python在继续之前等待第n个数字

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 位数字，然后再与上面代码中的 123 代码中的密码进行比较。

应该怎么做：

等待用户从键盘输入 3 位数字，例如 123，然后打印正确。

实际作用：

用户输入1位密码后会立即打印正确或错误的密码

Answer 1

以@furas 为例更新覆盆子：

# 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.

这可能适合您的情况。

---

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 一个值，因为正如你所说的用户输入它，调用 processKey() 将要求用户输入密码并根据检查中的“123”返回真/假。

这是如果您想输入密码，但如果以下答案不适合您的需求（没有完全理解您要完成的任务），请提供更聪明的示例。

编辑：

由于您希望严格输入 3 位数字并重新输入密码，以防输入错误，您可以执行以下操作：

在调用 processKey() 时，您可以：

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

修改后的代码以满足您的需求：

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

keypress 在每次按键后执行 - 这是很自然的。您必须将所有键保留在列表或字符串中并检查其长度。

code = ''

def processKey(key):
    global code

    code += key

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