Question

使用raspberry pi 3读取GPIO按钮状态 ----------------------------------------------

问题：当我运行下面的代码并按下我的pi 附加的按钮时，它会多次打印“按钮已被按下”（有时，它可以打印它100次）

有谁知道为什么会这样？ 在此先感谢您的合作

-----
code:
-----
 buttonPin = 17
    import RPi.GPIO as gpio
    gpio.setmode(gpio.BCM)
    gpio.setup(buttonPin, gpio.IN)
    count=0
    ButtonState=True    #means that the button is in the up position and has not yet been pressed.

    while True:
        input_value = gpio.input(17)
        if input_value == False:
            print('The button has been pressed...')
            print(count)

Answer 1

是的，您需要在print（）之后添加延迟。您的脚本移动太快，第一次迭代后仍然按下开关。

轻松修复：

buttonPin = 17
import RPi.GPIO as gpio

#Import time library for delays
import time

gpio.setmode(gpio.BCM)
gpio.setup(buttonPin, gpio.IN)
count=0
ButtonState=True    #means that the button is in the up position and has not yet been pressed.

while True:
    input_value = gpio.input(17)
    if input_value == False:
        print('The button has been pressed...')
        print(count)

        #Add a debounce delay "time.sleep(duration)"
        time.sleep(0.5)

这意味着您的脚本在按下按钮后0.5秒才会继续，为按钮释放时间。

GPIO按钮按下多次运行打印方法

1 个答案: