How can i print something on the serial monitor only once

Question

I have a button connect to my arduino board and i want the serial monitor to display pressed when the button is pressed and released when the button is not pressed. My problem is that i want it to be print only once but with my code it prints it non stop. I already tried writing the code in void setup but i cant seem to make it work. Does anyone have any suggestions? I would really appreciate the help.

const int pinButton = 8;

void setup() {
  pinMode(pinButton, INPUT);
  Serial.begin(9600);
}

void loop() {
  int stateButton = digitalRead(pinButton);
  if(stateButton == 1) {
     Serial.println("PRESSED"); 
  } else {
     Serial.println("RELEASED"); 
  }
  delay(20);
}

Answer 1

this is my first answer on stack overflow. Anyway, the solution I suggest is to save previous state of the button in another variable, compare it to the new state, if they are diffrent you print the message, else you don't. Here's a code example:

const int pinButton = 8;
int previous_state;

void setup() {
   pinMode(pinButton, INPUT);
   Serial.begin(9600);
   previous_state = digitalRead(pinButton);
}

void loop() {
    int new_state = digitalRead(pinButton);
    if(new_state == 1 && previous_state==0) {
       Serial.println("PRESSED"); 
    } if(new_state == 0 && previous_state==1) {
       Serial.println("RELEASED"); 
    }
    previous_state=new_state;
    delay(20);
}

This is not the optimal solution, but it should work. Chek out interruptions on Arduino to see how to do it better.