regex - check for decimal(javascript)

Question

I got this expression from stackoverflow itself - /^\d+(\.\d{0,9})?$/.

It takes care of:

but fails on .23. Can that be added to the above expression, or something that would take care of all of them?

Do you care about false positives? If not a simple \.? at the beginning will do it. — NorthGuard
– NorthGuard, Commented Jun 8, 2011 at 20:58
You just need to change the first "+" to "*". Plus sign means ONE(1) or more while star means NONE(0) or more. This also doesn't take into account the plus(+) sign or minus(-) sign that can be in front of numbers. That would be "[+-]?" or it might be "[\+\-]?" if they have special meaning inside of the brackets. Then you also have the "e+10" stuff. — Mark Manning
– Mark Manning, Commented Mar 3, 2015 at 15:08

Ben Roux · Accepted Answer · 2015-08-14 06:15:54Z

31

This will capture every case you posted as well as .23

Limit to 9 decimal places

var isDecimal = string.match( /^(\d+\.?\d{0,9}|\.\d{1,9})$/ );

No decimal limits:

var isDecimal = string.match( /^(\d+\.?\d*|\.\d+)$/ );

edited Aug 14, 2015 at 6:15

answered Jun 8, 2011 at 21:01

Ben Roux

7,4161 gold badge20 silver badges21 bronze badges

Sign up to request clarification or add additional context in comments.

9 Comments

Sean Vieira Over a year ago

Limit the decimal places to nine, and it's perfect.

KooiInc Over a year ago

SyntaxError: Invalid regular expression: /^(\d+\.?\d{0,9}|\.\d{1,9}$/: Unterminated group

Ben Roux Over a year ago

@Kooilnc: My bad! I messed it up editing in the decimal limit of 9. Try now

KooiInc Over a year ago

@Ben: yep, but then again it will not capture 3.141592653589793238462643383279502884197169399375105820974 ;~)

Justin Morgan Over a year ago

@KooiInc - The OP's pattern imposes the 9-decimal-places limit. He doesn't mention the limit specifically, but it looks intentional.

|

Justin Morgan · Accepted Answer · 2019-04-18 15:45:15Z

This covers all your examples, allows negatives, and enforces the 9-digit decimal limit:

/^[+-]?(?=.?\d)\d*(\.\d{0,9})?$/

Live demo: https://regexr.com/4chtk

To break that down:

[+-]?       # Optional plus/minus sign (drop this part if you don't want to allow negatives)
(?=.?\d)    # Must have at least one numeral (not an empty string or just `.`)
\d*         # Optional integer part of any length
(\.\d{0,9}) # Optional decimal part of up to 9 digits

Since both sides of the decimal point are optional, the (?=.?\d) makes sure at least one of them is present. So the number can have an integer part, a decimal part, or both, but not neither.

One thing I want to note is that this pattern allows 23., which was in your example. Personally, I'd call that an invalid number, but it's up to you. If you change your mind on that one, it gets a lot simpler (demo):

/^[+-]?\d*\.?\d{1,9}$/

Andrew Hare · Accepted Answer · 2011-06-08 20:57:53Z

1

Try this expression:

^\d*\.?\d*$

answered Jun 8, 2011 at 20:57

Andrew Hare

353k75 gold badges649 silver badges642 bronze badges

1 Comment

Paulius Over a year ago

It should be ^\d*\.?\d+$

Collectives™ on Stack Overflow

regex - check for decimal(javascript)

3 Answers 3

9 Comments

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

9 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related