I'm trying to make a small program more robust and I need some help with that.
Scanner kb = new Scanner(System.in);
int num1;
int num2 = 0;
System.out.print("Enter number 1: ");
num1 = kb.nextInt();
while(num2 < num1) {
System.out.print("Enter number 2: ");
num2 = kb.nextInt();
}
Number 2 has to be greater than number 1
Also I want the program to automatically check and ignore if the user enters a character instead of a number. Because right now when a user enters for example r
instead of a number the program just exits.
This question is related to
java
validation
java.util.scanner
String
and then try
ing Integer.parseInt()
and if you don't catch
an exception then it's a number, if you do, read a new one, maybe by setting num2 to Integer.MIN_VALUE and using the same type of logic in your example.Try this:
public static void main(String[] args)
{
Pattern p = Pattern.compile("^\\d+$");
Scanner kb = new Scanner(System.in);
int num1;
int num2 = 0;
String temp;
Matcher numberMatcher;
System.out.print("Enter number 1: ");
try
{
num1 = kb.nextInt();
}
catch (java.util.InputMismatchException e)
{
System.out.println("Invalid Input");
//
return;
}
while(num2<num1)
{
System.out.print("Enter number 2: ");
temp = kb.next();
numberMatcher = p.matcher(temp);
if (numberMatcher.matches())
{
num2 = Integer.parseInt(temp);
}
else
{
System.out.println("Invalid Number");
}
}
}
You could try to parse the string into an int
as well, but usually people try to avoid throwing exceptions.
What I have done is that I have defined a regular expression that defines a number, \d
means a numeric digit. The +
sign means that there has to be one or more numeric digits. The extra \
in front of the \d
is because in java, the \
is a special character, so it has to be escaped.
This should work:
import java.util.Scanner;
public class Test {
public static void main(String... args) throws Throwable {
Scanner kb = new Scanner(System.in);
int num1;
System.out.print("Enter number 1: ");
while (true)
try {
num1 = Integer.parseInt(kb.nextLine());
break;
} catch (NumberFormatException nfe) {
System.out.print("Try again: ");
}
int num2;
do {
System.out.print("Enter number 2: ");
while (true)
try {
num2 = Integer.parseInt(kb.nextLine());
break;
} catch (NumberFormatException nfe) {
System.out.print("Try again: ");
}
} while (num2 < num1);
}
}
What you could do is also to take the next token as a String, converts this string to a char array and test that each character in the array is a digit.
I think that's correct, if you don't want to deal with the exceptions.
I see that Character.isDigit perfectly suits the need, since the input will be just one symbol. Of course we don't have any info about this kb object but just in case it's a java.util.Scanner instance, I'd also suggest using java.io.InputStreamReader for command line input. Here's an example:
java.io.BufferedReader reader = new java.io.BufferedReader(new java.io.InputStreamReader(System.in));
try {
reader.read();
}
catch(Exception e) {
e.printStackTrace();
}
reader.close();
Source: Stackoverflow.com