獲取的功能相同的範圍值

我一直試圖做我的任務，但我碰到的使用Python 3獲取的功能相同的範圍值

print("Car Service Cost") 
def main(): 
    loan=int(input("What is your loan cost?:\t")) 
    maintenance=int(input("What is your maintenance cost?:\t")) 
    total= loan + maintenance 
    for rank in range(1,10000000): 
     print("Total cost of Customer #",rank, "is:\t", total) 
     checker() 
def checker(): 
    choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
    if choice not in ["y","Y","n","N"]: 
     print("Invalid Choice!") 
    else: 
     main() 
main()

並即時得到這個輸出邏輯error.I'm：

Car Service Cost 
What is your loan cost?: 45 
What is your maintenance cost?: 50 
Total cost of Customer # 1 is: 95 
Do you want to proceed with the next customer?(Y/N): y 
What is your loan cost?: 70 
What is your maintenance cost?: 12 
Total cost of Customer # 1 is: 82 
Do you want to proceed with the next customer?(Y/N): y 
What is your loan cost?: 45 
What is your maintenance cost?: 74 
Total cost of Customer # 1 is: 119 
Do you want to proceed with the next customer?(Y/N): here

我每次都有我的排名爲1。我究竟做錯了什麼？

來源

2016-11-28 Aslı Büşra Ç.

'main'調用'checker'進入循環，調用'main' - >你根本沒有循環，只需在循環的每次* first *迭代中重新啓動'main'（並且你將用完遞歸深度遲早）。 –

您不應該在checker中再次致電main()。你可以返回（你也可以使用break如果你把它在一個循環中）：

def checker(): 
    while True: 
     choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
     if choice not in ["y","Y","n","N"]: 
      print("Invalid Choice!") 
     else: 
      return

如果你想在main打出來的迴路，如果'n'或'N'輸入的，那麼你可以嘗試返回值：

def checker(): 
    while True: 
     choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
     if choice not in ["y","Y","n","N"]: 
      print("Invalid Choice!") 
     else: 
      return choice.lower()

然後檢查它是否'y'或main'n'。

編輯：如果你不想使用return你可以只取出環和else，但這樣一來，你就不能檢查，如果用戶想阻止：

def checker(): 
    choice=input("Do you want to proceed with the next customer?(Y/N):\t") 
    if choice not in ["y","Y","n","N"]: 
     print("Invalid Choice!")

來源

2016-11-28 12:55:25 abacles

您未使用for循環。從checker()你再打電話main()。

而不是

else: main()

你應該簡單地return。

我不確定你在做什麼，你打算在checker()。

來源

2016-11-28 12:52:52

我們還沒有學會返回函數，我們被告知我們不應該使用它。謝謝 –

獲取的功能相同的範圍值

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