运行 LED 闪烁和蜂鸣器同时发声的功能

Question

我正在做一个 Raspberry Pi 安全小项目，该项目有 LED、压电蜂鸣器、9 位数字键盘和 LCD 屏幕（完整资源在这里：https://github.com/kevbo423/RPHSP/blob/master/Keypad.py）。

简而言之，当用户使用键盘输入错误的密码时，我希望 LED 闪烁，同时蜂鸣器熄灭。我一辈子都想不出如何做到这一点。我已创建并单独工作的 LED 和蜂鸣器功能，但无法让它们同时熄灭。

# Function for error LED feedback
def errorLED():
    for i in range(3):
        GPIO.output(6,1)
        time.sleep(0.2)
        GPIO.output(6,0)
        GPIO.output(12,1)
        time.sleep(0.2)
        GPIO.output(12,0)

# Function for Buzzer Alarm Output
def buzzerAlarm():
    for i in range(3):
        GPIO.output(21,1)
        time.sleep(0.5)
        GPIO.output(21,0)
        time.sleep(0.5)

我研究过使用线程模块，但是运行当我尝试执行线程时蜂鸣器发出多次蜂鸣声时出现了问题。下面是我创建的测试代码，用于多次运行蜂鸣器线程...

# Buzzer Test
buzzerSound = threading.Thread(target=buzzerAlarm())
buzzerSound.start()
time.sleep(3)
buzzerSound.start()
time.sleep(3)
buzzerSound.start()
time.sleep(3)

...以及我收到的错误消息。

Traceback (most recent call last):
  File "Keypad.py", line 277, in <module>
    buzzerSound.start()
  File "/usr/lib/python2.7/threading.py", line 730, in start
    raise RuntimeError("threads can only be started once")
RuntimeError: threads can only be started once

我使用线程模块的方向是否正确？如果是这样，我应该终止 BuzzerAlarm() 函数内部的蜂鸣器线程还是不应该使用 .start() 到运行函数？还是我需要完全做其他事情？如果线程不是最好的方法，我愿意接受建议。

Answer 1

来自documentation：

start()

It must be called at most once per thread object.

因此，如果您想再次启动它，您应该创建一个新的 Thread 对象。例如：

buzzer_thread = threading.Thread(target=buzzerAlarm())
led_thread = threading.Thread(target=errorLED())
led_thread.start()
buzzer_thread.start()
time.sleep(3)

# Second time:
buzzer_thread = threading.Thread(target=buzzerAlarm())
led_thread = threading.Thread(target=errorLED())
led_thread.start()
buzzer_thread.start()

你可以考虑将其重构为一个函数：

def buzzer_and_led():
    buzzer_thread = threading.Thread(target=buzzerAlarm())
    led_thread = threading.Thread(target=errorLED())
    led_thread.start()
    buzzer_thread.start()
    # Wait until both the buzzer and the LED stop
    while led_thread.is_alive() or buzzer_thread.is_alive():
       pass

Answer 2

首先不是多线程答案，这个答案应该仍然有效。

好的，快回答。你可以做的是结合这两个功能

# Function for error LED feedback
def errorLED():
    for i in range(3):
        GPIO.output(6,1)
        time.sleep(0.2)
        GPIO.output(6,0)
        GPIO.output(12,1)
        time.sleep(0.2)
        GPIO.output(12,0)
# Function for Buzzer Alarm Output
def buzzerAlarm():
    for i in range(3):
        GPIO.output(21,1)
        time.sleep(0.5)
        GPIO.output(21,0)
        time.sleep(0.5)

这会变成

#Combined function
def combined():
    for i in range(3):
        GPIO.output(6,1) #Error LED
        GPIO.output(21,1) #Alarm Buzzer
        time.sleep(0.25) #Notice times changed by 0.05 so it can do the combination easyer
        GPIO.output(6,0)
        GPIO.output(12,1)
        time.sleep(0.25)
        GPIO.output(12,0)
        #Repeats Code 
        GPIO.output(6,1)
        GPIO.output(21,0) #Only difference is turning off the buzzer/alarm
        time.sleep(0.25) 
        GPIO.output(6,0)
        GPIO.output(12,1)
        time.sleep(0.25)
        GPIO.output(12,0)

告诉我这是否有效，我已经有一段时间没有使用我的 Raspberry PI

运行 LED 闪烁和蜂鸣器同时发声的功能

Run functions for LED flashing and Buzzer sound simultaneously

python

python-3.x

raspberry-pi

python-multithreading

gpio