如何使用 arduino 串行监视器打开和关闭 LED

Question

I am trying to toggle 3 LED's on and off. So basically input 1 into the serial monitor and LED one will turn on, I have this part done but then when I press 1 again LED one will turn off. The second part is what im having issues with, I need to create some sort of toggle.

Here is my code

const int greenPin = 2;
const int yellowPin = 3;
const int redPin = 4;

void setup()

{
pinMode(greenPin, OUTPUT);
  pinMode(yellowPin, OUTPUT);
  pinMode(redPin, OUTPUT);


  Serial.begin(9600);

  while (!Serial);

  Serial.println("Input 1 to Turn LED on and 2 to off");

}

void loop() {

  if (Serial.available())

  {

    int state = Serial.parseInt();

    if (state == 1)

    {

      digitalWrite(greenPin, HIGH);
      digitalWrite(yellowPin, LOW);
       digitalWrite(redPin, LOW);
      Serial.println("Command completed LED turned ON");

    }

    if (state == 2)

    {
  digitalWrite(greenPin, LOW);
      digitalWrite(yellowPin, HIGH);
       digitalWrite(redPin, LOW);

      Serial.println("Command completed LED turned OFF");

    }

      if (state == 3)

    {
     digitalWrite(greenPin, LOW);
      digitalWrite(yellowPin, LOW);
       digitalWrite(redPin, HIGH);

      Serial.println("Command completed LED turned OFF");

    }




  }

}

Answer 1

digitalWrite(ledPin,  !digitalRead(ledPin));

That way you always invert the current pin state.