如何让按钮执行一次功能？

Question

我目前正在开发一个连接到树莓派的键盘。按下每个按钮时，都需要将一个字符串附加到列表中一次，然后重新启动循环，以便可以放置下一个输入。代码如下所示：

message = [] #final list will be stored here
loop = 0 #determines the position of each number inputted
while True:
  if (GPIO.input(12) == GPIO.HIGH) and (GPIO.input(19) == GPIO.HIGH): #the character pad works in an array, when 2 "buttons" are pressed, they correspond do a location on the pad
    message.insert(loop, "1") #inserts the number into the list
    loop = loop + 1
  if (GPIO.input(12) == GPIO.HIGH) and (GPIO.input(15) == GPIO.HIGH):
    messege.insert(loop, "4")
    loop = loop + 1
  #this repeats for the other 14 buttons however the code is the same

如果按下“1”的按钮，然后按下“4”，则此代码的输出应该如下所示：

['1' '4']

但是，输出看起来像这样：

['1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1' '1...]

有没有办法让每个按钮每次按下激活一次？

这段代码中的按钮都是在pud_down中设置的

Answer 1

考虑这样的事情。关键是不要评估状态，除非您知道某些情况发生了变化：

message = [] #final list will be stored here
loop = 0 
states = {
    12: GPIO.LOW,
    15: GPIO.LOW,
    19: GPIO.LOW
}
while True:
    change = False
    for line in (12,15,19):
        if GPIO.input(line) != states[line]:
            states[line] = GPIO.input(line)
            change = True

    if change:
        change = False
        if states[12]==GPIO.HIGH and states[19]==GPIO.HIGH:
            message.insert(loop, "1") #inserts the number into the list
            loop = loop + 1
        if states[12]==GPIO.HIGH and states[15]==GPIO.HIGH:
            messege.insert(loop, "4")
            loop = loop + 1

Answer 2

这是一种面向对象的方法。你可以创建一个Button 类。每个按钮都可以处于pressed 或released 状态。每当最初按下按钮时，它都会将消息附加到类变量Button.messages。在释放之前，相同的按钮不会附加另一条消息。在主轮询循环中，我们遍历所有按钮并按下或释放它们，具体取决于哪些引脚为高电平：

class Button:

    messages = []

    def __init__(self, value):
        self.value = value
        self.is_pressed = False

    def hold(self):
        if not self.is_pressed:
            Button.messages.append(value)
            self.is_pressed = True

    def release(self):
        self.is_pressed = False


buttons = {
    (12, 19): Button("1"),
    (12, 15): Button("4"),
    # ...
}

row_pins = (12, 13, 14) # You'll have to change these. I don't know what these values should be
col_pins = (19, 15, 10) # You'll have to change these. I don't know what these values should be

while True:
    for row in row_pins:
        for col in col_pins:
            button = buttons[(row, col)]
            (button.release, button.hold)[all((GPIO.input(pin) for pin in (row, col)))]()

Answer 3

试试这个

message = [] #final list will be stored here
loop = 0 #determines the position of each number inputted
val i = 0
val k = 0
bool canPressAgain =true
while True:
  if (GPIO.input(i) == GPIO.LOW) and (GPIO.input(k) == GPIO.LOW):
    canPressAgain=true;
  if (GPIO.input(12) == GPIO.HIGH) and (GPIO.input(19) == GPIO.HIGH): 
    #the character pad works in an array, when 2 "buttons" are pressed, they correspond do a location on the pad
    i = 12
    k = 19
    message.insert(loop, "1") #inserts the number into the list
    loop = loop + 1
  if (GPIO.input(12) == GPIO.HIGH) and (GPIO.input(15) == GPIO.HIGH):
    i = 12
    k = 15
    messege.insert(loop, "4")
    loop = loop + 1
  canPressAgain = false; #Put this at the end of all 'if' statements
  #this repeats for the other 14 buttons however the code is the same