如何正确完成我的自卸车程序？

Question

我目前正在为这项工作编写一个 Tipper 程序，但我遇到了很多问题。结果一直说语法无效，我已经切换了很多东西，但似乎我永远无法让它恰到好处。举个例子，这是我当前的代码：

print("Now that your belly is full, how much is the fee?")
fee = input("Enter your fee: ")
print("Would you like to give us a tip?")
answer = input("Yes or no?")
if answer == no:
    why = input("Why do you not want to tip us? Also, you don't have to answer. When done, just press enter.")
    print("We understand. Here is your fee:", fee)
elif answer == yes:
    tip = input("Would you like to tip 15 percent or 20 percent? ")
    if tip == 15 :
        cost1 = fee/15 + fee
    elif tip == 20 :
            cost2 = fee/20 + fee
    else:
        print("I do not understand.")
else:
    print("I do not understand.")
print:("Have a wonderful day!")

所以，是的，老实说，我不确定为什么它不起作用，但有人可以回答吗？我真的被这件事难住了。

Answer 1

我修复了您代码中的错误并对其进行了测试，但整体代码结构仍有待改进。

那么是什么导致你的程序崩溃了呢？

• “No”和“Yes”都是字符串，所以你必须把它们放在“”之间
• 您必须使用int() 函数将用户输入的费用转换为整数，以便对其进行数学计算。
• 您计算的小费错误，我使用这些计算修复了它 15% >> 总计 = 1.15xfee
  20% >> 总计 = 1.20xfee
  注意：在下面的代码中，我使用了math.ceil() 函数将小费四舍五入为整数。
  为了稍微改善用户体验，我还添加了一行打印总成本（包括小费）返回给用户。 这是更正后的代码：

import math

print("Now that your belly is full, how much is the fee?")
fee = int(input("Enter your fee: "))
print("Would you like to give us a tip?")
answer = input("yes or no?")
if answer == "no":
    why = input("Why do you not want to tip us? Also, you don't have to answer. When done, just press enter.")
    print("We understand. Here is your fee:", fee)
elif answer == "yes":
    tip = int(input("Would you like to tip 15 percent or 20 percent? "))
    if tip == 15 :
        cost = math.ceil(1.15 * fee)
        print("Your total, including fee is, ", cost)
    elif tip == 20 :
            cost = math.ceil(1.20 * fee)
            print("Your total, including fee is, ", cost)
    else:
        print("I do not understand.")
else:
    print("I do not understand.")
print("Have a wonderful day!")

Answer 2

正如 cmets 中所述，您需要将要测试的答案用引号括起来（双引号 " 或单引号 ' 可以），否则 python 会认为它们是变量名而不是字符串文字，您要与之进行比较。

另外，在最后的打印中，您有一个不属于那里的冒号 (:)。

这是修复了这两个问题的代码。

print("Now that your belly is full, how much is the fee?")
fee = input("Enter your fee: ")
print("Would you like to give us a tip?")
answer = input("Yes or no?")
if answer == 'no':
    why = input("Why do you not want to tip us? Also, you don't have to answer. When done, just press enter.")
    print("We understand. Here is your fee:", fee)
elif answer == 'yes':
    tip = input("Would you like to tip 15 percent or 20 percent? ")
    if tip == 15:
        cost1 = fee / 15 + fee
    elif tip == 20:
        cost2 = fee / 20 + fee
    else:
        print("I do not understand.")
else:
    print("I do not understand.")
print("Have a wonderful day!")

此外，这种硬编码选项和跳过问题的简单方法（例如一旦回答“是”）会导致非常令人沮丧的用户体验，但是这是另一个问题。

编辑： 您还将字符串与数字进行比较，这始终是错误的，这使您免于错误，即尝试将字符串除以数字会产生

Answer 3

这是一个有效的版本。现在，仍然存在问题。首先，如果他们输入的垃圾不是账单或小费的数字，它不会被捕获。其次，它不圆小费，因为我不知道你使用的是什么货币。但至少你可以运行它并得到答案。

print("Now that your belly is full, how much is the bill?")
fee = float(input("Enter your bill: "))
print("Would you like to give us a tip?")
answer = input("yes or no?")
if answer.lower() == 'no':
    why = input("Why do you not want to tip us? Also, you don't have to answer. When done, just press enter.")
    print("We understand.")
elif answer.lower() == "yes":
    tip = int(input("What tip percentage do you use? "))
    fee += fee * tip / 100
else:
    print("I do not understand.")
print("Your total bill is", fee)