Here's my code:
const int AnalogInPin1 = A1;
const int AnalogInPin2 = A2;
int SerialPrint = 0;
int SerialMonitor = 0;
void setup() {
Serial.begin(9600);
}
void loop() {
SerialPrint = analogRead(AnalogInPin1);
Serial.print("Sensor = ");
Serial.println(SerialPrint);
if (SerialPrint > 400) {
digitalWrite(3, LOW);
digitalWrite(4, HIGH);
delay(500);
}
else{
digitalWrite(4, HIGH);
digitalWrite(3, LOW);
delay(500);
}
}
void setup1() {
Serial.begin();
}
void loop1() {
SerialMonitor = analogRead(AnalogInPin2);
Serial.print("Sensor = ");
Serial.println(SerialMonitor);
if (SerialMonitor > 400) {
digitalWrite(6, LOW);
digitalWrite(5, HIGH);
delay(500);
}
else{
digitalWrite(5, HIGH);
digitalWrite(6, LOW);
delay(500);
}
}
I am trying to make a car guidance system like the ones found in underground parking lots. and I am trying to have two serial monitors read from 2 sensors.
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The characters above are looping in the serial monitor whilst using 19200 baud but not in 9600 baud why is that?
If your question is about how to display two analogRead values in one SerialMonitor sketch, this should compile and work (if your SerialMonitor is set to 9600, of course):
const int AnalogInPin1 = A1;
const int AnalogInPin2 = A2;
int value1 = 0;
int value2 = 0;
void setup() {
Serial.begin(9600);
pinMode(3, OUTPUT);
pinMode(4, OUTPUT);
pinMode(5, OUTPUT);
pinMode(6, OUTPUT);
}
void loop() {
value1 = analogRead(AnalogInPin1);
Serial.print("Sensor1 = ");
Serial.print(value1);
if (value1 > 400) {
digitalWrite(3, LOW);
digitalWrite(4, HIGH);
}
else{
digitalWrite(4, HIGH);
digitalWrite(3, LOW);
}
value2 = analogRead(AnalogInPin2);
Serial.print("\t Sensor2 = ");
Serial.println(value2);
if (value2 > 400) {
digitalWrite(6, LOW);
digitalWrite(5, HIGH);
}
else{
digitalWrite(5, HIGH);
digitalWrite(6, LOW);
}
delay(500);
}
I left most of your code as is, just made it compile and work. If this is not what you want, please edit your question.