Java method which check is it a number

Question

I have wrote a method which takes a string and return true if its a valid single integer or floating number or false if its not.

My code:

public static boolean isDigit(String s)
    {
        boolean b;
        try
        {
            Integer.parseInt(s);
            b = true;
        }
        catch (NumberFormatException e)
        {
            b = false;
        }

        try
        {
            Double.parseDouble(s);
            b = true;
        }
        catch (NumberFormatException e)
        {
            b = false;
        }

        return b;
    }

I am sure there is a better way of writing it. Thank you

Possible duplicate of [How to check if a String is numeric in Java](https://stackoverflow.com/questions/1102891/how-to-check-if-a-string-is-numeric-in-java) — Saurabh Bhoomkar, Jun 24 '18 at 16:38
You can start by removing the first half of the method, since you don't do anything with the boolean value that it computes. — JB Nizet, Jun 24 '18 at 16:38
I would suggest to use a lib if you dont have to write yourself: https://stackoverflow.com/questions/1102891/how-to-check-if-a-string-is-numeric-in-java — hasan, Jun 24 '18 at 16:40
If you're using Guava, `return Doubles.tryParse(s) != null;` — shmosel, Jun 24 '18 at 17:30

score 2 · Answer 1 · answered Jun 24 '18 at 16:39

2

You do not need check if it is int, since int number can also be parsed to double. It can be simplified to this:

public static boolean isDigit(String s)
{
    boolean b;

    try
    {
        Double.parseDouble(s);
        b = true;
    }
    catch (NumberFormatException e)
    {
        b = false;
    }

    return b;
}

answered Jun 24 '18 at 16:39

xingbin

27,410
9
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103

Could be simplified further by dropping the `b` and `return`ing `true` and `false` directly. – tobias_k Jun 24 '18 at 16:44

Gaurav Srivastav · Answer 2 · 2018-06-24T16:48:28.303

0

You do this using Regular expression too.

 public static boolean isDigit(String s){
    String regex = "[0-9]*\\.?[0-9]*";
    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(s);
    boolean b = m.matches();
    return b;
  }

edited Jun 24 '18 at 16:48

answered Jun 24 '18 at 16:41

Gaurav Srivastav

2,381
1
15
18

GJCode · Answer 3 · 2018-06-24T17:28:08.427

0

you can simply do this:

 return s.matches("[+-]?\\d+(\\.\\d+)?");

if "." is the separator for decimals

edited Jun 24 '18 at 17:28

answered Jun 24 '18 at 16:41

GJCode

1,959
3
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30

what about negative numbers/unary positive numbers? – Reimeus Jun 24 '18 at 17:03
answer edited, now works for negative numbers too, thanks @Reimeus – GJCode Jun 24 '18 at 17:29

score 0 · Answer 4 · answered Jun 24 '18 at 16:52

0

The lightest solution is this one, also for code readability:

public boolean isDigit(String str) {
     try {
          Integer.parseInt(str)
          return true;
     } catch (NumberFormatException: e) { return false }
}

answered Jun 24 '18 at 16:52

ci0ccarellia

795
9
26

score 0 · Answer 5 · answered Jun 24 '18 at 17:21

0

Use Apache Commons StringUtils.isNumeric() to check if String is valid number

Examples:-

StringUtils.isNumeric("123") = true StringUtils.isNumeric(null) = false StringUtils.isNumeric("") = false StringUtils.isNumeric(" ") = false

answered Jun 24 '18 at 17:21

iqval singh

16
2

Java method which check is it a number

5 Answers5