raspberry pi 如何使用定时器控制 LED 的开关持续时间

Question

我想使用定时器来控制 LED 的开/关持续时间。我写了下面的代码，但它没有运行.

def main ():
    LED =27
    GPIO.setup(LED, GPIO.OUT)

    while (True):
            now = datetime.datetime.now()
            todayon = now.replace(hour = 17, minute=47, second =0, microsecond =0)
            todayoff = now.replace(hour = 17, minute=48, second =0, microsecond =0)
            turnon = now>todayon
            turnoff = now>todayoff

            if(turnon == True):
                    GPIO.output(LED, GPIO.HIGH)
                    time.sleep(1)
                    GPIO.output(LED, GPIO.LOW)
                    time.sleep(1)

            if(turnoff == True):
                    GPIO.output(LED, GPIO.HIGH)

Answer 1

如果它没有运行（什么都不做），可能是因为您从未调用 main

在这种情况下，在底部添加以下行：

if __name__ == '__main__': 
    main()

此外，您需要设置 GPIO 以使用端口号或引脚号。使用 GPIO.setmode(GPIO.BCM) 将模式设置为端口号，因此当您调用 GPIO.setup(LED, GPIO.OUT).

时，GPIO 端口号 27 将用作输出

完整的脚本现在应该如下所示：

#!/usr/bin/python
import time
import datetime
import RP1.GPIO as GPIO

def main ():
    LED =27
    GPIO.setmode(GPIO.BCM)
    GPIO.setup(LED, GPIO.OUT)

    while (True):
        now = datetime.datetime.now()
        todayon = now.replace(hour = 17, minute=47, second =0, microsecond =0)
        todayoff = now.replace(hour = 17, minute=48, second =0, microsecond =0)
        turnon = now>todayon and now<todayoff
        turnoff = now>todayoff

        if(turnon == True):
                GPIO.output(LED, GPIO.HIGH)
                time.sleep(1)
                GPIO.output(LED, GPIO.LOW)
                time.sleep(1)

        if(turnoff == True):
                GPIO.output(LED, GPIO.HIGH)

if __name__ == '__main__': 
    main()

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