Java How to loop until variable is equal to a specific type (String/int)

Question

Hi I'm currently developing a small booking program, I'm curios to know if i can use a Loop to run until a variable is equal to a specific type, notably a String or An Integer. For example i have the following below, and I want the user to always enter a number.

If not, am i right in presuming i should just develop a try/catch etc.?

Regards.

  public int SetHouseNo(){   
  System.out.println();        
  System.out.print("Please enter your house number:   ");    
  HouseNo=in.nextInt();
  return HouseNo; 
  }

`String` and `Integer` are not primitive types: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html — Bruno, Mar 06 '12 at 21:41
Thank you, pretty basic stuff so thanks for clearing that up. — Herten Ferford, Mar 06 '12 at 21:44
When you say "I want the user to always enter a number" is it (a) you want the user to enter the same specific number every time or (b) you want to make sure the user enters a number and not some other string? — Bruno, Mar 06 '12 at 21:50
The user needs to always enter a number(Integer), and not say 'blah'. — Herten Ferford, Mar 06 '12 at 21:51

score 4 · Accepted Answer · edited May 23 '17 at 12:21

4

How to validate that the user entered an integer, without throwing or catching an exception:

Scanner sc = new Scanner(System.in);

while (!sc.hasNextInt()) { // <-- 'peeks' at, doesn't remove, the next token
    System.out.println("Please enter a number!");
    sc.next(); // <-- skips over an invalid token
}

return sc.nextInt();

edited May 23 '17 at 12:21

Community

1
1

answered Mar 06 '12 at 22:00

calebds

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Thanks for that and the links, managed to adapt it to my code so now working perfectly, a really good way of doing it , need to wrap my head round precisely what's going on but works perfectly for what I'm doing. – Herten Ferford Mar 07 '12 at 12:14

score 3 · Answer 2 · answered Mar 06 '12 at 21:42

3

Try/catch is overkill. This is a job for do/while.

int house;
do {
  house = in.nextInt();
while (house isn't right for whatever reason);

answered Mar 06 '12 at 21:42

Louis Wasserman

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There will be an uncaught `InputMismatchException` if the token is not an `int`. – calebds Mar 06 '12 at 21:56
Ah. That _would_ justify a `try/catch`, but it sounded like the OP's issue was with the loop. – Louis Wasserman Mar 06 '12 at 21:58

Martijn Courteaux · Answer 3 · 2012-03-07T08:32:17.777

3

Use this simple loop:

public static int askForInteger(Scanner in, String msg)
{
     while (true)
     {
         System.out.print(msg + " ");
         String input = in.next();
         if (input.matches("\\-?\d"))
             return Integer.parseInt(input);

         System.out.println("Sorry, you didn't enter a valid number.");
     }
}

edited Mar 07 '12 at 08:32

answered Mar 06 '12 at 21:44

Martijn Courteaux

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@MartijnCourteaux `parseInteger` isn't a method either. It's `Integer.parseInt()` – calebds Mar 06 '12 at 22:12
@paislee: Omg, I was really confused. Thanks, again. – Martijn Courteaux Mar 07 '12 at 08:32

Juvanis · Answer 4 · 2012-03-06T22:00:09.463

1

Change 6 to your "specific" variable (Assuming HouseNo is a public variable and initialized toa value different than 6 before entering the method.):

public int SetHouseNo()
{   
  System.out.println();
  while(HouseNo != 6) 
  {   
    System.out.print("Please enter your house number:   "); 
    try
    {  
       HouseNo=in.nextInt();
    }
    catch(Exception e) { }
  } 
  return HouseNo; 
}

edited Mar 06 '12 at 22:00

answered Mar 06 '12 at 21:43

Juvanis

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@HertenFerford time lag? what is it? – Juvanis Mar 06 '12 at 21:49
@HertenFerford hmm ok. i hope my answer was helpful. – Juvanis Mar 06 '12 at 21:52
Following the latest clarification in the question's comment, the question is more about checking a valid number (and not some other text) was entered, not that a specific number was entered. – Bruno Mar 06 '12 at 21:54
There will be an uncaught InputMismatchException if the token is not an int – calebds Mar 06 '12 at 21:57
@paislee method returns houseNo, so it has to be defined as integer. when it comes to user input, you are right and i have added handling all exceptions. – Juvanis Mar 06 '12 at 22:01
@HertenFerford i have edited my code so that it can handle exceptions, for example user may enter a string while you are excpeting an integer. if the 10 minutes is done tick accept please :) – Juvanis Mar 06 '12 at 22:07

score 1 · Answer 5 · answered Mar 06 '12 at 21:47

1

You'll need to catch InputMismatchException from your reader if you want to allow numbers only. You could use a boolean to indicate whether you've found a suitable number or use auto boxing/unboxing and use a null value when no suitable value has been found:

Integer houseNumber = null;
do {
    try {
        System.out.print("Please enter your house number:");
        houseNumber = in.nextInt();
    } catch (InputMismatchException e) {
        houseNumber = null; // Not actually necessary, but perhaps clearer
    }
} while (houseNumber == null);

answered Mar 06 '12 at 21:47

Bruno

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Thanks for the help will give it a shot. – Herten Ferford Mar 06 '12 at 21:57
@HertenFerford you don't _have_ to catch the exception, there is a standard approach for this. See my answer – calebds Mar 06 '12 at 22:09
@HertenFerford: I'd agree with this, paislee's and MartijnCourteaux's answers sound better than mine. – Bruno Mar 06 '12 at 22:11

score 1 · Answer 6 · answered Mar 07 '12 at 21:56

Some people aren't agreeing with this, but here's what I would do:

public int SetHouseNo(){   
    System.out.println();        
    System.out.print("Please enter your house number:   ");    

    try
    {
        HouseNo=in.nextInt();
    }
    catch(NumberFormatException nfe)
    {
        System.out.println("Please enter a number...");
    }
    return HouseNo; 
}

And then somehow, possibly in the place where you call the method, make a while loop or something so it keeps asking until you entered a number. If you want, of course...

Java How to loop until variable is equal to a specific type (String/int)

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