How do i add exception handling for the parts where the answer is numeric?

Question

Really confused about this. How do i add exception handling for the parts where the answer is numeric? Don't know where to add the try and catch.

 do {
//zip code      
     do {
            System.out.println("Enter your zip code(or enter 0 to quit): ");
            zip = input.nextInt();
        } while ((zip > 99999 || zip < 10000) && zip != 0);
        if (zip == 0){
            break;
                    }
                    //age
    do {
            System.out.println("Enter your age: ");
            age = input.nextInt();
    } while (age < 10 || age > 110);


                    //items
     do {
            System.out.println("Enter number of items : ");
            numItems = input.nextInt();
            entries = entries + 1;
            if (entries == 3 && numItems < 1) {
                System.out.println("Invalid. Order was not counted");
                           break;
            }

http://docs.oracle.com/javase/tutorial/essential/exceptions/ — Brian Roach, Mar 02 '14 at 00:38
I am confused about your question: What do you actually want to do? — donfuxx, Mar 02 '14 at 00:38
I would like it so that if i enter a letter instead of an integer the program would not crash. I believe i need a try and catch with a mismatch exception — user3344218, Mar 02 '14 at 00:43

score 1 · Answer 1 · edited May 23 '17 at 12:10

The crude way to loop until a valid number is entered, is to ask for a string, attempt to convert it to a number, and keep doing so until this conversion doesn't fail:

   import  java.util.Scanner;
/**
   <P>{@code java StringToNumberWithTesting}</P>
 **/
public class StringToNumberWithTesting  {
   public static final void main(String[] ignored)  {

      int num = -1;

      do  {
         nfx2 = null;
         System.out.print("Number please: ");
         String strInput = (new Scanner(System.in)).next();
         try  {
            num = Integer.parseInt(strInput);
         }  catch(NumberFormatException nfx)  {
            nfx2 = nfx;
            System.out.println(strInput + " is not a number. Try again.");
         }
      }  while(nfx2 != null);

      System.out.println("Number: " + num);
   }
}

Output:

[C:\java_code\]java StringToNumberWithErrLoop
Number please: etuh
etuh is not a number. Try again.
Number please: 2
Number: 2

But using the absence of an exception as a substitute for logic is never a good idea in a real-world application. A better way is to confirm the string contains a number, for which you can use Commons' NumberUtils.isNumber(s)

    import  java.util.Scanner;
    import  org.apache.commons.lang.math.NumberUtils;
 /**
    <P>{@code java StringToNumberWithTesting}</P>
  **/
 public class StringToNumberWithTesting  {
    public static final void main(String[] ignored)  {

       int num = -1;
       boolean isNum = false;

       do  {
          System.out.print("Number please: ");
          String strInput = (new Scanner(System.in)).next();
          if(!NumberUtils.isNumber(strInput))  {
             System.out.println(strInput + " is not a number. Try again.");
          }  else  {
             //Safe to convert
             num = Integer.parseInt(strInput);
             isNum = true;
          }
       }  while(!isNum);

       System.out.println("Number: " + num);
    }
}

Output:

[C:\java_code\]java StringToNumberWithTesting
Number please: uthoeut
uthoeut is not a number. Try again.
Number please: 3
Number: 3

Some more information: How to check if a String is numeric in Java

If you feel this has helped you, please consider up-voting and accepting this as your answer. Best of luck. — aliteralmind, Mar 02 '14 at 02:41

How do i add exception handling for the parts where the answer is numeric?

1 Answers1

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