I want to validate a single character in a java application.
I don't want the user to be able to enter a character outside the range [a - p] (ignoring uppercase or lowercase) or numbers.
Scanner input = new Scanner(System.in);
System.out.print("Choose letter in range [a - p]");
letter = input.next().charAt(0);
Any ideas?
you can use regex to filter input
.matches("^[a-pA-P0-9]*$") //this will not allow outside range of [a-p A-P] or numbers
^ Assert position at start of the string
- Create a character range with the adjascent tokens
a-p A single character in the range between a and p (case sensitive)
A-P A single character in the range between A and P (case sensitive)
0-9 A single character in the range between 0 and 9
* Repeat previous token zero to infinite times, as many times as possible
$ Assert position at end of the string
like this:
Scanner input = new Scanner(System.in);
System.out.print("Choose letter in range [a - p]");
letter = input.next().charAt(0);
if (Character.toString(letter).matches("^[a-pA-P0-9]*$")) {
System.out.println("valid input");
}else{
System.out.println("invalid input");
}
you can encapsulate it with a while loop to check if the letter is in the range or not and just keep asking the user to input a letter
do {
System.out.print("Choose letter in range [a - p]"); letter = input.next().charAt(0);
} while (letter is not in range of [a-p]); // pseudo code
If you're not familiar with regex, simple checks on the input can take care of it.
System.out.print("Choose letter in range [a - p]");
String letter = input.nextLine();
if (letter.length() == 1 && ('a' <= letter.charAt(0) && letter.charAt(0) <= 'p')) {
System.out.println(letter);
} else {
System.out.println("Invalid input");
}
I'm using nextLine() in the event that multiple characters are accidentally entered, which is why I check that the length of letter == 1 and that the letter falls in the range that I want, otherwise it is invalid input.
