Thread: User selection problem

Hey,

So this is my code for my getChoice ( ) method.

private int getChoice ( int lower, int upper ) {
 
		System.out.println ( "--------------------------------------------------------------" );
		System.out.print   ( "Please enter your choice: ");
		String stringSelection = Keyboard.readInput ( ); 
	        System.out.println ( "--------------------------------------------------------------" );
	            if ( stringSelection.length ( ) == 0 ) {
	    	       return INVALID;
	             } else {
	    		    int selection = Integer.parseInt ( stringSelection );
	    		    if ( selection >= lower && selection <= upper ) { 
	    			return selection;
	    		    } else {	 
	    			return INVALID;
	    		}
	    } 
	}

It's meant to take an input, and then if it's not between or equal to lower and upper (in this case the method is being used for lower int = 1, and upper = 4)
then it will return my constant INVALID variable. My Keyboard.readInput() method returns a String as you can see.

At the moment I have it set up so that if the input is empty (now that I think of it I could have just said null instead of .length() ) then it will return invalid.
And if the INTEGER is not between lower and upper, then it will return invalid. But say if I was to type "a22asdf" or something else random, then it will give me an error.
So my question, is how can I make it return INVALID if anything entered as a choice is not between or equal to lower and upper.
Probably a very silly thing that I will understand as soon as someone helps me out. One of those things

(I was thinking of an array and a for loop that cycled through int i)

Thank you very much and regards,
Max

Originally Posted by andreas90

try {
        //code that may throw the exception, the parsing method
      } catch (ExceptionThatMayBeThrown e) {
                e.getStackTrace();             
                return INVALID;
      }

To understand which return statement is executed (since INVALID will be returned in multiple cases) I would also suggest you print a message before you return anything .

Getting there I think. This is my code now (I'm confused about the "exceptionthatmaybethrown")

private int getChoice ( int lower, int upper ) {
 
		System.out.println ( "--------------------------------------------------------------" );
		System.out.print   ( "Please enter your choice: ");
		String stringSelection = Keyboard.readInput ( ); 
	    System.out.println ( "--------------------------------------------------------------" );
	    if ( stringSelection.length ( ) == 0 ) {
	    	return INVALID;
	    } else {
	    		int selection = Integer.parseInt ( stringSelection );
	    		try {
	    			if ( selection >= lower && selection <= upper ) { 
		    			return selection;
		    		}
	    		} catch ( java.util.InputMismatchException e ) {
	    			e.getStackTrace ( );
	    			return INVALID;
	    		}
	    }

Now I am getting an error that this needs to return something, which it does, so somehow the returns are not being reached.

Sorry, this is the first time I have attempted error handling in Java. (other than a few simple if statements in previous works.)

Regards,
Max

EDIT: Updated code (I think I am getting closer) (but eclipse is still saying I need a return statement)

private int getChoice ( int lower, int upper ) {
 
		System.out.println ( "--------------------------------------------------------------" );
		System.out.print   ( "Please enter your choice: ");
		String stringSelection = Keyboard.readInput ( ); 
	    System.out.println ( "--------------------------------------------------------------" );
 
	    try {
	    	int selection = Integer.parseInt ( stringSelection );
 
	    	if ( selection >= lower && selection <= upper ) {
	    		return selection;
	    	}
	    } catch ( java.util.InputMismatchException e ) {
	    	return INVALID;
	    }
	}

Originally Posted by andreas90

"exceptionthatmaybethrown" was just fordemonstration. You should catch the exception that actually may be thrown which in your case is the NumberFormatException.


The try block should contain the statement that may throw the exception. In your code the exception will may be thrown from this statement

int selection = Integer.parseInt ( stringSelection );

For example if the user enters a letter the method won't parse it to an int since it doesn't have the appropriate format (see the details in the above link). The inner if...else block won't throw an exception so it doesn't need to be indide the try block.
You should read the java tutorials for exceptions handling.

If I was to not include the if statement in the try statement how would I check if the integers were between the lower and upper local variables?
It still requires a return type apparently.

I really appreciate your help, thanks,
Max

EDIT: I GOT IT! I always feel so good once a problem that I've been working on for a long time is solved.

Here is my updated code if anyone finds it useful:

	private int getChoice ( int lower, int upper ) {
 
		System.out.println ( "--------------------------------------------------------------" );
		System.out.print   ( "Please enter your choice: ");
		String stringSelection = Keyboard.readInput ( ); 
	    System.out.println ( "--------------------------------------------------------------" );
	    try {
	    	int selection = Integer.parseInt ( stringSelection );
		    if ( selection >= lower && selection <= upper ) {
	    		return selection;
	    	} else { 
	    		return INVALID;
	    	}
	    } catch ( NumberFormatException e ) {
	    	e.getStackTrace();
	    	return INVALID;
	    }	    	
 
	}

Thank you very much for your help

Regards and out,

Max