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After years of sponging knowledge from the SO community, I'm here to ask my first question.

I'm trying to create a regex expression for a JavaScript web project to find a whole number (positive or negative) with a thousands separator, or just a whole number, in a given string.

Some valid examples:

561,085 
3,894,320 
-59,099 
1,000 
1000 
-1000 
1 
-1 
0

Invalid examples:

-01
01393
01,300,00
-04,044

I've created this expression so far:

\b(?:[0])|(?:(((-)?[1-9]?\d{2}|\d)((,)?\d{3})|((-)?[1-9]+\d*)))\b

(?:[0]) - match a leading 0

|(?:(((-)?[1-9?\d{2}|\d)((,)?\d{3}) - or match a 1-9 starting number with a comma and 3 digits after

|((-)?[1-9]+\d*))) - or just match a whole 1-9 starting number without any commas

I think it works in all cases except for when I test a string with 5 digits trailing a minus sign, for example like so:
Value: -01110

In this case it accepts 01110 as a match.

Could anyone help me figure out what I've gotten wrong with the expression?
Also would not mind some general pointers/feedback on how I popped the ask-a-question-cherry here, cheers!

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1 Answer 1

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3

You could use

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

See a demo on regex101.com.

That means:

^                             # start of the string
(?:                           # open par
    0                         # zero
    |                         # or
    -?[1-9]\d*                # -100 or 234 or -121323232
    |                         # or
    -?[1-9]\d{0,2}(?:,\d{3})+ # separated by comma
)                             # close par
$                             # end of the string

To actually use it "in the wild" (that is in any given text), you need to adjust the anchors to lookarounds:

(?<=^|\s)(?:0|-?[1-9]\d*|-?[1-9]\d{0,2}(?:,\d{3})+)(?=\s|$)
# ^^1^^                                              ^^2^^

^^1^^ means "either the start of the string or some whitespaces in front of the expression", ^^2^^ means "either some whitespaces or the end of the string right after the expression".

See another demo on regex101.com.

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5 Comments

This is great! The expression is more neat than mine, too. I replaced "^" with "\b" to make it look through a given string, I'm assuming it will not produce any odd side effects? Thank you for your reply! -Edit: I need at least 15 rep to upvote your answer, will return later and upvote, haha
I'd would not use \b here, see regex101.com/r/aFvsFT/1 in comparison to regex101.com/r/2gWiyL/1 (the one at the end of the answer). Btw., you can accept answers which helped you (green tick on the left).
I don't think javascript supports \Z. Am I wrong?
@PoulBak: You're right, thanks for spotting it. I always test it in PCRE first.
@Jan gotcha! using your adjusted string now and it's perfect. accepted your answer, cheers

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