In this post, we will explore several ways to determine if a String is a valid number or not. The solution checks if a String contains only Unicode digits and doesn’t allow either positive or negative leading sign or a decimal point. A null string or an empty String should return false.
Java did not provide any standard function for this simple task. Nevertheless, there are several methods to check if the given String is numeric in Java –
1. Custom method
Simple solution is to write our own utility method for this simple task. The idea is to iterate over characters of a String and check each character to be numeric using Character.isDigit(char). This is demonstrated below:
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class IsNumeric { public static boolean isNumeric(String s) { if (s == null || s.equals("")) { return false; } for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c < '0' || c > '9') { return false; } } return true; } public static void main(String[] args) { String s = "100"; System.out.println("IsNumeric: " + isNumeric(s)); } } |
Output:
IsNumeric: true
2. Lambda expressions:
From Java 8 onwards, this can be efficiently done using lambda expressions:
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class IsNumeric { public static boolean isNumeric(String s) { if (s == null || s.equals("")) { return false; } return s.chars().allMatch(Character::isDigit); } public static void main(String[] args) { String s = "100"; System.out.println("IsNumeric: " + isNumeric(s)); } } |
Output:
IsNumeric: true
3. Regular Expression
We can use the regular expression "[0-9]+" or "\\d+" which checks if a String is numeric. This can be done using matches() method of String class which tells whether or not this string matches the given regular expression.
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class IsNumeric { public static void main(String[] args) { String s = "100"; boolean isNumeric = (s != null && s.matches("[0-9]+")); System.out.println("IsNumeric: " + isNumeric); } } |
Output:
IsNumeric: true
If the regular expression is frequently called, we should compile it first for performance boost:
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import java.util.regex.Pattern; class IsNumeric { private static Pattern p = Pattern.compile("[0-9]+"); public static void main(String[] args) { String s = "100"; boolean isNumeric = (s != null && p.matcher(s).find()); System.out.println("IsNumeric: " + isNumeric); } } |
Output:
IsNumeric: true
4. Apache Commons Lang:
We can also use Apache Commons Lang library which provides several methods to check for a numeric string –
1. StringUtils#isNumeric()
2. NumberUtils#isCreatable()
3. NumberUtils#isNumber()
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import org.apache.commons.lang3.StringUtils; class IsNumeric { public static void main(String[] args) { String s = "100"; System.out.println("IsNumeric: " + StringUtils.isNumeric(s)); } } |
5. Using Wrapper classes built-in methods:
The idea is to parse a String by Integer.parseInt() or Long.parseLong(), which generates a NumberFormatException for non-numeric strings. Notice that this method fails when the value is outside the range for int or long respectively.
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class IsNumeric { public static boolean isNumeric(String s) { try { Integer.parseInt(s); } catch (NumberFormatException ex) { return false; } return true; } public static void main(String[] args) { String s = "100"; System.out.println("IsNumeric: " + isNumeric(s)); } } |
Output:
IsNumeric: true
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