Java: Getting the Value of the User Input that Threw an Exception

Question

I'm trying to catch an InputMismatchException with the Scanner expecting a double, and if the input is equal to "c" or "q" I want to do run some specific code in the catch block. I'm wondering if there's a way to get the value of the user input after an exception has been thrown. So for example, if the code calls for a double and the user enters the letter "c", I want to be able to know that specifically "c" was input, and do something if true. Here's the code I'm trying to write, where "getValue()" is a made-up method name to describe what I'm trying to accomplish:

double value = 0.0;

try{
    System.out.print("\nEnter a number: ");
    value = input.nextDouble();
}
catch(InputMismatchException notADouble){
    if(notADouble.getValue().equalsIgnoreCase("c")){
        //run the specific code for "c"
    }
    else if(notADouble.getValue().equalsIgnoreCase("q")){
        //run the specific code for "q"
    }
    else{
        System.out.print("\nInvalid Input");
    }
}

Thanks in advance for your input :)

Answer 1

Use Scanner.hasNextDouble() to validate the input before Scanner tries to convert it to a number. Something like this should work:

  double value = 0.0;

  Scanner input = new Scanner(System.in);

     System.out.print("\nEnter a number: ");
     while(input.hasNext())
     {
        while(input.hasNextDouble())
        {
           value = input.nextDouble();
        }

        String next = input.next();

        if("c".equals(next))
        {
           //do something
        }
        else if("q".equals(next))
        {
           //do something
        }
        else
        {
           System.out.print("\nInvalid Input");
           //return or throw exception

        }
     }