Reputation: 677

JAVA: Trying to write a method that checks if a string is numeric. Always returning false

I'm trying to write a method that checks if a string has only numbers in it. For some reason it returns false even if I input a string containing "1234". I'm guessing my problem is with the if statement, but I'm not sure what to do to fix it.

public static boolean isNumeric(String input)
{
    input.trim();

    for (int count=0; count<=input.length(); count++)
    {

        if (input.substring(count) == "0" || input.substring(count) == "1"||
        input.substring(count) == "2" || input.substring(count) == "3" ||
        input.substring(count) == "4" || input.substring(count) == "5" ||
        input.substring(count) == "6" || input.substring(count) == "7" ||
        input.substring(count) == "8" || input.substring(count) == "9")
        {
            integerstate = true;
        }

        else
        {
            integerstate = false;
            break;
        }

    }

return integerstate;
}

Does anyone see what the problem is?

Upvotes: 0

Answers (10)

kundan bora

Reputation: 3889

The cause behind this is difference between == and equals().

This happen because == compare for exact equality means two object must be same. while equal() compare for meaningful equality means object may not me same but the value they contain are meaningfully same.

Upvotes: 0

toto233

Reputation: 21

public static boolean isNumeric(String input) {
    Pattern pattern = Pattern.compile("[0-9]*");
    Matcher isNum = pattern.matcher(input);
    if (isNum.matches()) {
        return true;
    }
    return false;
}

Upvotes: 2

Andrew Thompson

Reputation: 168835

You might let the tool-kit do the 'heavy lifting'.

class TestForIntegers {
    public static void main(String[] args) {
        String s = "blah";
        System.out.println(s + " is integer: " + isInteger(s));
        s = "1234";
        System.out.println(s + " is integer: " + isInteger(s));
        s = "max123";
        System.out.println(s + " is integer: " + isInteger(s));
    }

    public static boolean isInteger(String s) {
        try {
            Integer.parseInt(s);
            return true;
        } catch (Exception e) {
            return false;
        }
    }
}

Output

blah is integer: false
1234 is integer: true
max123 is integer: false
Press any key to continue . . .

Upvotes: 1

Ernest Friedman-Hill

Reputation: 81724

@Kal's answer is literally correct, but my goodness, @Petefic, this is a horribly inefficient way to do things. There's a method in the Character class that checks if a char is a digit, and you could just extract each character once instead of calling substring() so many times.

For instance, you might do

 public static boolean isNumeric(String input) {
     input = input.trim();
     for (char c: input.toCharArray()) {
         if (Character.isDigit(c))
           return true;
     }
     return false;
 }

For the number 99999, your code would produce over 50 new objects; this would produce one (the array from toCharArray().)

I see a few other answers that suggest using Integer.parseInt() and catching the exception. That's not a terrible idea, but it's actually not a great one if many of the strings will not be numbers. Throwing and catching an exception is very computationally expensive; many objects are created in the process. It should be avoided in cases where actual error recovery is not involved.

Upvotes: 5

Haz

Reputation: 2679

Like Kal said, the == operator is determining whether the objects are the same object, not if they have the same value. With Java, this will often get you the same result, which can make it more difficult to realize why using == to compare strings is wrong. It will sometimes work because Java interns Strings with a String pool, which means Java will generally only store one String of each value. In other words, if String x and String y both have the value of "asdf", the String pool will only store "asdf" once and both x and y will reference that String object. The only way to force Java to actually create a duplicate String object is with the "new" keyword. I think in this case, == is not working due to the substring. In any case, unless you're actually checking to see if the objects being referenced are the same, always use .equals().

Upvotes: 1

paulsm4

Reputation: 121799

// Bad
for (int count=0; count<=input.length(); count++)
{
  if (input.substring(count) == "0" || input.substring(count) == "1"||
      input.substring(count) == "2" || input.substring(count) == "3" ||
      input.substring(count) == "4" || input.substring(count) == "5" ||
      input.substring(count) == "6" || input.substring(count) == "7" ||
      input.substring(count) == "8" || input.substring(count) == "9")
  {
      integerstate = true;
      ...

// Better
for (int count=0; count<=input.length(); count++)
{
   if (input.substring(count).equals("0") || ...
      ...

// Much better
try {
  Integer.parseInt( input );  
  integerstate = true;
}  
catch ...

Upvotes: 0

Naphstor

Reputation: 2496

In java to check for equality of two strings or comparing them you need to use the equals() since equals() does the content comparison. But '==' does the reference comparison so it just checks if the two references on both side of '==' are pointing to same object or not.

Upvotes: 0

Chetter Hummin

Reputation: 6817

Another option would be to use charAt() instead of substring and compare against '0', '1' etc. Then you can use ==. A far more efficient way however would be to use regular expressions.

Upvotes: 1