Few Java examples to show you how to check if a String is numeric.
1. Character.isDigit()
Convert a String into a char array and check it with Character.isDigit()
NumericExample.java
package com.mkyong;
public class NumericExample {
public static void main(String[] args) {
System.out.println(isNumeric("")); // false
System.out.println(isNumeric(" ")); // false
System.out.println(isNumeric(null)); // false
System.out.println(isNumeric("1,200")); // false
System.out.println(isNumeric("1")); // true
System.out.println(isNumeric("200")); // true
System.out.println(isNumeric("3000.00")); // false
}
public static boolean isNumeric(final String str) {
// null or empty
if (str == null || str.length() == 0) {
return false;
}
for (char c : str.toCharArray()) {
if (!Character.isDigit(c)) {
return false;
}
}
return true;
}
}
Output
false
false
false
false
true
true
false
2. Java 8
This is much simpler now.
public static boolean isNumeric(final String str) {
// null or empty
if (str == null || str.length() == 0) {
return false;
}
return str.chars().allMatch(Character::isDigit);
}
3. Apache Commons Lang
If Apache Commons Lang is present in the claspath, try NumberUtils.isDigits()
pom.xml
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-lang3</artifactId>
<version>3.9</version>
</dependency>
import org.apache.commons.lang3.math.NumberUtils;
public static boolean isNumeric(final String str) {
return NumberUtils.isDigits(str);
}
4. NumberFormatException
This solution is working, but not recommend, performance issue.
public static boolean isNumeric(final String str) {
if (str == null || str.length() == 0) {
return false;
}
try {
Integer.parseInt(str);
return true;
} catch (NumberFormatException e) {
return false;
}
}
References
About Author
Founder of Mkyong.com, love Java and open source stuff. Follow him on Twitter. If you like my tutorials, consider make a donation to these charities.
Only Integer.parseInt(str) works because in other cases, if there are too many digits, it will return true but Integer.parseInt() will still fail. Also only Integer.parseInt() will work with negative numbers.
{code}
public static int parseInt(String s, int radix)
throws NumberFormatException
{
/*
* WARNING: This method may be invoked early during VM initialization
* before IntegerCache is initialized. Care must be taken to not use
* the valueOf method.
*/
if (s == null) {
throw new NumberFormatException(“null”);
}
if (radix Character.MAX_RADIX) {
throw new NumberFormatException(“radix ” + radix +
” greater than Character.MAX_RADIX”);
}
int result = 0;
boolean negative = false;
int i = 0, len = s.length();
int limit = -Integer.MAX_VALUE;
int multmin;
int digit;
if (len > 0) {
char firstChar = s.charAt(0);
if (firstChar < '0') { // Possible leading "+" or "-"
if (firstChar == '-') {
negative = true;
limit = Integer.MIN_VALUE;
} else if (firstChar != '+')
throw NumberFormatException.forInputString(s);
if (len == 1) // Cannot have lone "+" or "-"
throw NumberFormatException.forInputString(s);
i++;
}
multmin = limit / radix;
while (i < len) {
// Accumulating negatively avoids surprises near MAX_VALUE
digit = Character.digit(s.charAt(i++),radix);
if (digit < 0) {
throw NumberFormatException.forInputString(s);
}
if (result < multmin) {
throw NumberFormatException.forInputString(s);
}
result *= radix;
if (result < limit + digit) {
throw NumberFormatException.forInputString(s);
}
result -= digit;
}
} else {
throw NumberFormatException.forInputString(s);
}
return negative ? result : -result;
}
{code}
I didn’t know about Character.isDigit(). I’d normally uses a regex and String.matches (e.g. string.matches(“[0-9]”) or string.matches(“\d”)).
I would like to tweak the isNumeric method to trim the string before processing. The existing code will return false for a string “123456 “.
public static boolean isNumeric(final String str) {
// null or empty
if (str == null || str.length() == 0) {
return false;
}
//Remove leading and trailing spaces from a string
str = str.trim();
for (char c : str.toCharArray()) {
if (!Character.isDigit(c)) {
return false;
}
}
return true;
}
Using regex is faster than Charachter::isDigit. Besides, regex can also return true for floating point numbers like “3000.00”, as well as negative numeric values.
public class NumericExample {
public static void main(String[] args) {
long time1, time2;
time1 = System.currentTimeMillis();
System.out.println(“IsNumeric()”);
System.out.println(isNumeric(“”)); // false
System.out.println(isNumeric(” “)); // false
System.out.println(isNumeric(null)); // false
System.out.println(isNumeric(“1,200”)); // false
System.out.println(isNumeric(“1”)); // true
System.out.println(isNumeric(“200”)); // true
System.out.println(isNumeric(“3000.00”)); // false
time2 = System.currentTimeMillis();
System.out.println(“time: ” + (time2 – time1));
time1 = System.currentTimeMillis();
System.out.println(“\nIsNumericRegex()”);
System.out.println(isNumericRegex(“”)); // false
System.out.println(isNumericRegex(” “)); // false
System.out.println(isNumericRegex(null)); // false
System.out.println(isNumericRegex(“1,200”)); // false
System.out.println(isNumericRegex(“1”)); // true
System.out.println(isNumericRegex(“200”)); // true
System.out.println(isNumericRegex(“3000.00”)); // true
time2 = System.currentTimeMillis();
System.out.println(“time: ” + (time2 – time1));
}
public static boolean isNumeric(final String value) {
if (value == null || value.isEmpty()) {
return false;
}
return value.chars().allMatch(Character::isDigit);
}
public static boolean isNumericRegex(final String value) {
if (value == null || value.isEmpty()) {
return false;
}
return value.matches(“[-+\\d\\.]*”);
}
}