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

Question

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?

Java convention would be to name integerstate integerState. Also, you need to save off the results of your trim to a temporary variable. trim() does not change the calling object, it just returns a changed object. — AHungerArtist
– AHungerArtist, Commented Mar 18, 2012 at 4:09
@Petefic - Q: Do you want to check for a string that contains a valid integer number ... or an arbitrarily long string that contains only numbers? If the former, use "parseInt()" (as I suggested below). If the latter, then use a regex, as Amit Bhargava. IMHO ... — paulsm4
– paulsm4, Commented Mar 18, 2012 at 5:05

Ernest Friedman-Hill · Accepted Answer · 2012-03-18 04:15:54Z

5

@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.

edited Mar 18, 2012 at 4:15

answered Mar 18, 2012 at 4:10

Ernest Friedman-Hill

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2 Comments

Makoto Over a year ago

This is not only a more readable (and efficient) way to perform this test, but it also circumvents the issue of the boolean being overwritten before you return the value.

Kazekage Gaara Over a year ago

+1 about mentioning the extra computational overhead involved in using Integer.parseInt()(due to throwing and catching exceptions). I learned a new thing! :-)

Community · Accepted Answer · 2017-05-23 11:56:25Z

4

Use "0".equals(input.substring(count) ) instead of ==

== is checking for object equality and not string equality

See this question for how to do it the way @Ernest suggests.

edited May 23, 2017 at 11:56

CommunityBot

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answered Mar 18, 2012 at 4:07

Kal

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3 Comments

Kal Over a year ago

@Peterfic -- make sure there's no trailing spaces. input = input.trim();

paulsm4 Over a year ago

The basic algorithm is so inefficient that suggesting "substring.equals("0") (in and of itself a legitimate observation) is just putting lipstick on a pig. IMHO...

Petefic Over a year ago

I was having the same problem with your method but the link you included worked. Thanks.

toto233 · Accepted Answer · 2012-03-18 04:37:21Z

2

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

answered Mar 18, 2012 at 4:37

toto233

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Comments

Chetter Hummin · Accepted Answer · 2012-03-18 04:11:18Z

1

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.

answered Mar 18, 2012 at 4:11

Chetter Hummin

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Comments

Haz · Accepted Answer · 2012-03-18 04:13:49Z

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().

Andrew Thompson · Accepted Answer · 2012-03-18 04:16:31Z

1

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 . . .

answered Mar 18, 2012 at 4:16

Andrew Thompson

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Comments

Kazekage Gaara · Accepted Answer · 2012-03-18 04:10:43Z

0

You should use the equals method for String equality.

And if you want to use it the other way around,consider converting your substrings using Integer.parseInt() and then you can use == 0.

answered Mar 18, 2012 at 4:10

Kazekage Gaara

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Comments

Naphstor · Accepted Answer · 2012-03-18 04:12:41Z

0

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.

answered Mar 18, 2012 at 4:12

Naphstor

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Comments

paulsm4 · Accepted Answer · 2012-03-18 04:13:10Z

0

// 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 ...

answered Mar 18, 2012 at 4:13

paulsm4

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3 Comments

Kal Over a year ago

The input string could be much longer than an int value .. so Integer.parseInt() will not always work.

Ernest Friedman-Hill Over a year ago

@Kal, The idea is that if it's not an int, you get an exception; it doesn't matter what other junk there is. If there's junk, it's not an int, and that's all you need to know.

Kal Over a year ago

@Ernest -- I understand. But if you read the question, it says "check if a string contains numbers". A very very long string that contains numbers will throw an exception on Integer.parseInt() .. Thats what I was referring to.

kundan bora · Accepted Answer · 2012-03-18 05:17:25Z

0

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.

answered Mar 18, 2012 at 5:17

kundan bora

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