How to determine whether a string contains an integer?

Question

Say you have a string that you want to test to make sure that it contains an integer before you proceed with other the rest of the code. What would you use, in java, to find out whether or not it is an integer?

Your question isn't clear. Do you want to check whether it contains an integer or with it is an integer? abc123 or 123456? — Ash Burlaczenko, Dec 08 '10 at 14:30
@Ash **to find out whether or not it is an integer** - to me, the question is clear (the title is ambigious). — Andreas Dolk, Dec 08 '10 at 14:58
You may also want to look at this http://stackoverflow.com/questions/3517686/cannot-convert-string-to-integer-in-java — CoolBeans, Dec 08 '10 at 15:29

score 15 · Accepted Answer · answered Dec 08 '10 at 14:32

If you want to make sure that it is only an integer and convert it to one, I would use parseInt in a try/catch. However, if you want to check if the string contains a number then you would be better to use the String.matches with Regular Expressions: stringVariable.matches("\\d")

dimitrisli · Answer 2 · 2010-12-08T14:38:03.567

8

You can check whether the following is true: "yourStringHere".matches("\\d+")

edited Dec 08 '10 at 14:38

answered Dec 08 '10 at 14:30

dimitrisli

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As a clarification, this checks if your string matches the regular expression "\\d+" which is one or more digits [0-9]. – Seabass77 Jan 31 '19 at 21:27

score 5 · Answer 3 · edited Jul 12 '17 at 06:25

5

String s = "abc123";
for(char c : s.toCharArray()) {
    if(Character.isDigit(c)) {
        return true;
    }
}
return false;

edited Jul 12 '17 at 06:25

Luciano van der Veekens

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answered Dec 08 '10 at 14:48

user489041

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3

try "for(char c : s.toCharArray())" instead. – cytinus Dec 12 '12 at 15:30

score 1 · Answer 4 · answered Nov 13 '12 at 23:19

I use the method matches() from the String class:

    Scanner input = new Scanner(System.in)
    String lectura;
    int number;
    lectura = input.next();
    if(lectura.matches("[0-3]")){
         number = lectura;
    }

This way you can also validate that the range of the numbers is correct.

CoolBeans · Answer 5 · 2015-04-08T18:21:03.993

1

You can use apache StringUtils.isNumeric .

edited Apr 08 '15 at 18:21

answered Dec 08 '10 at 15:27

CoolBeans

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score 0 · Answer 6 · answered Nov 13 '13 at 14:02

0

You could always use Googles Guava

String text = "13567";
CharMatcher charMatcher = CharMatcher.DIGIT;
int output = charMatcher.countIn(text);

answered Nov 13 '13 at 14:02

Ryan Heitner

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fty4 · Answer 7 · 2017-03-10T12:21:46.377

0

That should work:

public static boolean isInteger(String p_str)
{
    if (p_str == null)
        return false;
    else
        return p_str.matches("^\\d*$");
}

edited Mar 10 '17 at 12:21

answered Mar 10 '17 at 12:13

fty4

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score 0 · Answer 8 · edited Dec 08 '10 at 14:50

0

User regular expression:

Pattern.compile("^\\s*\\d+\\s*$").matcher(myString).find();
Just wrap Integer.parse() by try/catch(NumberFormatException)

edited Dec 08 '10 at 14:50

Andreas Dolk

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answered Dec 08 '10 at 14:29

AlexR

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Will the regex recognize `١` as a digit? – Michael Konietzka Dec 08 '10 at 15:02
I am sorry, I do not know this character. What is this? – AlexR Dec 08 '10 at 15:06
It is ARABIC-INDIC DIGIT ONE. Try for yourself: `System.out.println(Integer.valueOf("١"));` – Michael Konietzka Dec 08 '10 at 17:45

score 0 · Answer 9 · answered Dec 08 '10 at 15:54

0

You might also want to have a look at java.util.Scanner

Example:

new Scanner("456").nextInt

answered Dec 08 '10 at 15:54

Pau

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score -1 · Answer 10 · answered Apr 27 '12 at 19:52

-1

int number = 0;
try { 
   number = Integer.parseInt(string); 
}
catch(NumberFormatException e) {}

answered Apr 27 '12 at 19:52

Jon

1

Welcome to stackoverflow. This question is old and was already answered. Typically, it is best not to resurrect stale threads unless your response contributes something significantly new or different over previous answers. – oers Oct 26 '12 at 12:24

score -1 · Answer 11 · edited Aug 10 '18 at 18:41

-1

Use the method Integer.parseInt() at http://docs.oracle.com/javase/10/docs/api/java/lang/Integer.html

edited Aug 10 '18 at 18:41

Nosrep

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answered Dec 08 '10 at 14:27

Freddie

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When linking to a version of the API, I would suggesting using a more up to date version. – jzd Dec 08 '10 at 14:37
`Integer.parseInt("8.0")` throws an exception, despite 8.0 can fit into an int without loss – Phlip Feb 28 '23 at 23:09

score -1 · Answer 12 · answered Dec 08 '10 at 14:57

-1

If you just want to test, if a String contains an integer value only, write a method like this:

public boolean isInteger(String s) {
  boolean result = false;
  try {
    Integer.parseInt("-1234");
    result = true;
  } catch (NumberFormatException nfe) {
    // no need to handle the exception
  }
  return result;
}

parseInt will return the int value (-1234 in this example) or throw an exception.

answered Dec 08 '10 at 14:57

Andreas Dolk

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`NumberFormatException`is a `RuntimeExcption`. Will catching a RuntimeException not be dangerous or "evil", for example when used in transactions, which maybe marked as rollback because a RuntimeException was thrown? – Michael Konietzka Dec 09 '10 at 06:55

How to determine whether a string contains an integer?

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