我怎样才能让这个继电器被我拥有的 3 个触摸传感器中的任何一个打开？

Question

我有一个继电器，我想用它打开灯，我有 2 个触摸传感器，但使用此代码我只能用 1 个打开它，我怎样才能让它工作？第三个是一个开关，但它应该仍然可以正常工作。我试过了，它使用不同的代码。但该代码用于伺服而不是继电器。如果你知道如何让它工作，你能发送完整的代码吗？

int touchPin = 2;
int relayPin = 3;

int val = 0;
int lightON = 0;
int touched = 0;

void setup() {
  Serial.begin(9600);
  pinMode(touchPin, INPUT); 
  pinMode(relayPin, OUTPUT);

}

void loop() {

  val = digitalRead(touchPin);

  if(val == HIGH && lightON == LOW){

    touched = 1-touched;
    delay(100);
  }    

  lightON = val;

      if(touched == HIGH){
        Serial.println("Light ON");
        digitalWrite(relayPin, LOW); 
       
      }else{
        Serial.println("Light OFF");
        digitalWrite(relayPin, HIGH);
   
      }     

  delay(100);
}

Answer 1

我建议阅读有关 Arduino 编程的文章，这样您就可以充分了解其他人的程序中发生了什么，然后对其进行修改以适应。网上有海量资源等着你:)

我会这样写程序：

 `/*
  * As I understand the relay is to turn on a light, 
  * should either of the touch sensors be triggered
  */

  const int touchPin_1 = 2;
  const int touchPin_2 = 3;
  const int relayPin = 4;

  bool isLightOn = false;

  void setup() {
      pinMode(touchPin_1, INPUT);
      pinMode(touchPin_2, INPUT);
      pinMode(relayPin, OUTPUT);

      Serial.begin(9600);
  }

  void loop() {

  int touchPin_1_val = digitalRead(touchPin_1);
  int touchPin_2_val = digitalRead(touchPin_2);

  if (touchPin_1_val == HIGH || touchPin_2_val == HIGH) { //if either of the sensors are triggered...
      isLightOn = !isLightOn; //toggle light state
  }

  if (isLightOn) { //if the light state is true, or set to 'on'...
      Serial.println("light on");
      digitalWrite(relayPin, HIGH); //on
  } else {
      Serial.println("light off");
      digitalWrite(relayPin, LOW); //off
  }

  delay(100);
  }`

Answer 2

Tnx 但我就是这样写的，效果很好

int touchPin = 6;
int touchPin2 = 7;
int touchPin3 = 5;
    int relayPin = 9;
    
    int val = 0;
int val2 = 0;
int val3 = 0;
    int lightON = 0;
    int touched = 0;
    
    void setup() {
      Serial.begin(9600);
      pinMode(touchPin, INPUT); 
            pinMode(touchPin2, INPUT); 
                  pinMode(touchPin3, INPUT); 
      pinMode(relayPin, OUTPUT);
    
    }
    
    void loop() {
    
      val = digitalRead(touchPin);
val2 = digitalRead(touchPin2);
val3 = digitalRead(touchPin3);
    
      if((val == HIGH && lightON == LOW) || (val2 == HIGH && lightON == LOW) || (val3 == HIGH && lightON == LOW)){
    
        touched = 1-touched;
        delay(100);
      }    
    
      lightON = val;
    
          if(touched == HIGH){
            Serial.println("Light ON");
            digitalWrite(relayPin, LOW); 
           
          }else{
            Serial.println("Light OFF");
            digitalWrite(relayPin, HIGH);
       
          }     
      delay(100);
    }

我怎样才能让这个继电器被我拥有的 3 个触摸传感器中的任何一个打开？

How can i make this relay be turned on by any of the 3 touch sensors i have?

arduino

touch

sensors