(new to java) i need some help here

[fiery wind]

<@&987246399047479336> please have a look, thanks.

[fiery wind]

While you are waiting for getting help, here are some tips to improve your experience:

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Here is an AI assisted attempt to answer your question 🤖. Maybe it helps! In any case, a human is on the way 👍. To continue talking to the AI, you can use </chatgpt:1108714622413963314>.

class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Enter an integer: ");
if (scanner.hasNextInt()) {
int num = scanner.nextInt();
System.out.println("You entered: " + num);
} else {
System.out.println("Here");
}
}
}

[wide jewel]

See, referring from the documentation.

Whenever it gets any value that's not a number it raises java.util.InputMismatchException, you can use a try/catch block to try to input and do something when the InputMismatchException is caught.

It gave you a solution which checks if the nextToken that it'll retrieve on doing nextInt() is actually an int,
The hasNextInt() method returns a boolean indicating if the input entered was an integer.

From the docs^

[sick spoke]

System.out.println("Input should be a number not text, Try again: ");

so in java the easiest way is to do a try { Integer.parseInt(age) } catch (NumberFormatException e) {}

a more complex is to do a regex match on digits

[wide jewel]

No, the hasNextInt() method is just for that.

[wide jewel]

that's wrong, as OP is getting input from a scanner. In the condition the input isn't a valid integer, it'll raise an exception at that part and it would not reach at your code fragment.

[sick spoke]

ah, ye haddnt seen using the hasNextInt, not a common one to use 🙂

[wide jewel]

Well they can use scanner.next() to store input in a string variable then do your logic but that'll be undoubtedly more complex.

[sick spoke]

its a problem with having been with java for so long in trivialising these problems when you first start

[wide jewel]

I didn't get you, what are you trying to say?

[sick spoke]

just that if you work with java for many years, a lot of things apear simple but are still complex to those learning the language

[wide jewel]

ah yes.