Question

如何使用python将旋转拨号脉冲转换为数字？现在，下面的代码将打印从旋转拨盘接收的脉冲数。因此，旋转拨盘上的拨号2可打印两个“1+ 1+”。旋转拨盘上的拨号3可打印三个“1+ 1+ 1+”，依此类推。

#!/usr/bin/python3
import RPi.GPIO as GPIO  
import math, sys, os
import subprocess
import socket

GPIO.setwarnings(False)
GPIO.setmode(GPIO.BCM)  
GPIO.setup(18, GPIO.IN, pull_up_down=GPIO.PUD_UP)
GPIO.setup(23, GPIO.IN, pull_up_down=GPIO.PUD_UP)

c=0
last = 1

def count(pin):
    global c 
    c = c + 1

GPIO.add_event_detect(18, GPIO.BOTH)

while True:
    try:
        if GPIO.event_detected(18):
            current = GPIO.input(18)
            if(last != current):
                if(current == 0):
                    GPIO.add_event_detect(23, GPIO.BOTH, callback=count, bouncetime=5)
                else:
                    GPIO.remove_event_detect(23)    
                    print "1+"
                last = GPIO.input(18)
    except KeyboardInterrupt:
        break

您如何在旋转拨盘上为每个号码提供自己的功能？像拨号1解锁前门“GPIO.output（26，True）”。拨号2打开车库门“GPIO.output（27，True）”。

if dial 1:
  GPIO.output(26, True)
  print "Door unlocked"

if dial 2:
  GPIO.output(27, True)
  print "Garage Open"

if dial 3:
  print "Open slot"

if dial 4:
  print "Open slot"

这显然不是python但是怎么可能做类似的事情呢？

Answer 1

不太熟悉Pi，但是您可以将动作映射到dict中的键，而不是打印脉冲，保存多少然后使用计数来调用访问相应的GPIO调用：

d = {1:lambda: GPIO.output(26, True), 2: lambda: GPIO.output(27, True), 3:lambda : GPIO.output(28, True)}
GPIO.add_event_detect(18, GPIO.BOTH)
var = 0
flag = False

while True:
    try:
        if GPIO.event_detected(18):
            current = GPIO.input(18)
            if last != current:
                flag = True
                if current == 0:
                    GPIO.add_event_detect(23, GPIO.BOTH, callback=count, bouncetime=5)
                else:
                    GPIO.remove_event_detect(23)
                    var += 1
                last = GPIO.input(18)
            elif flag:
                d[var]()
                var = 0
                flag = False
    except KeyboardInterrupt:
        break

我假设当last等于当前脉冲已经结束时，我设置一个标志，这样如果没有发生任何事情，你就不会继续将引脚设置为开启

Answer 2

我想对我的好朋友Bob发出嘘声让这个python脚本正常工作。与此同时，我不想对奥巴马做出任何帮助，使其无法帮助完成这项工作。

import RPi.GPIO as gpio
import time
gpio.setmode(gpio.BCM)
gpio.setup(18, gpio.IN, pull_up_down=gpio.PUD_UP)


num = 0
prnt = 0
last = 0

while True:
    input_value = gpio.input(18)

    if (input_value == 1) and (input_value != last):
        last = 1
        prnt = 1
        num += 1
        time.sleep(0.05)
        continue

    if (input_value == 0) and (input_value != last):
        last = 0
        time.sleep(0.05)
        continue

    if (input_value == 0) and (input_value == last):
        if (prnt == 1):
            if (num == 10):
                num = 0

            if (num == 0):
                print("a")
            if (num == 1):
                print("b")                
            if (num == 2):
                print("c")
            if (num == 3):
                print("d")
            if (num == 4):
                print("e")                
            if (num == 5):
                print("f")                
            if (num == 6):
                print("g")                
            if (num == 7):
                print("h")                
            if (num == 8):
                print("i")                
            if (num == 9):
                print("jklmnopqrstuvwxyz")                


            num = 0
            prnt = 0
            last = 0
        continue

旋转拨盘连接到gpio pi

2 个答案: