regex for validation of positve numbers including zeros and 2 places after decimal in java?

Question

("^\d{1,15}(\.\d{1,2})?$")

This is the regex i'm trying to use but java gives syntax error.

^(?!(?:0|0\.0|0\.00)$)[+]?\d+(\.\d|\.\d[0-9])?$

This one works well for numbers like 00.00, 124.03, 0.13 but not for 0.0 and 0.

please modify the regex so that it accepts the following kind of numbers :

123456.00,
12415366.88,
0.23,
0,
0.00,
0.0,
432547,

i.e. only postive numbers which include zero and upto 2 places after decimal

Possible duplicate of [This](http://stackoverflow.com/questions/8609714/regex-greater-than-zero-with-2-decimal-places) question except you have to do some modification for 0. — Vimal Bera, Aug 23 '13 at 04:57
The compilation error in your first example can be fixed by properly escaping the backslashes in `\d` to `\\d`. The second regex specifically excludes `0`, `0.0`, and `0.00`. You should attempt to understand your code before asking questions about it here. — FThompson, Aug 23 '13 at 04:57
why do you need to use regex? try something like this http://stackoverflow.com/questions/50532/how-do-i-format-a-number-in-java — Sean F, Aug 23 '13 at 04:58

score 1 · Answer 1 · answered Aug 23 '13 at 05:12

public class RegexTest {

public static void main(String[] args) {
    String regexExpression = "([0-9]+[.]?|[0-9]*[.][0-9]{0,2})";

    // True examples
    System.out.println("123456.00".matches(regexExpression));

    System.out.println("12415366.88".matches(regexExpression));

    System.out.println("0".matches(regexExpression));

    System.out.println("0.0".matches(regexExpression));
    System.out.println("0.00".matches(regexExpression));
    System.out.println("432547".matches(regexExpression));

    System.out.println("00.00".matches(regexExpression));

    System.out.println("124.03".matches(regexExpression));

    // False examples
    System.out.println("124.033".matches(regexExpression));
    System.out.println("-124.03".matches(regexExpression));
}

}

Matches `1.`, matches `.2`, matches `.` too! :) – Ravi K Thapliyal Aug 23 '13 at 05:28 — Ravi K Thapliyal, Aug 23 '13 at 05:28

Bohemian · Accepted Answer · 2013-08-23T05:49:25.620

1

Your first regex is the best, but remember that in java you must use a double backslash to code a literal backslash, so:

str.matches("\\d{1,15}(\\.\\d{1,2})?")

Note that with matches(), you don't need the leading ^ or trailing $, because the expression must match the whole string anyway to return true.

The syntax error was probably because \d is not a valid escape sequence, whereas \n etc is valid.

edited Aug 23 '13 at 05:49

answered Aug 23 '13 at 05:19

Bohemian

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Ruchira Gayan Ranaweera · Answer 3 · 2013-08-23T05:20:51.397

0

I don't know why regex here. You can try in this way

  String num="253.65";
    try{
        double d=Double.parseDouble(num);
        if(d==0.0){
            System.out.println("valid");
        }else if(d>0&&(num.split("\\.")[1].length()==2)){
            System.out.println(num+" is valid");
        }else{
            System.out.println(num+" is invalid");
        }
    } catch (NumberFormatException e){
        System.out.println(num+"is not a valid number");
    }

Live Demo

edited Aug 23 '13 at 05:20

answered Aug 23 '13 at 05:03

Ruchira Gayan Ranaweera

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If it's not a valid string representation of a double, an exception will be thrown. – FThompson Aug 23 '13 at 05:03
At this point, regex offers a far more clear and compact solution. – FThompson Aug 23 '13 at 05:07

regex for validation of positve numbers including zeros and 2 places after decimal in java?

3 Answers3