How to check if a String is numeric in Java

This is generally done with a simple user-defined function (i.e. Roll-your-own "isNumeric" function).

Something like:

public static boolean isNumeric(String str) { 
  try {  
    Double.parseDouble(str);  
    return true;
  } catch(NumberFormatException e){  
    return false;  
  }  
}

However, if you're calling this function a lot, and you expect many of the checks to fail due to not being a number then performance of this mechanism will not be great, since you're relying upon exceptions being thrown for each failure, which is a fairly expensive operation.

An alternative approach may be to use a regular expression to check for validity of being a number:

public static boolean isNumeric(String str) {
  return str.matches("-?\\d+(\\.\\d+)?");  //match a number with optional '-' and decimal.
}

Be careful with the above RegEx mechanism, though, as it will fail if you're using non-Arabic digits (i.e. numerals other than 0 through to 9). This is because the "\d" part of the RegEx will only match [0-9] and effectively isn't internationally numerically aware. (Thanks to OregonGhost for pointing this out!)

Or even another alternative is to use Java's built-in java.text.NumberFormat object to see if, after parsing the string the parser position is at the end of the string. If it is, we can assume the entire string is numeric:

public static boolean isNumeric(String str) {
  ParsePosition pos = new ParsePosition(0);
  NumberFormat.getInstance().parse(str, pos);
  return str.length() == pos.getIndex();
}

if you are on android, then you should use:

android.text.TextUtils.isDigitsOnly(CharSequence str)

documentation can be found here

keep it simple. mostly everybody can "re-program" (the same thing).

We can try replacing all the numbers from the given string with ("") ie blank space and if after that the length of the string is zero then we can say that given string contains only numbers. Example:

boolean isNumber(String str){
        if(str.length() == 0)
            return false; //To check if string is empty
        
        if(str.charAt(0) == '-')
            str = str.replaceFirst("-","");// for handling -ve numbers
    
        System.out.println(str);
        
        str = str.replaceFirst("\\.",""); //to check if it contains more than one decimal points
        
        if(str.length() == 0)
            return false; // to check if it is empty string after removing -ve sign and decimal point
        System.out.println(str);
        
        return str.replaceAll("[0-9]","").length() == 0;
    }

You can use NumberFormat#parse:

try
{
     NumberFormat.getInstance().parse(value);
}
catch(ParseException e)
{
    // Not a number.
}

I have illustrated some conditions to check numbers and decimals without using any API,

Check Fix Length 1 digit number

Character.isDigit(char)

Check Fix Length number (Assume length is 6)

String number = "132452";
if(number.matches("([0-9]{6})"))
System.out.println("6 digits number identified");

Check Varying Length number between (Assume 4 to 6 length)

//  {n,m}  n <= length <= m
String number = "132452";
if(number.matches("([0-9]{4,6})"))
System.out.println("Number Identified between 4 to 6 length");

String number = "132";
if(!number.matches("([0-9]{4,6})"))
System.out.println("Number not in length range or different format");

Check Varying Length decimal number between (Assume 4 to 7 length)

//  It will not count the '.' (Period) in length
String decimal = "132.45";
if(decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Numbers Identified between 4 to 7");

String decimal = "1.12";
if(decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Numbers Identified between 4 to 7");

String decimal = "1234";
if(decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Numbers Identified between 4 to 7");

String decimal = "-10.123";
if(decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Numbers Identified between 4 to 7");

String decimal = "123..4";
if(!decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Decimal not in range or different format");

String decimal = "132";
if(!decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Decimal not in range or different format");

String decimal = "1.1";
if(!decimal.matches("(-?[0-9]+(\.)?[0-9]*){4,6}"))
System.out.println("Decimal not in range or different format");

Hope it will help manyone.

Based off of other answers I wrote my own and it doesn't use patterns or parsing with exception checking.

It checks for a maximum of one minus sign and checks for a maximum of one decimal point.

Here are some examples and their results:

"1", "-1", "-1.5" and "-1.556" return true

"1..5", "1A.5", "1.5D", "-" and "--1" return false

Note: If needed you can modify this to accept a Locale parameter and pass that into the DecimalFormatSymbols.getInstance() calls to use a specific Locale instead of the current one.

 public static boolean isNumeric(final String input) {
    //Check for null or blank string
    if(input == null || input.isBlank()) return false;

    //Retrieve the minus sign and decimal separator characters from the current Locale
    final var localeMinusSign = DecimalFormatSymbols.getInstance().getMinusSign();
    final var localeDecimalSeparator = DecimalFormatSymbols.getInstance().getDecimalSeparator();

    //Check if first character is a minus sign
    final var isNegative = input.charAt(0) == localeMinusSign;
    //Check if string is not just a minus sign
    if (isNegative && input.length() == 1) return false;

    var isDecimalSeparatorFound = false;

    //If the string has a minus sign ignore the first character
    final var startCharIndex = isNegative ? 1 : 0;

    //Check if each character is a number or a decimal separator
    //and make sure string only has a maximum of one decimal separator
    for (var i = startCharIndex; i < input.length(); i++) {
        if(!Character.isDigit(input.charAt(i))) {
            if(input.charAt(i) == localeDecimalSeparator && !isDecimalSeparatorFound) {
                isDecimalSeparatorFound = true;
            } else return false;
        }
    }
    return true;
}

Parse it (i.e. with Integer#parseInt ) and simply catch the exception. =)

To clarify: The parseInt function checks if it can parse the number in any case (obviously) and if you want to parse it anyway, you are not going to take any performance hit by actually doing the parsing.

If you would not want to parse it (or parse it very, very rarely) you might wish to do it differently of course.

That's why I like the Try* approach in .NET. In addition to the traditional Parse method that's like the Java one, you also have a TryParse method. I'm not good in Java syntax (out parameters?), so please treat the following as some kind of pseudo-code. It should make the concept clear though.

boolean parseInteger(String s, out int number)
{
    try {
        number = Integer.parseInt(myString);
        return true;
    } catch(NumberFormatException e) {
        return false;
    }
}

Usage:

int num;
if (parseInteger("23", out num)) {
    // Do something with num.
}

I modified CraigTP's solution to accept scientific notation and both dot and comma as decimal separators as well

^-?\d+([,\.]\d+)?([eE]-?\d+)?$

example

var re = new RegExp("^-?\d+([,\.]\d+)?([eE]-?\d+)?$");
re.test("-6546"); // true
re.test("-6546355e-4456"); // true
re.test("-6546.355e-4456"); // true, though debatable
re.test("-6546.35.5e-4456"); // false
re.test("-6546.35.5e-4456.6"); // false

I think the only way to reliably tell if a string is a number, is to parse it. So I would just parse it, and if it's a number, you get the number in an int for free!

You could use BigDecimal if the string may contain decimals:

try {
    new java.math.BigInteger(testString);
} catch(NumberFormatException e) {
    throw new RuntimeException("Not a valid number");
}

Parallel checking for very long strings using IntStream

In Java 8, the following tests if all characters of the given string are within '0' to '9'. Mind that the empty string is accepted:

string.chars().unordered().parallel().allMatch( i -> '0' <= i && '9' >= i )

If you guys using the following method to check:

public static boolean isNumeric(String str) {
    NumberFormat formatter = NumberFormat.getInstance();
    ParsePosition pos = new ParsePosition(0);
    formatter.parse(str, pos);
    return str.length() == pos.getIndex();
}

Then what happend with the input of very long String, such as I call this method:

System.out.println(isNumeric("94328948243242352525243242524243425452342343948923"));

The result is "true", also it is a too-large-size number! The same thing will happen if you using regex to check! So I'd rather using the "parsing" method to check, like this:

public static boolean isNumeric(String str) {
    try {
        int number = Integer.parseInt(str);
        return true;
    } catch (Exception e) {
        return false;
    }
}

And the result is what I expected!

private static Pattern p = Pattern.compile("^[0-9]*$");

public static boolean isNumeric(String strNum) {
    if (strNum == null) {
        return false;
    }
    return p.matcher(strNum).find();
}

For non-negative number use this

public boolean isNonNegativeNumber(String str) {
  return str.matches("\\d+");
}

For any number use this

public boolean isNumber(String str) {
  return str.matches("-?\\d+");
}