python上的作业，麻烦与循环

Question

我目前在我的大学初学者的Python课程。我的作业部分是创建一个类似着名儿童歌曲“99瓶流行音乐”的节目。任务是：

1. 用户输入一个项目。
2. 该程序打印出如下陈述：“在墙上弹出x瓶，弹出x瓶，取下一个，通过它，（x-1）瓶在墙上弹出”直到它达到0 一旦达到0，程序必须停下来，并说墙上没有瓶子弹出。
3. 上限为99。如果用户输入任何超过99，则显示一个错误并迫使计数器开始在99

有道理正确？所以，这里是我的代码：

#Bottles of Pop on the Wall 
#User inputs a number to start with 
#Program returns with singing the popular song down to 0 
# 

print("Bottles of Pop on the Wall Program") 

userBottle = int(input("How many bottles? ")) 
bottleCount = userBottle 

while bottleCount > 1: 
     newBottle = userBottle - 1 
     bottleCount -= 1 

    if bottleCount > 99: 
     print("ALERT: no more than 99 bottles allowed, reverting to 99 bottles") 

     userBottle = 99 
     bottleCount = 99 
     newBottle = 99 


    print(userBottle , "bottles of pop on the wall, ", userBottle , "bottles of pop" , 
     "\ntake one down, pass it around, " , newBottle , "bottles of pop on the wall") 
    userBottle -= 1 

if bottleCount == 1: 
    print(userBottle , "bottle of pop on the wall, ", userBottle , "bottle of pop" , 
     "\ntake one down, pass it around, " , "no bottles of pop on the wall") 



input("\nThank you for playing! Press Enter to exit.") 

因此，如果用户输入任何低于100的数字，程序将完美工作。但是，如果用户输入超过99的任何东西，那就是我遇到问题的地方。

会发生什么是循环将运行到1，但不会结束。它会重复一次，它会返回：

1 bottles of pop on the wall, 1 bottles of pop 
take one down, pass it around, 0 bottles of pop on the wall 
0 bottle of pop on the wall, 0 bottle of pop 
take one down, pass it around, no bottles of pop on the wall 

Thank you for playing! Press Enter to exit. 

显然这是不正确的。我的循环出了什么问题，所以我可以确保当用户输入大于99的数字时不会发生这种情况？

非常感谢，我非常感谢您的帮助！

Answer 1

你应该只使用一个变量来跟踪瓶

print("Bottles of Pop on the Wall Program") 

bottle_count = int(input("How many bottles? ")) 

if bottle_count > 99: 
    print("ALERT: no more than 99 bottles allowed, reverting to 99 bottles") 
    bottle_count = 99 

while bottle_count > 1: 
    print(bottle_count , "bottles of pop on the wall, ", bottle_count , "bottles of pop") 
    bottle_count -= 1 
    print("take one down, pass it around, ", bottle_count, "bottles of pop on the wall") 

print(bottle_count , "bottle of pop on the wall, ", bottle_count , "bottle of pop") 
print("take one down, pass it around, no bottles of pop on the wall") 

input("\nThank you for playing! Press Enter to exit.") 

对于下一步，您应该使用的格式字符串和.format方法的数量。例如。

print("{bottle_count} bottles of pop on the wall, " 
     "{bottle_count} bottles of pop".format(bottle_count=bottle_count)) 

Answer 2

看起来你有一个别名问题。

在循环之外，你要求用户输入一个数字，比如说他们输入101，然后你设置bottleCount = userBottle。然后，在循环中，你的userBottle值重置为99，但注意以下事项：

In [6]: userBottle = 101 

In [7]: bottleCount = userBottle 

In [8]: userBottle = 99 

In [9]: userBottle 
Out[9]: 99 

In [10]: bottleCount 
Out[10]: 101 

以下是为你的程序。您可能已更改了userBottle值，但尚未更改bottleCount值。

要做的事情是编写一个函数来以受控方式获取userBottle值，并且函数只是返回itslef，直到值正确为止。一个例子可能是：

def get_user_bottles(input_message, error_message1, error_message2): 
    #This makes sure the user inputs a number 
    try: 
     userBottle = int(raw_input(input_message)) 
    except ValueError: 
     print error_message1 
     return get_user_bottles(input_message, error_message1, error_message2) 
    #This makes sure the user inputs a number between 1 and 99 
    if userBottle not in range(1, 100): 
     print error_message2 
     return get_user_bottles(input_message, error_message1, error_message2) 
    else: 
     return userBottle 

userBottle = get_user_bottles('How many bottles?', 
           'The value must be an integer between 1 and 99', 
           'That value is out of bounds, it must be between 1 and 99') 

Answer 3

这可以避免循环瓶子。它使用列表理解来代替：

n = int(input("How many bottles? ")) 
if n > 99: 
    print("Sorry. You cannot afford %s bottles. Starting with 99." % n) 
    n = 99 
song = '\n'.join('%s bottles of pop on the wall, %s bottles of pop\ntake one down, pass it around, %s bottles of pop on the wall' % (i,i,i-1) for i in range(n, 0, -1)) 
print(song + "\nno bottles of pop on the wall, no bottles of pop\n") 
input("Thank you for playing! Press Enter to exit.") 

输出示例：

How many bottles? 2 
2 bottles of pop on the wall, 2 bottles of pop 
take one down, pass it around, 1 bottles of pop on the wall 
1 bottles of pop on the wall, 1 bottles of pop 
take one down, pass it around, 0 bottles of pop on the wall 
no bottles of pop on the wall, no bottles of pop 

Thank you for playing! Press Enter to exit. 

我把修改歌曲的最后一行的自由。在你有0瓶的流行音乐之后，我不认为有必要从墙上再拿下一个。如果您愿意，可以轻松将其更改回来。

Answer 4

简单的改变： if bottleCount == 1:

这样： if userBottle == 1:，因为这是你在递减每一步的变量。