Java regex on doubles

Question

I'm trying to check for a double that has a maximum of 13 digits before the decimal point and the decimal point and numbers following it are optional. So the user could write a whole number or a number with decimals.

To start with I had this:

if (check.matches("[0-9](.[0-9]*)?"))

I've been through several pages on Google and haven't had any luck getting it working despite various efforts. My thought was to do it like this but it doesn't work:

[0-9]{1,13}(.[0-9]*)?

How can I do it?

Why don't you just parse the double and check for value < 10000000000000? — Alvin Wong
– Alvin Wong, Commented Jan 28, 2013 at 13:32

John Kugelman · Accepted Answer · 2013-01-28 13:30:29Z

14

Don't forget to escape the dot.

if (check.matches("[0-9]{1,13}(\\.[0-9]*)?"))

answered Jan 28, 2013 at 13:30

John Kugelman

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1 Comment

Duane Over a year ago

Perfect thank you. I tried something similar but I was a bit naive to try use a regex checker online which can't accept "\\". Should have just tried it in the program.

Ivaylo Strandjev · Accepted Answer · 2013-01-28 13:32:41Z

5

First of all you need to escape the dot(in java this would be [0-9]{1,13}(\\.[0-9]*)?). Second of all don't forget there is also another popular representation of doubles - scientific. So this 1.5e+4 is again a valid double number. And lastly don't forget a double number may be negative, or may not have a whole part at all. E.g. -1.3 and .56 are valid doubles.

answered Jan 28, 2013 at 13:32

Ivaylo Strandjev

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2 Comments

Duane Over a year ago

It's for an accounting system exercise where we can't have negative numbers.

Rahul Over a year ago

Any solution that handles all these cases ?

stema · Accepted Answer · 2013-01-28 13:31:40Z

2

You need to escape the dot and you need at least on digit after the dot

[0-9]{1,13}(\\.[0-9]+)?

See it here on Regexr

answered Jan 28, 2013 at 13:31

stema

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Comments

Trenton D. Adams · Accepted Answer · 2014-12-28 05:32:44Z

1

John's answer is close. But a '-' needs to be added, in case it's negative, if you accept negative values. So, it would be modified to -?[0-9]{1,13}(\.[0-9]*)?

answered Dec 28, 2014 at 5:32

Trenton D. Adams

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august0490 · Accepted Answer · 2015-02-17 13:52:31Z

1

if you need to validate decimal with commas and negatives:

Object testObject = "-1.5";
boolean isDecimal = Pattern.matches("^[\\+\\-]{0,1}[0-9]+[\\.\\,]{1}[0-9]+$", (CharSequence) testObject);

Good luck.

answered Feb 17, 2015 at 13:52

august0490

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Deepak · Accepted Answer · 2020-07-09 13:07:03Z

0

You can use \d+(\.\d*)? as below in kotlin language-

    val rawString = "Hi..  2345.97 ..Bye"
    val DOUBLE_PATTERN = Pattern.compile("\\d+(\\.\\d*)?")
    val matcher = DOUBLE_PATTERN.matcher(rawString)
    var filterNumber = 0.0
    if (matcher.find())
        filterNumber = matcher.group() // extracted number- 2345.97

answered Jul 9, 2020 at 13:07

Deepak

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