How to check if a String is numeric in Java

Question

How would you check if a String was a number before parsing it?

Organist

All the solutions proposed with regular expresions will not work for hexadecimal numbers. – Astrometry

and passing null string in matches(...) function will throw NullPointer exception. – Charmer

See Max Malysh's answer for a concise Java 8 solution without third-party libraries. – Mar

@HiteshSahu null strings seem to be gracefully handled in latest version (including Java 6.x and 7.x) – Amends

All the solutions proposed to use Integer.parseInt() will fail to parse mobile numbers with NumberFormatException. – Etsukoetta

@OscarCastiblanco All strings are numbers in a given base. – Gybe

Answer 1

With Apache Commons Lang 3.5 and above: NumberUtils.isCreatable or StringUtils.isNumeric.

With Apache Commons Lang 3.4 and below: NumberUtils.isNumber or StringUtils.isNumeric.

You can also use StringUtils.isNumericSpace which returns true for empty strings and ignores internal spaces in the string. Another way is to use NumberUtils.isParsable which basically checks the number is parsable according to Java. (The linked javadocs contain detailed examples for each method.)

Shepard

StringUtils.isNumeric() probably wouldn't be appropriate here since it only checks if the string is a sequence of digits. Would be fine for most ints but not so for numbers with decimals, group separators, etc. – Hectic

Pretty old question but just for anyone who is reading for help - StringUtils.isNumeric won't work also if it is empty String. It will return true for empty String for StringUtils.isNumeric(). – Osric

@JUG: thx for the heads up. Strange enough, the doc linked says the opposite, maybe because of a newer version (note that the answer was edited just a few days after your comment was posted) – Chronometry

@JUG: searched into it & that's what I thought, some previous versions ie. Commons Lang version 2.4 stringutils.isNumeric() behaves as you said (return true for empty String) – Chronometry

reinvent the wheel because you don't include a whole library because you need a 3 line function in one place. – Busty

Is it really worth adding a whole library for this feature though? Obviously if it's used with other stuff that's great, but it's probably overkill considering people have solved this in one line of code. – Merlenemerlin

Why does isNumeric return true for question marks? commons.apache.org/proper/commons-lang/apidocs/org/apache/… – Moreau

@pete: Those question marks seem to be Devanagari digits (which probably are not supported by our fonts/browser). See: issues.apache.org/jira/browse/LANG-1017 – Shepard

Doesn't work with negatives. And half of all numbers are negative, so..... – Potiche

@PaulDraper: You're right, StringUtils does not support leading signs but you should check NumberUtils.isCreatable, it supports negatives properly. – Shepard

Adding a dependency to check if a string is a number... Wow – Friday

NumberUtils.isParsable(String str) method could be used to check if the given string is numeric, which would work for all the scenarios. – Galan

Here is all the ways: baeldung.com/java-check-string-number – Gaskell

In this case I would add a library dependency even though not enough programmers scrutinize doing so (if you have to support exponents). For those who really don't want to, you could copy and paste the source directly: github.com/apache/commons-lang/blob/master/src/main/java/org/… – Jetsam

Answer 2

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();
}

Felisha

Does \d in Java Regex match only latin digits? If it's like .NET regexes, you'll run into a problem with other (e.g. arabic) digits, as explained here: blogs.msdn.com/oldnewthing/archive/2004/03/09/86555.aspx – Dialytic

@Dialytic - Oops.. Good catch. You're right. \d will only match latin digits (i.e. 0 through to 9 effectively) and fail with arabic ones. I've edited my posted to make this clear. Thanks! – Felisha

the numberFormatter solution is probably only marginally better than catching the NumberFormatException one. I suspect the best way is to use regex one. – Rejuvenate

Note that the . in your regex will match any character, not just the decimal separator character. – Foretaste

+1 for realizing the expense of try/catch. This is actually a horrible approach to use in the long run for repeated use, but really we are stuck with that in Java. – Abstention

I would use the regex of (\\+|-)?\\d+(\\.\\d+)? as "+1.20" is a valid numeric value. – Warfarin

In the last proposition (NumberFormat), you are missing a null check at the parse method. Otherwise isNumeric("") will return true as position index will stay unchanged ! – Halfhardy

To handle exponent (+1.3E10): return str.matches("\\d") && str.matches("^[+-]?\\d*\\.?\\d*([eE]\\d)?\\d*$"); – Nashom

Keep in mind that: Double.parseDouble("NaN") <-- doesn't throw and Exception. – Geelong

also, don't forget .006 and 0.- those are valid too. – Brandiebrandise

Both solutions are bad if used extensibly. Throwing exception is costly and creating a regular expressions is costly as well. The regular expression must be created once and reused. – Countdown

@user1096901 I'm well aware of the consequences of using exceptions to validate data, and I even say why it's not necessarily the best approach within my answer. I then go on to offer alternatives thatare probably better options. All approaches are detailed to give as complete and balanced an answer as possible – Felisha

@fchen, right idea, but the 'E' in a scientific notation may also be followed by a '+' or '-'. – Mallarme

In this case the RegEx is 10 times slower than processing the exception. See for yourself with the code in my answer below. – Cleveland

Brilliant, even working for foreign numbers, like Gujarati (૧૦૦) , Asturian (২০১৯), Tibetan (৬৫), Bengali (႑), Hebrew (８), Japanese (１５), Khmer (២០១៥),,, let's say most – Sapid

what if the string is 2.2E8 – Almonry

What if the string uses a different decimal separator such as ,? – Apathy

@Eddnav The first approach (using Double.parseDouble) will not work for numbers with different decimal separators, however, the last approach shown in my answer (using the java.text.NumberFormat class) does work correctly for alternative decimal separators (at least it works for a comma). See here: ideone.com/vaOEae – Felisha

Note that there are no such things as "latin numerals", and the numerals 0-9 are in fact Arabic numerals. People are probably familar with Roman numerals, which were used by people who spoke Latin, in the form I, II, III, IV, V, VI, etc. en.wikipedia.org/wiki/Arabic_numerals; en.wikipedia.org/wiki/Roman_numerals – Upsurge

Ran this (the Double.parseDouble and regex) in comparison to converting string to char[] and running isDigit over it, and running the char[] vs. an array of acceptable chars. Comparison was done via running time with System.nanoTime(). They all came out to about the same with a 2500 character long string with either no non-digits, or at least 1 non digit. – Osmen

the first is numeric with: double d = Double.parseDouble(str); returns true if the string is like this "0F" :-) – Bruit

It is self0evidently self-evidently pointless to call Integer.parseInt() twice on the same input. The correct solution is to call it once, when needed, and catch and deal with the NumberFormatException as it arises. – Persimmon

This(your first approach) is exactly how I did it when creating a mathematical parser about 11 years ago. You can find the project at github.com/gbenroscience/ParserNG – Casilde

Answer 3

Java 8 lambda expressions.

String someString = "123123";
boolean isNumeric = someString.chars().allMatch( Character::isDigit );

Osis

You can use a method reference, too: someString.chars().allMatch(Character::isDigit) – Granary

Nice but still it's reinventing the wheel as almost all "solutions" here. Also, fails on 'null' (as almost all the others). – Kissee

This answer is concise, simple and readable. You can almost read it like English -- "chars all match digits". It requires no third-party libraries. It does not use exceptions in non-exceptional cases. This should become the accepted answer. – Mar

What will it produce for "-1"? – Rowboat

@AndyThomas for a 5135 character length string of numbers it takes 169,424,664 nanoseconds, or 169.424664 milliseconds (on my machine). This is way too inefficient. You can get 15 ms times or less running a nested for-loop to compare characters against a char[] of acceptable characters. – Osmen

@Adrian here is a benchmark: gist.github.com/maxmalysh/8f656be2843b7c35849a2eae2683df0e . 9ms for a 1 million character length string. 169ms for a 5k character length string is something way too off the mark. – Osis

@Adrian - In a universe of 10^80 atoms, there aren't that many useful 5135-character numbers. If candidate strings can be long, a length test could be applied beforehand. Optimization can be necessary. It can also be premature. Correctness may be a larger issue. It's worth noting that this answer (and some others) only detect positive integers. It will fail if required to detect negatives, reals, hexadecimal, scientific notation ... – Mar

Not the right answer. A numeric string can have non-numeric characters (ex. "." or "-") and still be perfectly numerical. For example 0.5, -1, and 1,000 will all fail with this answer and yet they are perfectly numerical. – Inhibit

It won't work for Negative Numbers, nor for decimal numbers. – Tuatara

Answer 4

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).

Enact

@kape123 :) sure "123.456" doesn´t contain digits. – Enact

Note: this results in NPE for null input. Also, doesn't work with negative numbers or decimals. – Anachronistic

I like it!! I think this is absolutely for digits. Not for ., - – Deibel

This is just what I was looking for. Something simple to check for only digits 0-9. I set a filter in the declaration of my EditText but just in case that get's changed or replaced down the road it's nice to have a simple programmatic check as well. – Quassia

Why this method returns true for empty strings? – Somniloquy

Answer 5

As @CraigTP had mentioned in his excellent answer, I also have similar performance concerns on using Exceptions to test whether the string is numerical or not. So I end up splitting the string and use java.lang.Character.isDigit().

public static boolean isNumeric(String str)
{
    for (char c : str.toCharArray())
    {
        if (!Character.isDigit(c)) return false;
    }
    return true;
}

According to the Javadoc, Character.isDigit(char) will correctly recognizes non-Latin digits. Performance-wise, I think a simple N number of comparisons where N is the number of characters in the string would be more computationally efficient than doing a regex matching.

UPDATE: As pointed by Jean-François Corbett in the comment, the above code would only validate positive integers, which covers the majority of my use case. Below is the updated code that correctly validates decimal numbers according to the default locale used in your system, with the assumption that decimal separator only occur once in the string.

public static boolean isStringNumeric( String str )
{
    DecimalFormatSymbols currentLocaleSymbols = DecimalFormatSymbols.getInstance();
    char localeMinusSign = currentLocaleSymbols.getMinusSign();

    if ( !Character.isDigit( str.charAt( 0 ) ) && str.charAt( 0 ) != localeMinusSign ) return false;

    boolean isDecimalSeparatorFound = false;
    char localeDecimalSeparator = currentLocaleSymbols.getDecimalSeparator();

    for ( char c : str.substring( 1 ).toCharArray() )
    {
        if ( !Character.isDigit( c ) )
        {
            if ( c == localeDecimalSeparator && !isDecimalSeparatorFound )
            {
                isDecimalSeparatorFound = true;
                continue;
            }
            return false;
        }
    }
    return true;
}

Chloe

Does -ve sign fail this function? – Fecal

I think this should be the accepted answer because it is the lightest solution. Using an exception or a regex are both really heavy to check if a string is numeric. Iterating over the characters is nice and simple! – Frink

The above code accepts a single '-' as numeric and will return true. change first if to something like: boolean isMinus = str.charAt(0) == localeMinusSign; if ((isMinus && str.length() < 2) || ((!isMinus) && !Character.isDigit(str.charAt(0)))) { return false; } – Bucher

Calling toCharArray() will create a copy of the array in the String object because Strings are immutable. Probably faster to use the charAt(int index) method on the String object directly. – Export

Will generate StringIndexOutOfBoundsException when passed a string with length 0. Can be fixed with if(str.length() == 0) return false; – Rutan

Answer 6

Google's Guava library provides a nice helper method to do this: Ints.tryParse. You use it like Integer.parseInt but it returns null rather than throw an Exception if the string does not parse to a valid integer. Note that it returns Integer, not int, so you have to convert/autobox it back to int.

Example:

String s1 = "22";
String s2 = "22.2";
Integer oInt1 = Ints.tryParse(s1);
Integer oInt2 = Ints.tryParse(s2);

int i1 = -1;
if (oInt1 != null) {
    i1 = oInt1.intValue();
}
int i2 = -1;
if (oInt2 != null) {
    i2 = oInt2.intValue();
}

System.out.println(i1);  // prints 22
System.out.println(i2);  // prints -1

However, as of the current release -- Guava r11 -- it is still marked @Beta.

I haven't benchmarked it. Looking at the source code there is some overhead from a lot of sanity checking but in the end they use Character.digit(string.charAt(idx)), similar, but slightly different from, the answer from @Ibrahim above. There is no exception handling overhead under the covers in their implementation.

Poon

Beware that this would throw NPE in case argument is null. – Keitloa

Answer 7

Do not use Exceptions to validate your values. Use Util libs instead like apache NumberUtils:

NumberUtils.isNumber(myStringValue);

Edit:

Please notice that, if your string starts with an 0, NumberUtils will interpret your value as hexadecimal.

NumberUtils.isNumber("07") //true
NumberUtils.isNumber("08") //false

Zephyr

The accepted answer, three years earlier, already covered Number.isNumber(). – Mar

I don't think so. It was updated or op changed the accepted answer. I remember the accepted answer didn't covered NumberUtils thats why I added my answer. But thanks for the comment – Zephyr

@Goot - The history of the accepted answer shows that Number.isNumber() was present from the first version of the answer, dated Sep 24 '12 at 17:01. – Mar

@Goot, this is pretty good as it also covers the decimal value check, unlike StringUtils. – Fabe

Answer 8

Why is everyone pushing for exception/regex solutions?

While I can understand most people are fine with using try/catch, if you want to do it frequently... it can be extremely taxing.

What I did here was take the regex, the parseNumber() methods, and the array searching method to see which was the most efficient. This time, I only looked at integer numbers.

public static boolean isNumericRegex(String str) {
    if (str == null)
        return false;
    return str.matches("-?\\d+");
}

public static boolean isNumericArray(String str) {
    if (str == null)
        return false;
    char[] data = str.toCharArray();
    if (data.length <= 0)
        return false;
    int index = 0;
    if (data[0] == '-' && data.length > 1)
        index = 1;
    for (; index < data.length; index++) {
        if (data[index] < '0' || data[index] > '9') // Character.isDigit() can go here too.
            return false;
    }
    return true;
}

public static boolean isNumericException(String str) {
    if (str == null)
        return false;
    try {  
        /* int i = */ Integer.parseInt(str);
    } catch (NumberFormatException nfe) {  
        return false;  
    }
    return true;
}

The results in speed I got were:

Done with: for (int i = 0; i < 10000000; i++)...

With only valid numbers ("59815833" and "-59815833"):
    Array numeric took 395.808192 ms [39.5808192 ns each]
    Regex took 2609.262595 ms [260.9262595 ns each]
    Exception numeric took 428.050207 ms [42.8050207 ns each]
    // Negative sign
    Array numeric took 355.788273 ms [35.5788273 ns each]
    Regex took 2746.278466 ms [274.6278466 ns each]
    Exception numeric took 518.989902 ms [51.8989902 ns each]
    // Single value ("1")
    Array numeric took 317.861267 ms [31.7861267 ns each]
    Regex took 2505.313201 ms [250.5313201 ns each]
    Exception numeric took 239.956955 ms [23.9956955 ns each]
    // With Character.isDigit()
    Array numeric took 400.734616 ms [40.0734616 ns each]
    Regex took 2663.052417 ms [266.3052417 ns each]
    Exception numeric took 401.235906 ms [40.1235906 ns each]

With invalid characters ("5981a5833" and "a"):
    Array numeric took 343.205793 ms [34.3205793 ns each]
    Regex took 2608.739933 ms [260.8739933 ns each]
    Exception numeric took 7317.201775 ms [731.7201775 ns each]
    // With a single character ("a")
    Array numeric took 291.695519 ms [29.1695519 ns each]
    Regex took 2287.25378 ms [228.725378 ns each]
    Exception numeric took 7095.969481 ms [709.5969481 ns each]

With null:
    Array numeric took 214.663834 ms [21.4663834 ns each]
    Regex took 201.395992 ms [20.1395992 ns each]
    Exception numeric took 233.049327 ms [23.3049327 ns each]
    Exception numeric took 6603.669427 ms [660.3669427 ns each] if there is no if/null check

Disclaimer: I'm not claiming these methods are 100% optimized, they're just for demonstration of the data

Exceptions won if and only if the number is 4 characters or less, and every string is always a number... in which case, why even have a check?

In short, it is extremely painful if you run into invalid numbers frequently with the try/catch, which makes sense. An important rule I always follow is NEVER use try/catch for program flow. This is an example why.

Interestingly, the simple if char <0 || >9 was extremely simple to write, easy to remember (and should work in multiple languages) and wins almost all the test scenarios.

The only downside is that I'm guessing Integer.parseInt() might handle non ASCII numbers, whereas the array searching method does not.

---

For those wondering why I said it's easy to remember the character array one, if you know there's no negative signs, you can easily get away with something condensed as this:

public static boolean isNumericArray(String str) {
    if (str == null)
        return false;
    for (char c : str.toCharArray())
        if (c < '0' || c > '9')
            return false;
    return true;

---

Lastly as a final note, I was curious about the assigment operator in the accepted example with all the votes up. Adding in the assignment of

double d = Double.parseDouble(...)

is not only useless since you don't even use the value, but it wastes processing time and increased the runtime by a few nanoseconds (which led to a 100-200 ms increase in the tests). I can't see why anyone would do that since it actually is extra work to reduce performance.

You'd think that would be optimized out... though maybe I should check the bytecode and see what the compiler is doing. That doesn't explain why it always showed up as lengthier for me though if it somehow is optimized out... therefore I wonder what's going on. As a note: By lengthier, I mean running the test for 10000000 iterations, and running that program multiple times (10x+) always showed it to be slower.

EDIT: Updated a test for Character.isDigit()

Merlenemerlin

Doesn't this compile a new regular expression every time? That doesn't seem very efficient. – Juglandaceous

@SamuelEdwinWard Thats the whole reason I made this post... the regex example used other people's provided answers and showed how inefficient it is. Even if you attempt regex with pre-compiling it ahead of time and only using that, the time differences are: 2587 ms for the regex I posted from other provided people, 950 ms when compiled ahead of time, 144 ms when doing it as a numeric array (for 1 mil iterations of the same string). Compiling ahead of time obviously would help, but sadly it's still quite inferior to the array way... unless there's some insane optimization I don't know of. – Merlenemerlin

Believing that Regex makes things faster is almost a fallacy. If its a one off search, yeah, I get it... but I have noticed efficiently written code actually outdoes regex enough to shock you! Great post @Merlenemerlin – Eladiaelaeoptene

Answer 9

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

CraigTP's regular expression (shown above) produces some false positives. E.g. "23y4" will be counted as a number because '.' matches any character not the decimal point.

Also it will reject any number with a leading '+'

An alternative which avoids these two minor problems is

public static boolean isNumeric(String str)
{
    return str.matches("[+-]?\\d*(\\.\\d+)?");
}

Gayn

this will return true for a single plus "+" or minus "-", and false for "0." – Mindful

Good catch on the single plus or minus. Is "0." a valid number ? – Gayn

"0." is valid for Double.parseDouble() and is a valid literal according the JLS (§3.10.2)! – Mindful

Creating a regular expressions is costly as well. The regular expression must be created once and reused – Countdown

you should change it to matches("-?\\d+([.]\\d+)?") – Macle

Also, this matches an empty string. [+-]?(\\d\\.)?\\d+ should work better – Alligator

Why is every Regex solution supposing the Local Decimal format to be a '.'(dot) and not a ','(comma) like in some EU countries?! – Eladiaelaeoptene

Answer 10

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;
    }

Minneapolis

So "" is a number but "3.14" and "-1" aren't? – Lauranlaurance

Evidently doesn't apply to all number forms, but here's an upvote for thinking differently...if the original thought was yours, that is. – Casilde

Answer 11

You can use NumberFormat#parse:

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

Verwoerd

Offered an edit - .getInstance() was missing. +1 as this was the answer I went with when finding this question. – Zampino

Costly if used extensibly – Countdown

It also will pass if there are garbage characters at the end of value. – Asthenia

It would create a sonar issue if you don't log the exception – Albaalbacete

This worked for the number format 0x0001 where Double.parseDouble wasn't working. +1 – Benz

Answer 12

If you using java to develop Android app, you could using TextUtils.isDigitsOnly function.

Diorite

already posted here: https://mcmap.net/q/53123/-how-to-check-if-a-string-is-numeric-in-java see the comments for some issues with this solution – Cartwheel

Answer 13

Here was my answer to the problem.

A catch all convenience method which you can use to parse any String with any type of parser: isParsable(Object parser, String str). The parser can be a Class or an object. This will also allows you to use custom parsers you've written and should work for ever scenario, eg:

isParsable(Integer.class, "11");
isParsable(Double.class, "11.11");
Object dateFormater = new java.text.SimpleDateFormat("yyyy.MM.dd G 'at' HH:mm:ss z");
isParsable(dateFormater, "2001.07.04 AD at 12:08:56 PDT");

Here's my code complete with method descriptions.

import java.lang.reflect.*;

/**
 * METHOD: isParsable<p><p>
 * 
 * This method will look through the methods of the specified <code>from</code> parameter
 * looking for a public method name starting with "parse" which has only one String
 * parameter.<p>
 * 
 * The <code>parser</code> parameter can be a class or an instantiated object, eg:
 * <code>Integer.class</code> or <code>new Integer(1)</code>. If you use a
 * <code>Class</code> type then only static methods are considered.<p>
 * 
 * When looping through potential methods, it first looks at the <code>Class</code> associated
 * with the <code>parser</code> parameter, then looks through the methods of the parent's class
 * followed by subsequent ancestors, using the first method that matches the criteria specified
 * above.<p>
 * 
 * This method will hide any normal parse exceptions, but throws any exceptions due to
 * programmatic errors, eg: NullPointerExceptions, etc. If you specify a <code>parser</code>
 * parameter which has no matching parse methods, a NoSuchMethodException will be thrown
 * embedded within a RuntimeException.<p><p>
 * 
 * Example:<br>
 * <code>isParsable(Boolean.class, "true");<br>
 * isParsable(Integer.class, "11");<br>
 * isParsable(Double.class, "11.11");<br>
 * Object dateFormater = new java.text.SimpleDateFormat("yyyy.MM.dd G 'at' HH:mm:ss z");<br>
 * isParsable(dateFormater, "2001.07.04 AD at 12:08:56 PDT");<br></code>
 * <p>
 * 
 * @param parser    The Class type or instantiated Object to find a parse method in.
 * @param str   The String you want to parse
 * 
 * @return true if a parse method was found and completed without exception
 * @throws java.lang.NoSuchMethodException If no such method is accessible 
 */
public static boolean isParsable(Object parser, String str) {
    Class theClass = (parser instanceof Class? (Class)parser: parser.getClass());
    boolean staticOnly = (parser == theClass), foundAtLeastOne = false;
    Method[] methods = theClass.getMethods();

    // Loop over methods
    for (int index = 0; index < methods.length; index++) {
        Method method = methods[index];

        // If method starts with parse, is public and has one String parameter.
        // If the parser parameter was a Class, then also ensure the method is static. 
        if(method.getName().startsWith("parse") &&
            (!staticOnly || Modifier.isStatic(method.getModifiers())) &&
            Modifier.isPublic(method.getModifiers()) &&
            method.getGenericParameterTypes().length == 1 &&
            method.getGenericParameterTypes()[0] == String.class)
        {
            try {
                foundAtLeastOne = true;
                method.invoke(parser, str);
                return true; // Successfully parsed without exception
            } catch (Exception exception) {
                // If invoke problem, try a different method
                /*if(!(exception instanceof IllegalArgumentException) &&
                   !(exception instanceof IllegalAccessException) &&
                   !(exception instanceof InvocationTargetException))
                        continue; // Look for other parse methods*/

                // Parse method refuses to parse, look for another different method
                continue; // Look for other parse methods
            }
        }
    }

    // No more accessible parse method could be found.
    if(foundAtLeastOne) return false;
    else throw new RuntimeException(new NoSuchMethodException());
}


/**
 * METHOD: willParse<p><p>
 * 
 * A convienence method which calls the isParseable method, but does not throw any exceptions
 * which could be thrown through programatic errors.<p>
 * 
 * Use of {@link #isParseable(Object, String) isParseable} is recommended for use so programatic
 * errors can be caught in development, unless the value of the <code>parser</code> parameter is
 * unpredictable, or normal programtic exceptions should be ignored.<p>
 * 
 * See {@link #isParseable(Object, String) isParseable} for full description of method
 * usability.<p>
 * 
 * @param parser    The Class type or instantiated Object to find a parse method in.
 * @param str   The String you want to parse
 * 
 * @return true if a parse method was found and completed without exception
 * @see #isParseable(Object, String) for full description of method usability 
 */
public static boolean willParse(Object parser, String str) {
    try {
        return isParsable(parser, str);
    } catch(Throwable exception) {
        return false;
    }
}

Siusan

Answer 14

A well-performing approach avoiding try-catch and handling negative numbers and scientific notation.

Pattern PATTERN = Pattern.compile( "^(-?0|-?[1-9]\\d*)(\\.\\d+)?(E\\d+)?$" );

public static boolean isNumeric( String value ) 
{
    return value != null && PATTERN.matcher( value ).matches();
}

Teth

Answer 15

To match only positive base-ten integers, that contains only ASCII digits, use:

public static boolean isNumeric(String maybeNumeric) {
    return maybeNumeric != null && maybeNumeric.matches("[0-9]+");
}

Roll

Answer 16

Here is my class for checking if a string is numeric. It also fixes numerical strings:

Features:

1. Removes unnecessary zeros ["12.0000000" -> "12"]
2. Removes unnecessary zeros ["12.0580000" -> "12.058"]
3. Removes non numerical characters ["12.00sdfsdf00" -> "12"]
4. Handles negative string values ["-12,020000" -> "-12.02"]
5. Removes multiple dots ["-12.0.20.000" -> "-12.02"]
6. No extra libraries, just standard Java

Here you go...

public class NumUtils {
    /**
     * Transforms a string to an integer. If no numerical chars returns a String "0".
     *
     * @param str
     * @return retStr
     */
    static String makeToInteger(String str) {
        String s = str;
        double d;
        d = Double.parseDouble(makeToDouble(s));
        int i = (int) (d + 0.5D);
        String retStr = String.valueOf(i);
        System.out.printf(retStr + "   ");
        return retStr;
    }

    /**
     * Transforms a string to an double. If no numerical chars returns a String "0".
     *
     * @param str
     * @return retStr
     */
    static String makeToDouble(String str) {

        Boolean dotWasFound = false;
        String orgStr = str;
        String retStr;
        int firstDotPos = 0;
        Boolean negative = false;

        //check if str is null
        if(str.length()==0){
            str="0";
        }

        //check if first sign is "-"
        if (str.charAt(0) == '-') {
            negative = true;
        }

        //check if str containg any number or else set the string to '0'
        if (!str.matches(".*\\d+.*")) {
            str = "0";
        }

        //Replace ',' with '.'  (for some european users who use the ',' as decimal separator)
        str = str.replaceAll(",", ".");
        str = str.replaceAll("[^\\d.]", "");

        //Removes the any second dots
        for (int i_char = 0; i_char < str.length(); i_char++) {
            if (str.charAt(i_char) == '.') {
                dotWasFound = true;
                firstDotPos = i_char;
                break;
            }
        }
        if (dotWasFound) {
            String befDot = str.substring(0, firstDotPos + 1);
            String aftDot = str.substring(firstDotPos + 1, str.length());
            aftDot = aftDot.replaceAll("\\.", "");
            str = befDot + aftDot;
        }

        //Removes zeros from the begining
        double uglyMethod = Double.parseDouble(str);
        str = String.valueOf(uglyMethod);

        //Removes the .0
        str = str.replaceAll("([0-9])\\.0+([^0-9]|$)", "$1$2");

        retStr = str;

        if (negative) {
            retStr = "-"+retStr;
        }

        return retStr;

    }

    static boolean isNumeric(String str) {
        try {
            double d = Double.parseDouble(str);
        } catch (NumberFormatException nfe) {
            return false;
        }
        return true;
    }

}

Trippet

Answer 17

Regex Matching

Here is another example upgraded "CraigTP" regex matching with more validations.

public static boolean isNumeric(String str)
{
    return str.matches("^(?:(?:\\-{1})?\\d+(?:\\.{1}\\d+)?)$");
}

1. Only one negative sign - allowed and must be in beginning.
2. After negative sign there must be digit.
3. Only one decimal sign . allowed.
4. After decimal sign there must be digit.

Regex Test

1                  --                   **VALID**
1.                 --                   INVALID
1..                --                   INVALID
1.1                --                   **VALID**
1.1.1              --                   INVALID

-1                 --                   **VALID**
--1                --                   INVALID
-1.                --                   INVALID
-1.1               --                   **VALID**
-1.1.1             --                   INVALID

Rayshell

Answer 18

Exceptions are expensive, but in this case the RegEx takes much longer. The code below shows a simple test of two functions -- one using exceptions and one using regex. On my machine the RegEx version is 10 times slower than the exception.

import java.util.Date;


public class IsNumeric {

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

public static boolean isNumericTwo(String s) {
    try {
        Double.parseDouble(s);
        return true;
    } catch (Exception e) {
        return false;
    }
}

public static void main(String [] args) {

    String test = "12345.F";

    long before = new Date().getTime();     
    for(int x=0;x<1000000;++x) {
        //isNumericTwo(test);
        isNumericOne(test);
    }
    long after = new Date().getTime();

    System.out.println(after-before);

}

}

Cleveland

Generally, I think this sort of code would be used to check things like typed input. In that case speed is not a consideration and doing something as ugly as throwing an exception to check for number or non-number is wrong. – Gayn

Maybe not. Typed input is generally checked by the UI component where errors can be immediately shown before submitting the value. It might be more common to be validating strings from large input text files -- where performance matters. The goal in my answer here is to address the "exceptions are slow" statement in the accepted answer. Complex regex is much more expensive. And there is no "ugly throw" in my code at all -- just a faster way to detect violations. With a check-first-then-calculate approach you make two passes through the input: one to verify and then another to convert. – Cleveland

"On my machine the RegEx version is 10 times slower than the exception." - that's only because you test value that is numeric, so exception in never thrown. Test this on not numeric value, and version with exception will be slower than regex one. – Pomerania

Excellent point. I thought that adding an "F" on the end would make it not numeric, but the java "parseDouble" likes it. I stand corrected. – Cleveland

Answer 19

// please check below code

public static boolean isDigitsOnly(CharSequence str) {
    final int len = str.length();
    for (int i = 0; i < len; i++) {
        if (!Character.isDigit(str.charAt(i))) {
            return false;
        }
    }
    return true;
}

Jurist

The question says "numeric" which could include non-integer values. – Arlinda

Answer 20

You can use the java.util.Scanner object.

public static boolean isNumeric(String inputData) {
      Scanner sc = new Scanner(inputData);
      return sc.hasNextInt();
    }

Simard

Answer 21

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.

Procumbent

Answer 22

// only int
public static boolean isNumber(int num) 
{
    return (num >= 48 && c <= 57); // 0 - 9
}

// is type of number including . - e E 
public static boolean isNumber(String s) 
{
    boolean isNumber = true;
    for(int i = 0; i < s.length() && isNumber; i++) 
    {
        char c = s.charAt(i);
        isNumber = isNumber & (
            (c >= '0' && c <= '9') || (c == '.') || (c == 'e') || (c == 'E') || (c == '')
        );
    }
    return isInteger;
}

// is type of number 
public static boolean isInteger(String s) 
{
    boolean isInteger = true;
    for(int i = 0; i < s.length() && isInteger; i++) 
    {
        char c = s.charAt(i);
        isInteger = isInteger & ((c >= '0' && c <= '9'));
    }
    return isInteger;
}

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

Kendalkendall

Answer 23

This a simple example for this check:

public static boolean isNumericString(String input) {
    boolean result = false;

    if(input != null && input.length() > 0) {
        char[] charArray = input.toCharArray();

        for(char c : charArray) {
            if(c >= '0' && c <= '9') {
                // it is a digit
                result = true;
            } else {
                result = false;
                break;
            }
        }
    }

    return result;
}

Hermilahermina

Answer 24

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;
}

Brahmaputra

Answer 25

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+");
}

Waylan

Answer 26

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.

Barratry

Costly if used extensibly – Countdown

It would create a sonar issue if you don't log the exception – Albaalbacete

Double.parseDouble – Diaster

Answer 27

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.
}

Dialytic

yep, there's no "out parameters" in Java and since the Integer wrapper is immutable (thus cannot be used as a valid reference to store the output), the sensible idiomatic option would be to return an Integer object that could be null if the parse failed. An uglier option could be to pass an int[1] as output parameter. – Sinhalese

Yes, I remember a discussion about why Java has no output parameters. but returning an Integer (as null, if needed) would be fine too, I guess, though I don't know about Java's performance with regard to boxing/unboxing. – Dialytic

I like C# as much as the next guy, but its no use adding a .NET C# code snippet for a Java question when the features don't exist in Java – Lithuanian

It would create a sonar issue if you don't log the exception – Albaalbacete

Answer 28

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

Jaleesa

Answer 29

Java 8 Stream, lambda expression, functional interface

All cases handled (string null, string empty etc)

String someString = null; // something="", something="123abc", something="123123"

boolean isNumeric = Stream.of(someString)
            .filter(s -> s != null && !s.isEmpty())
            .filter(Pattern.compile("\\D").asPredicate().negate())
            .mapToLong(Long::valueOf)
            .boxed()
            .findAny()
            .isPresent();

Caius

Answer 30

You can use NumberUtils.isCreatable() from Apache Commons Lang.

Since NumberUtils.isNumber will be deprecated in 4.0, so use NumberUtils.isCreatable() instead.

Besant

Answer 31

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!

Foretaste

Answer 32

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");
}

Kataway

Welcome to stackoverflow. Generally, it is best to avoid resurrecting old threads unless the response adds something significantly different to the thread. While valid, that approach was already mentioned in the accepted answer. – Lublin

It would create a sonar issue if you don't log the exception – Albaalbacete

Answer 33

Here are two methods that might work. (Without using Exceptions). Note : Java is a pass-by-value by default and a String's value is the address of the String's object data. So , when you are doing

stringNumber = stringNumber.replaceAll(" ", "");

You have changed the input value to have no spaces. You can remove that line if you want.

private boolean isValidStringNumber(String stringNumber)
{
    if(stringNumber.isEmpty())
    {
        return false;
    }

    stringNumber = stringNumber.replaceAll(" ", "");

    char [] charNumber = stringNumber.toCharArray();
    for(int i =0 ; i<charNumber.length ;i++)
    {
        if(!Character.isDigit(charNumber[i]))
        {
            return false;
        }
    }
    return true;
}

Here is another method in case you want to allow floats This method allegedly allows numbers in the form to pass 1,123,123,123,123,123.123 i have just made it , and i think it needs further testing to ensure it is working.

private boolean isValidStringTrueNumber(String stringNumber)
{
    if(stringNumber.isEmpty())
    {
        return false;
    }

    stringNumber = stringNumber.replaceAll(" ", "");
    int countOfDecimalPoint = 0;
    boolean decimalPointPassed = false;
    boolean commaFound = false;
    int countOfDigitsBeforeDecimalPoint = 0;
    int countOfDigitsAfterDecimalPoint =0 ;
    int commaCounter=0;
    int countOfDigitsBeforeFirstComma = 0;

    char [] charNumber = stringNumber.toCharArray();
    for(int i =0 ; i<charNumber.length ;i++)
    {
        if((commaCounter>3)||(commaCounter<0))
        {
            return false;
        }
        if(!Character.isDigit(charNumber[i]))//Char is not a digit.
        {
            if(charNumber[i]==',')
            {
                if(decimalPointPassed)
                {
                    return false;
                }
                commaFound = true;
                //check that next three chars are only digits.
                commaCounter +=3;
            }
            else if(charNumber[i]=='.')
            {
                decimalPointPassed = true;
                countOfDecimalPoint++;
            }
            else
            {
                return false;
            }
        }
        else //Char is a digit.
        {
            if ((commaCounter>=0)&&(commaFound))
            {
                if(!decimalPointPassed)
                {
                    commaCounter--;
                }
            }

            if(!commaFound)
            {
                countOfDigitsBeforeFirstComma++;
            }

            if(!decimalPointPassed)
            {
                countOfDigitsBeforeDecimalPoint++;
            }
            else
            {
                countOfDigitsAfterDecimalPoint++;
            }
        }
    }
    if((commaFound)&&(countOfDigitsBeforeFirstComma>3))
    {
        return false;
    }
    if(countOfDecimalPoint>1)
    {
        return false;
    }

    if((decimalPointPassed)&&((countOfDigitsBeforeDecimalPoint==0)||(countOfDigitsAfterDecimalPoint==0)))
    {
        return false;
    }
    return true;
}

Aggappora

Oh , Good question. I guess this one only works normal type integers. The method were initially created to filter input phone numbers , and count numbers. – Aggappora

Answer 34

If you want to do the check using a regex you should create a final static Pattern object, that way the regex only needs to be compiled once. Compiling the regex takes about as long as performing the match so by taking this precaution you'll cut the execution time of the method in half.

final static Pattern NUMBER_PATTERN = Pattern.compile("[+-]?\\d*\\.?\\d+");

static boolean isNumber(String input) {
    Matcher m = NUMBER_PATTERN.matcher(input);
    return m.matches();
}

I'm assuming a number is a string with nothing but decimal digits in it, possibly a + or - sign at the start and at most one decimal point (not at the end) and no other characters (including commas, spaces, numbers in other counting systems, Roman numerals, hieroglyphs).

This solution is succinct and pretty fast but you can shave a couple of milliseconds per million invocations by doing it like this

static boolean isNumber(String s) {
    final int len = s.length();
    if (len == 0) {
        return false;
    }
    int dotCount = 0;
    for (int i = 0; i < len; i++) {
        char c = s.charAt(i);
        if (c < '0' || c > '9') {
            if (i == len - 1) {//last character must be digit
                return false;
            } else if (c == '.') {
                if (++dotCount > 1) {
                    return false;
                }
            } else if (i != 0 || c != '+' && c != '-') {//+ or - allowed at start
                return false;
            }

        }
    }
    return true;
}

Transfer

Answer 35

Try this:

public  boolean isNumber(String str)
{       
    short count = 0;
    char chc[]  = {'0','1','2','3','4','5','6','7','8','9','.','-','+'};
    for (char c : str.toCharArray())
    {   
        for (int i = 0;i < chc.length;i++)
        {
            if( c  == chc[i]){
                count++;        
            }
         }                      
    }
    if (count != str.length() ) 
        return false;
    else
        return true;
}

Impassion

Answer 36

This is the fastest way i know to check if String is Number or not:

public static boolean isNumber(String str){
  int i=0, len=str.length();
  boolean a=false,b=false,c=false, d=false;
  if(i<len && (str.charAt(i)=='+' || str.charAt(i)=='-')) i++;
  while( i<len && isDigit(str.charAt(i)) ){ i++; a=true; }
  if(i<len && (str.charAt(i)=='.')) i++;
  while( i<len && isDigit(str.charAt(i)) ){ i++; b=true; }
  if(i<len && (str.charAt(i)=='e' || str.charAt(i)=='E') && (a || b)){ i++; c=true; }
  if(i<len && (str.charAt(i)=='+' || str.charAt(i)=='-') && c) i++;
  while( i<len && isDigit(str.charAt(i)) ){ i++; d=true;}
  return i==len && (a||b) && (!c || (c && d));
}
static boolean isDigit(char c){
  return c>='0' && c<='9';
}

Gorky

Instead of having these many checks we can simply have if (ch >= '0' && ch <= '9') {} – Toenail

Answer 37

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 )

Vezza

Answer 38

String text="hello 123";
if(Pattern.matches([0-9]+))==true
System.out.println("String"+text);

Gregoriagregorian

Answer 39

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!

Sateen

"it is a too-large-size number" → There is no such thing as a "too-large-size number". Numbers are infinite. 7645 is a number and 92847294729492875982452012471041141990140811894142729051 is also a number. The first can be represented as an Integer and the second can be represented as a BigDecimal. And even if they could not be represented as an object in Java, they are still numbers -- which is what the question asks. – Pitchblack

Also, don't misuse exceptions. Java isn't python. At least use more specific NumberFormatException – January

Answer 40

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

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

Enterpriser

While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply. – Teodorateodorico

Answer 41

import java.util.Scanner;

public class TestDemo {
    public static void main(String[] args) {
        boolean flag = true;
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the String:");
        String str = sc.nextLine();

        for (int i = 0; i < str.length(); i++) {
            if(str.charAt(i) > 48 && str.charAt(i) < 58) {
                flag = false;
                break;
            }
        }

        if(flag == true) {
            System.out.println("String is a valid String.");
        } else {
            System.out.println("String contains number.");
        }
    }
}

Collaborationist