有沒有辦法縮短這個？

初學者程序員在這裏學習的過程。我只是想知道我輸入的這個簡單的代碼是否是最佳的方式。有沒有辦法縮短這個？

with open('guest_book.txt', 'a') as file_object: 
    while True: 
     name=input("What is your name?") 
     print("Welcome " + name + ", have a nice day!") 
     file_object.write(name + " has visited! \n") 
     another = input("Do you need to add another name?(Y/N)") 
     if another == "y": 
      continue 
     elif another == "n": 
      break 
     else: 
      print("That was not a proper input!") 
      while True: 
       another = input("Do you need to add another name?(Y/N)") 
       if another == "y": 
        a = "t" 
        break 
       if another == "n": 
        a = "f" 
        break 
      if a == "t": 
       continue 
      else: 
       break

我的問題是在if語句中。當我詢問輸入內容時（「是否需要添加另一個名稱？（y/n）」），我輸入了重新提問的最佳方式，如果我得到的答案不是y或n，基本上我想。要重複的問題，如果我沒有得到任何一個是或否的回答，我找到了解決辦法似乎並不像最優化的解決方案

來源

2016-11-24 lmurdock12

既然你是新的，在代碼審查上http://codereview.stackexchange.com/通常做，請有此張貼。 – ishaan

在「這不是正確的輸入」打印後面添加一個「繼續」。 – sal

[請求用戶輸入，直到他們給出有效響應]的可能重複（http://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-迴應） – tripleee

你基本上那裏你可以簡單地說：

with open('guest_book.txt', 'a') as file_object: 
    while True: 
     name=input("What is your name?") 
     print("Welcome " + name + ", have a nice day!") 
     file_object.write(name + " has visited! \n") 
     another = input("Do you need to add another name?(Y/N)") 
     if another == "y": 
      continue 
     elif another == "n": 
      break 
     else: 
      print("That was not a proper input!") 
      continue

來源

2016-11-24 06:20:09 sal

您可以使用函數在一個地方寫你的所有邏輯。

def calculate(file_object): 
    name=raw_input("What is your name?") 
    print("Welcome " + name + ", have a nice day!") 
    file_object.write(name + " has visited! \n") 
    another = raw_input("Do you need to add another name?(Y/N)") 
    if another == "y": 
     calculate(file_object) 
    elif another == "n": 
     return 
    else: 
     print("That was not a proper input!") 
     calculate(file_object) 

if __name__=='__main__':  
    with open('guest_book.txt', 'a') as file_object: 
     calculate(file_object)

來源

2016-11-24 06:31:43

你能做到這樣，但有將不會有任何無效的意見，說不。它只會檢查說y

with open('guest_book.txt', 'a') as file_object: 
    another = 'y' 
    while another.lower() == 'y': 
     name=input("What is your name?") 
     print("Welcome " + name + ", have a nice day!") 
     another = input("Do you need to add another name?(Y/N)")

來源

2016-11-24 06:46:14

有沒有辦法縮短這個？

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