檢查用戶輸入python

Question

在我的inputCheck函數中，當用戶輸入被檢查通過後，它是一個可接受的輸入，應該由打印消息確認，然後運行另一個函數 - 但它不這樣做，我不能爲什麼 - 你能告訴如何解決問題？非常感謝！

def main(): 
    print('WELCOME TO THE WULFULGASTER ENCRYPTOR 9000') 
    print('==========================================') 
    print('Choose an option...') 
    print('1. Enter text to Encrypt') 
    print('2. Encrypt text entered') 
    print('3. Display Encrypted Text!') 
    menuChoice() 

def menuChoice(): 
    valid = ['1','2','3'] 
    userChoice = str(input('What Would You Like To Do? ')) 
    if userChoice in valid: 
     inputCheck(userChoice) 
    else: 
     print('Sorry But You Didnt Choose an available option... Try Again') 
     menuChoice() 

def inputCheck(userChoice): 
    if userChoice == 1: 
     print('You Have Chosen to Enter Text to Encrypt!') 
     enterText() 
    if userChoice == 2: 
     print('You Have Chosen to Encypt Entered Text!') 
     encryptText() 
    if userChoice == 3: 
     print('You Have Chosen to Display Encypted Text!') 
     displayText() 

def enterText(): 
    print('Enter Text') 

def encryptText(): 
    print('Encrypt Text') 

def displayText(): 
    print('Display Text') 


main() 

Answer 1

你用戶的輸入轉換爲字符串（str(input('What ...'))），但在inputCheck把它比作整數。由於inputCheck中沒有else路徑，因此當您輸入「有效」選項時沒有任何反應。另外，如果你使用的是Python 2，使用input不是你想要的，raw_input是要走的路（例如參見What's the difference between raw_input() and input() in python3.x?）。

除此之外，每當用戶輸入非法選擇時遞歸地調用menuChoice可能是一個糟糕的主意：幾百次或幾千次輸入非法選擇，並且程序崩潰（除了浪費大量內存外）。你應該把代碼放在一個循環中：

while True: 
    userChoice = str(raw_input('What Would You Like To Do? ')) 
    if userChoice in valid: 
     inputCheck(userChoice) 
     break 
    else: 
     print('Sorry But You Didnt Choose an available option... Try Again')