I *think* I know what the problem is, but rather than post my comments
on the solution, I would suggest two things:

a) post what is known as a SSCCE http://mindprod.com/jgloss/sscce.html 
so that we can see your code in context thus allowing a complete picture.
b) check out what the API says about the String class.  Pay particular
attention to the way the API explains how to do comparisons.

http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html

Honestly, a better way to handle it is to check the user's input
before you start doing your logic.  Something like this.

// get user input as a String - call it variable "userInput"

try {
 int inputVar = Integer.intValue(userInput);

 if(intputVar == 1) {
    // do stuff
 }
 // other if statements
} catch(NumberFormatException e) {
 System.out.println("Please enter an integer value!");
}

I realize using an exception as a form of control isn't always the
best idea.  But, I wanted to be able to easily demonstrate one way to
handle user input.

I'm not entirely sure I understand your problem but I will try to help
none the less.

char.  You are then trying to see if userInput is a 1, a 2, etc.  I
think the simple solution would be to do this:

if(userInput == '1') {
  string //not sure what this means...
}

...and so on through your code.

Placing the single quotation marks around the 1 tells Java that it is
no longer an int, but instead a char.  This idea should have been
covered in the first few chapters of the book you used in your
previous class.

If I did just answer your quesiton, a possible follow-up question of
you might have is "Why didn't the compiler throw any errors when you
can't compare to primitive data types?"  The answer is that char's
have a sort of "corresponding" int value.

Try compiling the line

System.out.println('a' == 97);

It should print true.

Hopefully I didn't completely miss your question.