Java String - See if a string contains only numbers and not letters

Question

I have a string that I load throughout my application, and it changes from numbers to letters and such. I have a simple if statement to see if it contains letters or numbers but, something isn't quite working correctly. Here is a snippet.

String text = "abc"; 
String number; 

if (text.contains("[a-zA-Z]+") == false && text.length() > 2) {
    number = text; 
}

Although the text variable does contain letters, the condition returns as true. The and && should eval as both conditions having to be true in order to process the number = text;

==============================

Solution:

I was able to solve this by using this following code provided by a comment on this question. All other post are valid as well!

What I used that worked came from the first comment. Although all the example code provided seems to be valid as well!

String text = "abc"; 
String number; 

if (Pattern.matches("[a-zA-Z]+", text) == false && text.length() > 2) {
    number = text; 
}

Answer 1

If you'll be processing the number as text, then change:

if (text.contains("[a-zA-Z]+") == false && text.length() > 2){

to:

if (text.matches("[0-9]+") && text.length() > 2) {

Instead of checking that the string doesn't contain alphabetic characters, check to be sure it contains only numerics.

If you actually want to use the numeric value, use Integer.parseInt() or Double.parseDouble() as others have explained below.

---

As a side note, it's generally considered bad practice to compare boolean values to true or false. Just use if (condition) or if (!condition).

Answer 2

In order to simply check whether the string only contains ALPHABETS use the following code:

if (text.matches("[a-zA-Z]+")){
   // your operations
}

In order to simply check whether the string only contains NUMBER use the following code:

if (text.matches("[0-9]+")){
   // your operations
}

Hope this will help someone!

Answer 3

This is how I would do it:

if(text.matches("^[0-9]*$") && text.length() > 2){
    //...
}

The $ will avoid a partial match e.g; 1B.

Answer 4

You can also use NumberUtil.isCreatable(String str) from Apache Commons

Answer 5

Performance-wise parseInt and such are much worser than other solutions, because at least require exception handling.

I've run jmh tests and have found that iterating over String using charAt and comparing chars with boundary chars is the fastest way to test if string contains only digits.

JMH testing

Tests compare performance of Character.isDigit vs Pattern.matcher().matches vs Long.parseLong vs checking char values.

These ways can produce different result for non-ascii strings and strings containing +/- signs.

Tests run in Throughput mode (greater is better) with 5 warmup iterations and 5 test iterations.

Results

Note that parseLong is almost 100 times slower than isDigit for first test load.

## Test load with 25% valid strings (75% strings contain non-digit symbols)

Benchmark       Mode  Cnt  Score   Error  Units
testIsDigit    thrpt    5  9.275 ± 2.348  ops/s
testPattern    thrpt    5  2.135 ± 0.697  ops/s
testParseLong  thrpt    5  0.166 ± 0.021  ops/s

## Test load with 50% valid strings (50% strings contain non-digit symbols)

Benchmark              Mode  Cnt  Score   Error  Units
testCharBetween       thrpt    5  16.773 ± 0.401  ops/s
testCharAtIsDigit     thrpt    5  8.917 ± 0.767  ops/s
testCharArrayIsDigit  thrpt    5  6.553 ± 0.425  ops/s
testPattern           thrpt    5  1.287 ± 0.057  ops/s
testIntStreamCodes    thrpt    5  0.966 ± 0.051  ops/s
testParseLong         thrpt    5  0.174 ± 0.013  ops/s
testParseInt          thrpt    5  0.078 ± 0.001  ops/s

Test suite

@State(Scope.Benchmark)
public class StringIsNumberBenchmark {
    private static final long CYCLES = 1_000_000L;
    private static final String[] STRINGS = {"12345678901","98765432177","58745896328","35741596328", "123456789a1", "1a345678901", "1234567890 "};
    private static final Pattern PATTERN = Pattern.compile("\\d+");

    @Benchmark
    public void testPattern() {
        for (int i = 0; i < CYCLES; i++) {
            for (String s : STRINGS) {
                boolean b = false;
                b = PATTERN.matcher(s).matches();
            }
        }
    }

    @Benchmark
    public void testParseLong() {
        for (int i = 0; i < CYCLES; i++) {
            for (String s : STRINGS) {
                boolean b = false;
                try {
                    Long.parseLong(s);
                    b = true;
                } catch (NumberFormatException e) {
                    // no-op
                }
            }
        }
    }

    @Benchmark
    public void testCharArrayIsDigit() {
        for (int i = 0; i < CYCLES; i++) {
            for (String s : STRINGS) {
                boolean b = false;
                for (char c : s.toCharArray()) {
                    b = Character.isDigit(c);
                    if (!b) {
                        break;
                    }
                }
            }
        }
    }

    @Benchmark
    public void testCharAtIsDigit() {
        for (int i = 0; i < CYCLES; i++) {
            for (String s : STRINGS) {
                boolean b = false;
                for (int j = 0; j < s.length(); j++) {
                    b = Character.isDigit(s.charAt(j));
                    if (!b) {
                        break;
                    }
                }
            }
        }
    }

    @Benchmark
    public void testIntStreamCodes() {
        for (int i = 0; i < CYCLES; i++) {
            for (String s : STRINGS) {
                boolean b = false;
                b = s.chars().allMatch(c -> c > 47 && c < 58);
            }
        }
    }

    @Benchmark
    public void testCharBetween() {
        for (int i = 0; i < CYCLES; i++) {
            for (String s : STRINGS) {
                boolean b = false;
                for (int j = 0; j < s.length(); j++) {
                    char charr = s.charAt(j);
                    b = '0' <= charr && charr <= '9';
                    if (!b) {
                        break;
                    }
                }
            }
        }
    }
}

Updated on Feb 23, 2018

• Add two more cases - one using charAt instead of creating extra array and another using IntStream of char codes
• Add immediate break if non-digit found for looped test cases
• Return false for empty string for looped test cases

Updated on Feb 23, 2018

• Add one more test case (the fastest!) that compares char value without using stream

Answer 6

Apache Commons Lang provides org.apache.commons.lang.StringUtils.isNumeric(CharSequence cs), which takes as an argument a String and checks if it consists of purely numeric characters (including numbers from non-Latin scripts). That method returns false if there are characters such as space, minus, plus, and decimal separators such as comma and dot.

Other methods of that class allow for further numeric checks.

Answer 7

A solution with Java 8 streams and lambda

String data = "12345";
boolean isOnlyNumbers = data.chars().allMatch(Character::isDigit);

Answer 8

StringUtils.isNumeric("1234")

this works fine.

Answer 9

boolean isNum = text.chars().allMatch(c -> c >= 48 && c <= 57)

Answer 10

Below regexs can be used to check if a string has only number or not:

if (str.matches(".*[^0-9].*")) or if (str.matches(".*\\D.*"))

Both conditions above will return true if String containts non-numbers. On false, string has only numbers.

Answer 11

You can use Regex.Match

if(text.matches("\\d*")&& text.length() > 2){
    System.out.println("number");
}

Or you could use onversions like Integer.parseInt(String) or better Long.parseLong(String) for bigger numbers like for example:

private boolean onlyContainsNumbers(String text) {
    try {
        Long.parseLong(text);
        return true;
    } catch (NumberFormatException ex) {
        return false;
    }
} 

And then test with:

if (onlyContainsNumbers(text) && text.length() > 2) {
    // do Stuff
}

Answer 12

There are lots of facilities to obtain numbers from Strings in Java (and vice versa). You may want to skip the regex part to spare yourself the complication of that.

For example, you could try and see what Double.parseDouble(String s) returns for you. It should throw a NumberFormatException if it does not find an appropriate value in the string. I would suggest this technique because you could actually make use of the value represented by the String as a numeric type.

Answer 13

This code is already written. If you don't mind the (extremely) minor performance hit--which is probably no worse than doing a regex match--use Integer.parseInt() or Double.parseDouble(). That'll tell you right away if a String is only numbers (or is a number, as appropriate). If you need to handle longer strings of numbers, both BigInteger and BigDecimal sport constructors that accept Strings. Any of these will throw a NumberFormatException if you try to pass it a non-number (integral or decimal, based on the one you choose, of course). Alternately, depending on your requirements, just iterate the characters in the String and check Character.isDigit() and/or Character.isLetter().

Answer 14

import java.util.*;

class Class1 {
    public static void main(String[] argh) {
        boolean ans = CheckNumbers("123");
        if (ans == true) {
            System.out.println("String contains numbers only");
        } else {
            System.out.println("String contains other values as well");

        }
    }


    public static boolean CheckNumbers(String input) {
        for (int ctr = 0; ctr < input.length(); ctr++) {
            if ("1234567890".contains(Character.valueOf(input.charAt(ctr)).toString())) {
                continue;
            } else {
                return false;
            }
        }
        return true;
    }
}

Answer 15

Here is my code, hope this will help you !

 public boolean isDigitOnly(String text){

    boolean isDigit = false;

    if (text.matches("[0-9]+") && text.length() > 2) {
        isDigit = true;
    }else {
        isDigit = false;
    }

    return isDigit;
}

Answer 16

Character first_letter_or_number = query.charAt(0);
                //------------------------------------------------------------------------------
                if (Character.isDigit())
                {

                }
                else if (Character.isLetter())
                {

                }

Answer 17

Working test example

import java.util.regex.Matcher;
import java.util.regex.Pattern;

import org.apache.commons.lang3.StringUtils;

public class PaserNo {

    public static void main(String args[]) {

        String text = "gg";

        if (!StringUtils.isBlank(text)) {
            if (stringContainsNumber(text)) {
                int no=Integer.parseInt(text.trim());
                System.out.println("inside"+no);

            } else {
                System.out.println("Outside");
            }
        }
        System.out.println("Done");
    }

    public static boolean stringContainsNumber(String s) {
        Pattern p = Pattern.compile("[0-9]");
        Matcher m = p.matcher(s);
        return m.find();
    }
}

Still your code can be break by "1a" etc so you need to check exception

if (!StringUtils.isBlank(studentNbr)) {
                try{
                    if (isStringContainsNumber(studentNbr)){
                    _account.setStudentNbr(Integer.parseInt(studentNbr.trim()));
                }
                }catch(Exception e){
                    e.printStackTrace();
                    logger.info("Exception during parse studentNbr"+e.getMessage());
                }
            }

Method for checking no is string or not

private boolean isStringContainsNumber(String s) {
        Pattern p = Pattern.compile("[0-9]");
        Matcher m = p.matcher(s);
        return m.find();
    }

Answer 18

It is a bad practice to involve any exception throwing/handling into such a typical scenario.

Therefore a parseInt() is not nice, but a regex is an elegant solution for this, but take care of the following:
-fractions
-negative numbers
-decimal separator might differ in contries (e.g. ',' or '.')
-sometimes it is allowed to have a so called thousand separator, like a space or a comma e.g. 12,324,1000.355

To handle all the necessary cases in your application you have to be careful, but this regex covers the typical scenarios (positive/negative and fractional, separated by a dot): ^[-+]?\d*.?\d+$
For testing, I recommend regexr.com.

Answer 19

Slightly modified version of Adam Bodrogi's:

public class NumericStr {


public static void main(String[] args) {
    System.out.println("Matches: "+NumericStr.isNumeric("20"));         // Should be true
    System.out.println("Matches: "+NumericStr.isNumeric("20,00"));          // Should be true
    System.out.println("Matches: "+NumericStr.isNumeric("30.01"));          // Should be true
    System.out.println("Matches: "+NumericStr.isNumeric("30,000.01"));          // Should be true
    System.out.println("Matches: "+NumericStr.isNumeric("-2980"));          // Should be true
    System.out.println("Matches: "+NumericStr.isNumeric("$20"));            // Should be true
    System.out.println("Matches: "+NumericStr.isNumeric("jdl"));            // Should be false
    System.out.println("Matches: "+NumericStr.isNumeric("2lk0"));           // Should be false
}

public static boolean isNumeric(String stringVal) {
    if (stringVal.matches("^[\\$]?[-+]?[\\d\\.,]*[\\.,]?\\d+$")) {
        return true;
    }

    return false;
}
}

Had to use this today so just posted my modifications. Includes currency, thousands comma or period notation, and some validations. Does not include other currency notations (euro, cent), verification commas are every third digit.

Answer 20

public class Test{  
public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String str;
    boolean status=false;
    System.out.println("Enter the String : ");
    str = sc.nextLine();
    
    char ch[] = str.toCharArray();
    
    for(int i=0;i<ch.length;i++) {
        if(ch[i]=='1'||ch[i]=='2'||ch[i]=='3'||ch[i]=='4'||ch[i]=='5'||ch[i]=='6'||ch[i]=='7'||ch[i]=='8'||ch[i]=='9'||ch[i]=='0') {
            ch[i] = 0;
        }
    }
    
    for(int i=0;i<ch.length;i++) {
        if(ch[i] != 0) {
            System.out.println("Mixture of letters and Digits");
            status = false;
            break;
        }
        else
            status = true;
    }
    
    if(status == true){
        System.out.println("Only Digits are present");
    }
}

}

Answer 21

Here is a sample. Find only the digits in a String and Process formation as needed.

text.replaceAll("\\d(?!$)", "$0 ");

For more info check google Docs https://developer.android.com/reference/java/util/regex/Pattern Where you can use Pattern

Answer 22

boolean flag = false;
        System.out.print("Enter String : ");
        String str = new Scanner(System.in).next();

        for (int i = 0; i < str.length(); i++)
        {
            if (str.length() <= 0)
            {
                System.out.println("String length Can't be zero.");
                return;
            }
            char ch = str.charAt(i);
            int c = ch;
            if (c >= 48 && c <= 58)
            {
                flag = true;
            } else
            {
                flag = false;
                break;
            }
        }
        if (flag)
        {
            System.out.println("input [" + str + "] contains number only.");
        } else
            System.out.println("input [" + str + "] have some non string values in it.");

Answer 23

You could use [a-zA-Z]{2,} as well.

where [a-zA-Z] checks for alphabets only and {2,} checks the length, should be greater than 2

Answer 24

I realize this is an old question, but I consider it still relevant today.

Although the String matching solution is the most elegant one (at least in my opinion), it is far from being performant. So, if you plan to use this method at scale, you might consider alternative implementations.

I wrote a few short classes for comparing the run times of various implementations. For each implementation, I ran 100 million iterations, half with an all-numeric input (so the method returns true) and half with letters (so the method returns false) and following are the results of the runs on my machine.

Alternatives

String matching solution

text.matches("\\d*");

Total runtime: 19206 ms

Pattern matching solution

Similar to the String matching solution, but with better running time because pattern compilation is done only once:

PATTERN.matcher(text).matches();

where PATTERN is defined only once as such: final Pattern PATTERN = Pattern.compile("\\d*");

Total runtime: 9193 ms

IntStream solution

Less concise than the matching solutions, but faster:

text.isEmpty() || IntStream.range(0, text.length()).allMatch(i -> Character.isDigit(text.charAt(i)));

Total runtime: 5568 ms

Simple loop solution

Using the same concept as the IntStream solution, but more verbose:

boolean allDigits = true;
if (!text.isEmpty()) {
    for (int i = 0; i < text.length(); i++) {
        if (!Character.isDigit(text.charAt(i))) {
            allDigits = false;
            break;
        }
    }
}
return allDigits;

Total runtime: 235 ms

Commons lang solution

There is also the commons lang solution which doesn't require you to write any code, but to depend on an external library instead:

StringUtils.isNumeric(text);

Total runtime: 1433 ms

Conclusion

What this test shows is that - leaving aside the external dependency solution - the more elegant the code, the less performant it is, with the most "primitive" solution being also the fastest running one.

That being said, I'm an advocate of clean code, so if you only need to call this method a few times, definitely go for one of the more elegant solutions; in many cases code quality is more important than performance. However, if you know you will be calling it a lot or if it's going to be part of a library, consider going for one of the less elegant but more performant solutions.