Java regex to check if string is valid number format (comma and decimal point placing)

Question

1000 - valid
1,000 - valid
1,000.00 - valid
1000.00 - valid
1000.00.00 - invalid
1,0.00 - invalid
1,000,00.00 - invalid
1,000,000.12 - valid

no of decimal places can be unlimited

I've been trying to find the right regex pattern, can't seem to find one that will accomodate all validations. Can anyone help

the pattern ^[1-9]\d{0,2}(.\d{3})*(,\d+)?$ did not work for me, based from the similar thread here

Possible duplicate of Decimal or numeric values in regular expression validation — royB
– royB, Commented Mar 15, 2017 at 23:32
the pattern ^[1-9]\d{0,2}(.\d{3})*(,\d+)?$ did not work for me, based from that similar thread — Hingle McJingleberry
– Hingle McJingleberry, Commented Mar 15, 2017 at 23:38
@Alain Del Rosario try using ^[1-9]\d*(\,\d+)?$ . You can try the regex script with sample input on this site regexplanet.com/advanced/java/index.html — Arthur Decker
– Arthur Decker, Commented Mar 15, 2017 at 23:41
i think validations from that link treats commas as required not optional — Hingle McJingleberry
– Hingle McJingleberry, Commented Mar 15, 2017 at 23:54

JUAN CALVOPINA M · Accepted Answer · 2017-03-16 23:47:29Z

You should try this expression:

^\d{1,3}|\d(([ ,]?\d{3})*([.,]\d{2}+)?$)

With this expression is covered with the scenarios raised.

Here the complete example:

public class Decimal {

    private static String REGEX = "^\\d{1,3}|\\d(([ ,]?\\d{3})*([.,]\\d{2}+)?$)";

    public static void main(String[] args) {
        String data[] = {"1000", "1,000", "1,000.00", "1000.00", "1000.00.00", "1,0.00", "1,000,00.00", "1,000,000.12"};

        Pattern.compile(REGEX);

        for (int i = 0; i < data.length; i++) {
            if (data[i].matches(REGEX)) {
                System.out.println(data[i] + " - valid");
            } else {
                System.out.println(data[i] + " - invalid");
            }
        }
    }

}

The output:

1000 - valid
1,000 - valid
1,000.00 - valid
1000.00 - valid
1000.00.00 - invalid
1,0.00 - invalid
1,000,00.00 - invalid
1,000,000.12 - valid

streetturtle · Accepted Answer · 2017-03-16 00:40:48Z

3

This is one of possible regexes you are looking for:

^\d{1,3}([ ,]?\d{3})*([.,]\d+)?$

Demo: https://regex101.com/r/T8tcDP/1

answered Mar 16, 2017 at 0:40

streetturtle

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2 Comments

johnw182 Over a year ago

This pattern shows "10,00" as valid and ".05" as invalid.

johnw182 Over a year ago

Also negatives dont work with this.

user557597 · Accepted Answer · 2017-03-16 00:14:29Z

0

This would match your numbers there (?:\d+(?:,\d{3})*(?:\.\d{2})?|\.\d{2})
It would also match .00 just incase. If you don't want it to, just remove
the |\.\d{2} part.

Add your own boundary constructs as needed ^$ or \b

Expanded

 (?:
      \d+ 
      (?: , \d{3} )*
      (?: \. \d{2} )?
   |  \. \d{2} 
 )

answered Mar 16, 2017 at 0:14

user557597

Comments

Anonymous · Accepted Answer · 2017-03-16 00:15:00Z

0

My suggestion is:

^[1-9]((\d{0,2}(,\d{3})*(\.(\d{3},)*\d{1,3})?)|(\d*(\.\d+)?))$

A number either has separators for every 3 digits or it hasn’t (no middle forms). It either has a decimal point and at least one digit after it, or it hasn’t. In all cases does it start with a non-zero digit.

Remember that a dot (period, .) has a special meaning in regex and therefore needs to be escaped to get a literal point.

answered Mar 16, 2017 at 0:15

Anonymous

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Comments

Daniil Bannov · Accepted Answer · 2023-10-09 20:32:24Z

0

this solution work also for one digit after dot:

^\d{1,3}(,{1}\d{3})*(\.\d{1,2})?$

100.1 - valid
1.11 - valid
1,000123,123.1 - invalid

good luck.

answered Oct 9, 2023 at 20:32

Daniil Bannov

162 bronze badges

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