Extract digits from string - StringUtils Java

Question

I have a String and I want to extract the (only) sequence of digits in the string.

Example: helloThisIsA1234Sample. I want the 1234

It's a given that the sequence of digits will occur only once within the string but not in the same position.

(for those who will ask, I have a server name and need to extract a specific number within it)

I would like to use the StringUtils class from Apache commomns.

Thanks!

Apache [StringUtils](http://commons.apache.org/lang/api-2.5/org/apache/commons/lang/StringUtils.html). Not sure why you wanna use it though. — Rahul, Feb 20 '13 at 07:02
That's what I have, was wondering if maybe there is a function already doing something like I need, but I am not so familiar with it — NinaNa, Feb 20 '13 at 07:04
Cuz, when you can do it with `String` itself, you won't need the `StringUtils` from Apache. — Rahul, Feb 20 '13 at 07:11

score 203 · Accepted Answer · answered Feb 20 '13 at 07:05

203

Use this code numberOnly will contain your desired output.

   String str="sdfvsdf68fsdfsf8999fsdf09";
   String numberOnly= str.replaceAll("[^0-9]", "");

answered Feb 20 '13 at 07:05

5

String.replaceAll() would compile that pattern each time this is run. Really unoptimal. Check its source. Use Guava's CharMathcer.DIGIT.retainFrom() as seen below. – Agoston Horvath Nov 22 '16 at 09:13
1

In more recent Java, the Regex can be simplified to "\\D" – CeePlusPlus Dec 04 '22 at 23:30

score 38 · Answer 2 · edited Jun 14 '19 at 22:50

38

I always like using Guava String utils or similar for these kind of problems:

String theDigits = CharMatcher.inRange('0', '9').retainFrom("abc12 3def"); // 123

edited Jun 14 '19 at 22:50

hd1

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answered Jan 08 '15 at 13:37

Michael Bavin

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5

This is a lot faster than regular expression. This should be the accepted answer. – Agoston Horvath Nov 22 '16 at 09:13
2

use CharMatcher.digit().retainFrom("xx123xx"); instead, DIGIT is scheduled to be removed in June 2018. – Scott 混合理论 May 17 '18 at 06:52
3

`CharMather.digit()` is also deprecated. I used `CharMatcher.inRange('0', '9').retainFrom("xx123xx");` – lamyarus Sep 05 '18 at 08:08

rnrneverdies · Answer 3 · 2016-03-11T15:02:13.307

28

Just one line:

int value = Integer.parseInt(string.replaceAll("[^0-9]", ""));

edited Mar 11 '16 at 15:02

answered Mar 11 '16 at 13:51

rnrneverdies

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score 12 · Answer 4 · edited Apr 11 '21 at 23:30

12

You can also use java.util.Scanner:

new Scanner(str).useDelimiter("[^\\d]+").nextInt()

You can use next() instead of nextInt() to get the digits as a String. Note that calling Integer.parseInt on the result may be many times faster than calling nextInt().

You can check for the presence of number using hasNextInt() on the Scanner.

edited Apr 11 '21 at 23:30

Dave Jarvis

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answered Feb 20 '13 at 07:18

Marimuthu Madasamy

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In more recent Java, the Regex can be simplified to "\\D" – CeePlusPlus Dec 04 '22 at 23:30

score 7 · Answer 5 · answered Feb 20 '13 at 07:01

7

Use a regex such as [^0-9] to remove all non-digits.

From there, just use Integer.parseInt(String);

answered Feb 20 '13 at 07:01

Sounds interesting, are you able to give me some more code? say my string var is called myString, how would I use it? (I am just a little new to Java...) – NinaNa Feb 20 '13 at 07:06

Shessuky · Answer 6 · 2017-10-27T17:08:40.020

4

try this :

String s = "helloThisIsA1234Sample";
s = s.replaceAll("\\D+","");

This means: replace all occurrences of digital characters (0 -9) by an empty string !

edited Oct 27 '17 at 17:08

answered Jan 30 '15 at 18:51

Shessuky

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It would be nice if you could add a little explanation as to what your code does. :) – Bhargav Rao Jan 30 '15 at 19:22

score 4 · Answer 7 · edited Apr 11 '21 at 23:25

4

Guava's CharMatcher class extracts Integers from a String.

String text="Hello1010";
System.out.println(CharMatcher.digit().retainFrom(text));

Yields:

edited Apr 11 '21 at 23:25

Dave Jarvis

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answered Mar 03 '20 at 06:45

Prasanna

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Deprecated as of 2022 – Sebastian Nielsen May 25 '22 at 08:39

score 3 · Answer 8 · answered May 03 '14 at 11:03

I've created a JUnit Test class(as a additional knowledge/info) for the same issue. Hope you'll be finding this helpful.

   public class StringHelper {
    //Separate words from String which has gigits
        public String drawDigitsFromString(String strValue){
            String str = strValue.trim();
            String digits="";
            for (int i = 0; i < str.length(); i++) {
                char chrs = str.charAt(i);              
                if (Character.isDigit(chrs))
                    digits = digits+chrs;
            }
            return digits;
        }
    }

And JUnit Test case is:

 public class StringHelperTest {
    StringHelper helper;

        @Before
        public void before(){
            helper = new StringHelper();
        }

        @Test
    public void testDrawDigitsFromString(){
        assertEquals("187111", helper.drawDigitsFromString("TCS187TCS111"));
    }
 }

score 3 · Answer 9 · answered May 10 '20 at 18:04

3

Extending the best answer for finding floating point numbers

       String str="2.53GHz";
       String decimal_values= str.replaceAll("[^0-9\\.]", "");
       System.out.println(decimal_values);

answered May 10 '20 at 18:04

Dinesh Kesavan

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score 3 · Answer 10 · answered Apr 13 '22 at 06:58

3

If you have access to org.apache.commons.lang3. you can use StringUtils.getDigits method

public static void main(String[] args) {
    String value = "helloThisIsA1234Sample";
    System.out.println(StringUtils.getDigits(value));
}
output: 12345

answered Apr 13 '22 at 06:58

Clover

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score 2 · Answer 11 · answered Feb 20 '13 at 07:01

2

You can use the following regular expression.

string.split(/ /)[0].replace(/[^\d]/g, '')

answered Feb 20 '13 at 07:01

Siva Arunachalam

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score 2 · Answer 12 · answered May 06 '16 at 05:51

2

        String line = "This order was32354 placed for QT ! OK?";
        String regex = "[^\\d]+";

        String[] str = line.split(regex);

        System.out.println(str[1]);

answered May 06 '16 at 05:51

Az.MaYo

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score 2 · Answer 13 · answered Mar 09 '17 at 02:38

2

You can use str = str.replaceAll("\\D+","");

answered Mar 09 '17 at 02:38

sendon1982

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score 2 · Answer 14 · answered Mar 09 '17 at 03:00

You can split the string and compare with each character

public static String extractNumberFromString(String source) {
    StringBuilder result = new StringBuilder(100);
    for (char ch : source.toCharArray()) {
        if (ch >= '0' && ch <= '9') {
            result.append(ch);
        }
    }

    return result.toString();
}

Testing Code

    @Test
    public void test_extractNumberFromString() {
    String numberString = NumberUtil.extractNumberFromString("+61 415 987 636");
    assertThat(numberString, equalTo("61415987636"));

    numberString = NumberUtil.extractNumberFromString("(02)9295-987-636");
    assertThat(numberString, equalTo("029295987636"));

    numberString = NumberUtil.extractNumberFromString("(02)~!@#$%^&*()+_<>?,.:';9295-{}[=]987-636");
    assertThat(numberString, equalTo("029295987636"));
}

score 1 · Answer 15 · answered Feb 20 '13 at 07:16

1

Try this approach if you have symbols and you want just numbers:

    String s  = "@##9823l;Azad9927##$)(^738#";
    System.out.println(s=s.replaceAll("[^0-9]", ""));
    StringTokenizer tok = new StringTokenizer(s,"`~!@#$%^&*()-_+=\\.,><?");
    String s1 = "";
    while(tok.hasMoreTokens()){
        s1+= tok.nextToken();
    }
    System.out.println(s1);

answered Feb 20 '13 at 07:16

Azad

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This is now a slow solution, because String is immutable and in every cycle iteration is created new instance of String, replace at least with StringBuilder. Still it isn't write nice – Perlos Jul 12 '16 at 05:17

score 0 · Answer 16 · answered Dec 15 '17 at 04:33

A very simple solution, if separated by comma or if not separated by comma

public static void main(String[] args) {

    String input = "a,1,b,2,c,3,d,4";
    input = input.replaceAll(",", "");

    String alpha ="";
    String num = "";

    char[] c_arr = input.toCharArray();

    for(char c: c_arr) {
        if(Character.isDigit(c)) {
            alpha = alpha + c;
        }
        else {
            num = num+c;
        }
    }

    System.out.println("Alphabet: "+ alpha);
    System.out.println("num: "+ num);

}

score -1 · Answer 17 · edited Mar 17 '18 at 17:29

-1

   `String s="as234dfd423";
for(int i=0;i<s.length();i++)
 {
    char c=s.charAt(i);``
    char d=s.charAt(i);
     if ('a' <= c && c <= 'z')
         System.out.println("String:-"+c);
     else  if ('0' <= d && d <= '9')
           System.out.println("number:-"+d);
    }

output:-

number:-4
number:-3
number:-4
String:-d
String:-f
String:-d
number:-2
number:-3

edited Mar 17 '18 at 17:29

OneCricketeer

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answered Feb 26 '18 at 05:47

Amol Nile

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Your output doesn't appear to match your test input. – Andrew Mortimer Feb 26 '18 at 05:54

score -1 · Answer 18 · edited Nov 03 '19 at 17:27

-1

You can try this:

  String str="java123java456";
  String out="";
  for(int i=0;i<str.length();i++)
  {
    int a=str.codePointAt(i);
     if(a>=49&&a<=57)
       {
          out=out+str.charAt(i);
       }
   }
 System.out.println(out);

edited Nov 03 '19 at 17:27

Dino

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answered Nov 03 '19 at 11:52

Gunapati Akhilchandra

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From Review: Hi, please don't answer just with source code. Try to provide a nice description about how your solution works. See: [How do I write a good answer?](https://stackoverflow.com/help/how-to-answer). Thanks – sɐunıɔןɐqɐp Nov 03 '19 at 12:17
what about Capital letters or another languages? – Viktor Yakunin Dec 23 '19 at 14:41

score -6 · Answer 19 · edited Dec 23 '19 at 19:37

-6

Simple python code for separating the digits in string

  s="rollnumber99mixedin447"
  list(filter(lambda c: c >= '0' and c <= '9', [x for x in s]))

edited Dec 23 '19 at 19:37

Glenn

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answered Nov 22 '19 at 09:07

rameshbabu reddy

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this question has explicit Java tag – Viktor Yakunin Dec 23 '19 at 14:41

Extract digits from string - StringUtils Java

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