why it is not compilling

Projects Programming
1

HELLO.I HAD A PROBLEM WHEN I WAS DOING CODING FOR MY PROJECT.I DONT KNOW WAT IS MY MISTAKE BECAUSE I'M NEW FOR ARDUINO.I HAVE ATTACHED THE CODING BELOW CAN YOU GUYS PLEASE HELP ME BECAUSE IT'S A HURRY.TQ FOR UR SUPPORT.

sketch_sep23a.ino (1.05 KB)

2

Could you at least tell us the error?

3

Please don't SHOUT.
Please do read the posting guidelines.
Tell us what the error is; don't make us guess.
Post code.
Use code tags.

#include  <LiquidCrystal> lcd(8,9,4,5,6,7);
const int tempPin = A0;

int value=0;           
float volts=0.0;      
float temp=0.0;      
float tempF=0.0;

void setup()
{
  pinMode(3,INPUT);      
  Serial.begin(9600);   
  lcd.begin(16,2); 
  pinMode(1, OUTPUT);
  pinMode(2, OUTPUT);
  pinMode(3, OUTPUT);     
 
}
void loop()
{
  value=analogRead(A0);         
  volts=(value/1024.0)*5.0;      
  temp= volts*100.0;            
  tempF=temp*9/5+32; 

   


  Serial.print("temperature= "); 
  Serial.println(temp);
  lcd.setCursor(0,0);
  lcd.print("TEMP= ");
  lcd.print(temp);
  lcd.print(" C");
  lcd.setCursor(0,1);
  lcd.print("TEMP= ");
  lcd.print(tempF);
  lcd.print(" F");

  if (temPin < 36.5)
  {
    digitalWrite(8, HIGH);
    digitalWrite(9, LOW);
    digitalWrite(10, LOW);
    delay(1000)
  }
  else (tempPin > 37.5)
  {
    digitalWrite(8, LOW);
    digitalWrite(9, HIGH);
    digitalWrite(10, LOW 
    delay(1000)
  }
  if (tempPin <=37.5 - 36.5),
  {
    digitalWrite(8, LOW);
    digitalWrite(9, LOW);
    digitalWrite(10, HIGH); 
    delay(1000)  
  }
4 
  if (tempPin <=37.5 - 36.5),

I'm guessing the compiler didn't like the comma.
Neither do I

5

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What is the , for?

if (tempPin <=37.5 - 36.5),
6

You made various mistakes because of missing parts like semicolons, braces and so on.

You can read the errors in the Arduino Software and post the error, so it is easier for helping you!

#include <LiquidCrystal.h>
LiquidCrystal lcd(8,9,4,5,6,7);
const int tempPin = A0;

int value=0;           
float volts=0.0;      
float temp=0.0;      
float tempF=0.0;

void setup()
{
  pinMode(3,INPUT);      
  Serial.begin(9600);   
  lcd.begin(16,2); 
  pinMode(1, OUTPUT);
  pinMode(2, OUTPUT);
  pinMode(3, OUTPUT);     
 
}
void loop()
{
  value=analogRead(A0);         
  volts=(value/1024.0)*5.0;      
  temp= volts*100.0;            
  tempF=temp*9/5+32; 

   


  Serial.print("temperature= "); 
  Serial.println(temp);
  lcd.setCursor(0,0);
  lcd.print("TEMP= ");
  lcd.print(temp);
  lcd.print(" C");
  lcd.setCursor(0,1);
  lcd.print("TEMP= ");
  lcd.print(tempF);
  lcd.print(" F");

  if (tempPin < 36.5)
  {
    digitalWrite(8, HIGH);
    digitalWrite(9, LOW);
    digitalWrite(10, LOW);
    delay(1000);
  }
  else if(tempPin > 37.5)
  {
    digitalWrite(8, LOW);
    digitalWrite(9, HIGH);
    digitalWrite(10, LOW);
    delay(1000);
  }
  if (tempPin <= 37.5 - 36.5)
  {
    digitalWrite(8, LOW);
    digitalWrite(9, LOW);
    digitalWrite(10, HIGH); 
    delay(1000);
  }
}

sketch_sep23a.ino (1.07 KB)

7 
  pinMode(3,INPUT);      
  Serial.begin(9600);  
  lcd.begin(16,2);
  pinMode(1, OUTPUT);
  pinMode(2, OUTPUT);
  pinMode(3, OUTPUT);

What mode should pin 3 really have?

Why are you using pin 1, which is one of the hardware serial pins, while also using Serial?

const int tempPin = A0;

  if (tempPin < 36.5)
  {

I'm almost certain that that will be true.

  else if(tempPin > 37.5)

And that that will be false

  if (tempPin <= 37.5 - 36.5)

As will that.

It seems to me that you want to compare the temperature read from the pin, not the pin number, to those values.