Java, If Scanner Input == letter?

Question

I have a question.
For example: I'm asking for the age of a person by using a scanner. And putting the input in an int (For example: x).
With an if statement I'm checking if the int x is smaller than 16, then he's not allowed to play. But now: when someone inputs a letter like abcdefg..... im getting an exception.
I know why, because I'm asking for a number without validating if the person actually inputs one. So when he puts in a letter, its not a number and I'm getting an exception.
Any way to catch the execption and asking the person to put in a number and not a letter?
I'm using the SCANNERNAME.nextInt() method.
Thanks for the help.

Answer 1

have you tried putting it in a while loop and catching the exception? if you catch an exception maybe offer a print output suggesting to enter a number and then letting the loop go back to reenter a number. if no exception is thrown, let them out of the loop
while {
try {
SCANNERNAME.nextint();
break;
}
catch (Exception e) {
System.out.println("Enter a number");
continue;
}
}
something like that

Answer 2

Thanks :)

Answer 3

You can use a try catch statement. It would look something like this:
try{
//line that throws exception (in your case scanner.nextInt())
}catch (Exception e){
//handle exception (re-prompt user for input)
}
You can also put your try/catch in a while loop until input is valid like this:
boolean valid;
While(!valid){
try{
//statement
valid = true;
}catch (Exception e){
//handle exception
}
}

Answer 4

You can catch exception using :-
try { int input = scannerName.nextInt(); // remaining logic } catch (InputMismatchException e) { System.out.println("uh oh"); }
or take the advantage of scanner class pre-existing function :-
if(scannerName.hasNextInt()){
//Input is integer
}
else
{
//input is not an integer
}