How can i print something on the serial monitor only once - arduino

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.