python在继续之前等待第n位数

Question

from pad4pi import rpi_gpio 

# Setup Keypad 
KEYPAD = [ 
     ["1","2","3","A"], 
     ["4","5","6","B"], 
     ["7","8","9","C"], 
     ["*","0","#","D"] 
] 

ROW_PINS = [5,6,13,19] # BCM numbering 
COL_PINS = [26,16,20,21] # BCM numbering 

factory = rpi_gpio.KeypadFactory() 

keypad = factory.create_keypad(keypad=KEYPAD, row_pins=ROW_PINS, col_pins=COL_PINS) 

def processKey(key): 
print("enter 3 digit") 
print(key) 
if key == 123: 
    print("correct") 
else: 
    print("wrong password") 

keypad.registerKeyPressHandler(processKey) 

我希望代码等待用户输入例如3位数，然后再与上面代码中123代码中的密码进行比较。

它应该做的：

等待用户从键盘输入3位，例如123，然后打印正确。

什么它实际上做：

它将打印正确或不正确的密码后，用户马上进入1个代码

Answer 1

更新树莓采取@furas例如：

# Initial keypad setup 
code = '' 

def processKey(key): 
    print("Enter your 3 digit PWD: \n") 
    global code 
    MAX_ALLOWED_CHAR = 3 
    code += key 
    if (len(code) == MAX_ALLOWED_CHAR): 
     if (code == "123"): 
      print("You entered the correct code.") 
      dostuff() 
     else: 
      code = ''   
      print("The passcode you entered is wrong, retry.") 

def dostuff(): 
    # do your things here since passcode is correct. 

这可能会为你的情况做。

---

def processKey(): 

    key = input("enter 3 digit") 
    if (key == "123"): 
     print("Correct password.") 
     return True 
    else: 
     print("You typed {0} wich is incorrect.".format(key)) 
     return False 

所以，现在你不给processKey值，因为正如你所说的用户输入它，调用processKey（）会要求用户输入密码，并返回真/假基于“ 123“在检查。

这是如果你想输入密码，但如果下面的答案不适合你的需求（没有完全理解你想完成的）只是提供更聪明的例子。

编辑：

既然你想严格有3位输入的情况下，重新输入密码，他们输入了错误的一个，你可以做到以下几点：

在调用processKey（）您可以：

while (processKey() == False): 
    processKey() 

Revisioned代码来满足您的需求：

def processKey(): 
    MAX_ALLOWED_CHAR = 3 
    key = input("Enter 3 digit PWD: \n") 
    if (key == 123): 
     print("Correct password.") 
     return True 
    elif (len(str(key)) > MAX_ALLOWED_CHAR): 
     print("The max allowed character is {0}, instead you entered {1}.".format(MAX_ALLOWED_CHAR,key)) 
     return False 
    else: 
     print("You typed {0} wich is incorrect.".format(key)) 
     return False 

while (processKey() == False): 
    processKey() 

输出：

Enter 3 digit PWD: 
3333 
The max allowed character is 3, instead you entered 3333. 
Enter 3 digit PWD: 
321 
You typed 321 wich is incorrect. 
Enter 3 digit PWD: 
123 
Correct password. 

Answer 2

按键每按一次键执行 - 这是自然的。你必须把所有的钥匙放在列表或字符串中，并检查它的长度。

code = '' 

def processKey(key): 
    global code 

    code += key 

    if len(code) == 3: 
     if code == "123": 
      print("correct") 
     else: 
      print("wrong password, try again") 
      code = ''