Best implementation for an isNumber(string) method

Question

In my limited experience, I\'ve been on several projects that have had some sort of string utility class with methods to determine if a given string is a number. The idea h

Answer 1

I needed to refactor code like yours to get rid of NumberFormatException. The refactored Code:

public static Integer parseInteger(final String str) {
    if (str == null || str.isEmpty()) {
        return null;
    }
    final Scanner sc = new Scanner(str);
    return Integer.valueOf(sc.nextInt());
}

As a Java 1.4 guy, I didn't know about java.util.Scanner. I found this interesting article:

http://rosettacode.org/wiki/Determine_if_a_string_is_numeric#Java

I personaly liked the solution with the scanner, very compact and still readable.

Answer 2

public static boolean CheckString(String myString) {

char[] digits;

    digits = myString.toCharArray();
    for (char div : digits) {// for each element div of type char in the digits collection (digits is a collection containing div elements).
        try {
            Double.parseDouble(myString);
            System.out.println("All are numbers");
            return true;
        } catch (NumberFormatException e) {

            if (Character.isDigit(div)) {
                System.out.println("Not all are chars");

                return false;
            }
        }
    }

    System.out.println("All are chars");
    return true;
}

Answer 3

A modified version of my previous answer:

public static boolean isInteger(String in)
{
    if (in != null)
    {
        char c;
        int i = 0;
        int l = in.length();
        if (l > 0 && in.charAt(0) == '-')
        {
            i = 1;
        }
        if (l > i)
        {
            for (; i < l; i++)
            {
                c = in.charAt(i);
                if (c < '0' || c > '9')
                    return false;
            }
            return true;
        }
    }
    return false;
}

Answer 4

That's my implementation to check whether a string is made of digits:

public static boolean isNumeric(String string)
{
    if (string == null)
    {
        throw new NullPointerException("The string must not be null!");
    }
    final int len = string.length();
    if (len == 0)
    {
        return false;
    }
    for (int i = 0; i < len; ++i)
    {
        if (!Character.isDigit(string.charAt(i)))
        {
            return false;
        }
    }
    return true;
}

Answer 5

Here is our way of doing this:

public boolean isNumeric(String string) throws IllegalArgumentException
{
   boolean isnumeric = false;

   if (string != null && !string.equals(""))
   {
      isnumeric = true;
      char chars[] = string.toCharArray();

      for(int d = 0; d < chars.length; d++)
      {
         isnumeric &= Character.isDigit(chars[d]);

         if(!isnumeric)
         break;
      }
   }
   return isnumeric;
}

Answer 6

For long numbers use this: (JAVA)

public static boolean isNumber(String string) {
    try {
        Long.parseLong(string);
    } catch (Exception e) {
        return false;
    }
    return true;
}