使用树莓派 3 从 4x4 键盘读取多个数字

Question

我正在做一些与出勤相关的项目。我在哪里使用 4x4 键盘和 LCD 显示器。所以问题是，如何从 4x4 矩阵键盘中读取多个数字？ 我正在使用 pad4pi 库。我一次只能读取一个数字或数字。现在我想读一个像 1234 或 12345 这样的数字。你能帮帮我吗？

Answer 1

感谢您向我展示。该代码不会做任何事情。在注册键之后，您必须对存储的数据结构进行实际操作。

例如：

#change store key function to do something on submission of a certain key that indicated send, will use pound for example.
def store_key(self,key):
     If(key==‘#’):
          #im printing but you should do whatever it is you intend to do with the sequence of keys. 
          print(self.pressed_keys)
     else:
          self.pressed_keys.append(key)

该代码将在按下 # 时打印内部数据结构。当然我假设传递的密钥是一个字符串，但我真的不知道我不熟悉这个库。

Answer 2

我这样做了，但我还是卡住了

from pad4pi import rpi_gpio
import time


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

# same as calling: factory.create_4_by_4_keypad, still we put here fyi:
ROW_PINS = [16, 20, 21, 5] # BCM numbering; Board numbering is: 7,8,10,11 (see pinout.xyz/)
COL_PINS = [6, 13, 19, 26] # BCM numbering; Board numbering is: 12,13,15,16 (see pinout.xyz/)

factory = rpi_gpio.KeypadFactory()

# Try keypad = factory.create_4_by_3_keypad() or 
# Try keypad = factory.create_4_by_4_keypad() #for reasonable defaults
# or define your own:
keypad = factory.create_keypad(keypad=KEYPAD, row_pins=ROW_PINS, col_pins=COL_PINS)


class KeyStore:
    def __init__(self):
        #list to store them
        self.pressed_keys =[]

    #function to add to list
    def store_key(self, key):
        self.pressed_keys.append(key)

    #function to clear list
    def clear_keys(self):
        self.pressed_keys.clear()

keys = KeyStore()

# store_key will be called each time a keypad button is pressed
keypad.registerKeyPressHandler(keys.store_key)
for key in keys.pressed_keys:
    print(key)

try:
    while(True):
        time.sleep(0.2)
except:
    keypad.cleanup()

但它什么也没做。

Answer 3

这是我根据更改的代码。

from pad4pi import rpi_gpio
import time


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

# same as calling: factory.create_4_by_4_keypad, still we put here fyi:
ROW_PINS = [16, 20, 21, 5] # BCM numbering; Board numbering is: 7,8,10,11 (see pinout.xyz/)
COL_PINS = [6, 13, 19, 26] # BCM numbering; Board numbering is: 12,13,15,16 (see pinout.xyz/)

factory = rpi_gpio.KeypadFactory()

# Try keypad = factory.create_4_by_3_keypad() or 
# Try keypad = factory.create_4_by_4_keypad() #for reasonable defaults
# or define your own:
keypad = factory.create_keypad(keypad=KEYPAD, row_pins=ROW_PINS, col_pins=COL_PINS)


class KeyStore:
    def __init__(self):
        #list to store them
        self.pressed_keys =''

    #function to clear string
    def clear_keys(self):
        self.pressed_keys = self.pressed_keys.replace(self.pressed_keys,'')

    def store_key(self,key):
        if key=='#':
            #printing the sequence of keys.
            print(self.pressed_keys)
            self.clear_keys()
        else:
            self.pressed_keys += key

keys = KeyStore()

# store_key will be called each time a keypad button is pressed
keypad.registerKeyPressHandler(keys.store_key)


try:
    while(True):
        time.sleep(0.2)
except:
    keypad.cleanup()

Answer 4

可以通过使用数据结构来存储按下的值来促进存储键序列。然后在需要重置时清除它。

class KeyStore:
    def __init__(self):
        #list to store them
        self.pressed_keys = []

    #function to add to list
   def store_key(self,key):
       if(key==‘#’):
      #im printing but you should do whatever it is you intend to do with the sequence of keys. 
      print(self.pressed_keys)
 else:
      self.pressed_keys.append(key)

    #function to clear list
    def clear_keys(self):
        self.pressed_keys.clear()

keys = KeyStore()

# store_key will be called each time a keypad button is pressed
keypad.registerKeyPressHandler(keys.store_key)