Regex that matches integers in between whitespace or start/end of string only

Question

I'm currently using the pattern: \b\d+\b, testing it with these entries:

numb3r 
2
3454 
3.214
test

I only want it to catch 2, and 3454. It works great for catching number words, except that the boundary flags (\b) include "." as consideration as a separate word. I tried excluding the period, but had troubles writing the pattern.

Basically I want to remove integer words, and just them alone.

In which context do you use regex ? grep ? vim ? perl ? other... — Luc M, Jan 28 '12 at 06:57

score 118 · Answer 1 · answered Jan 28 '12 at 07:03

118

All you want is the below regex:

^\d+$

answered Jan 28 '12 at 07:03

manojlds

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Works until you need to handle integers that are negative. – davidjb Jul 04 '13 at 02:22
11

This also will match leading zeros. Which i wouldnt expect you would want. – samsamm777 Aug 27 '15 at 11:11
1

Leading zeros aren't invalid, are they? They're meaningless, yes, but I'm not sure that they're invalid. – felwithe Aug 08 '20 at 13:56

ghiscoding · Answer 2 · 2016-12-15T14:17:20.367

68

Similar to manojlds but includes the optional negative/positive numbers:

var regex = /^[-+]?\d+$/;

EDIT

If you don't want to allow zeros in the front (023 becomes invalid), you could write it this way:

var regex = /^[-+]?[1-9]\d*$/;

EDIT 2

As @DmitriyLezhnev pointed out, if you want to allow the number 0 to be valid by itself but still invalid when in front of other numbers (example: 0 is valid, but 023 is invalid). Then you could use

var regex = /^([+-]?[1-9]\d*|0)$/

edited Dec 15 '16 at 14:17

answered Feb 02 '14 at 01:28

ghiscoding

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Still allows "023", doesn't it? – Andrei Duma Aug 07 '16 at 23:18
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I'd improve it liek this: `var regex = /^([+-]?[1-9]\d*|0)$/` – Dmitriy Lezhnev Oct 11 '16 at 14:50
DmitriyLezhnev's comment (directly above mine) is the best since it allows just plain 0 as well. – Kosta Kontos Dec 15 '16 at 13:53

score 43 · Accepted Answer · answered Jan 28 '12 at 06:57

43

You could use lookaround instead if all you want to match is whitespace:

(?<=\s|^)\d+(?=\s|$)

answered Jan 28 '12 at 06:57

Amber

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To catch negative numbers you can add [-+]? to catch negative numbers too. The regex becomes: (?<=\s|^)[-+]?\d+(?=\s|$) – Brad Campbell Jul 06 '12 at 17:30
1

overcomplicated – Sliq Aug 26 '19 at 09:09
Linter does not like this – Devin B. Jul 04 '20 at 21:59

ntjess · Answer 4 · 2018-07-24T18:37:36.373

I would add this as a comment to the other good answers, but I need more reputation to do so. Be sure to allow for scientific notation if necessary, i.e. 3e4 = 30000. This is default behavior in many languages. I found the following regex to work:

/^[-+]?\d+([Ee][+-]?\d+)?$/;
//          ^^              If 'e' is present to denote exp notation, get it
//             ^^^^^          along with optional sign of exponent
//                  ^^^       and the exponent itself
//        ^            ^^     The entire exponent expression is optional

score 4 · Answer 5 · answered Mar 28 '19 at 12:05

4

This just allow positive integers.

^[0-9]*[1-9][0-9]*$

answered Mar 28 '19 at 12:05

Muzaffar Mahmood

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score 3 · Answer 6 · answered Jul 14 '20 at 06:02

3

This solution matches integers:

Negative integers are matched (-1,-2,etc)
Single zeroes are matched (0)
Negative zeroes are not (-0, -01, -02)
Empty spaces are not matched ('')

/^(0|-*[1-9]+[0-9]*)$/

answered Jul 14 '20 at 06:02

egremont of yorke

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score 1 · Answer 7 · answered Feb 13 '17 at 08:43

1

^([+-]?[0-9]\d*|0)$

will accept numbers with leading "+", leading "-" and leadings "0"

answered Feb 13 '17 at 08:43

Zam

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score 1 · Answer 8 · answered Sep 26 '18 at 09:59

1

Try /^(?:-?[1-9]\d*$)|(?:^0)$/.

It matches positive, negative numbers as well as zeros.

It doesn't match input like 00, -0, +0, -00, +00, 01.

Online testing available at http://rubular.com/r/FlnXVL6SOq

answered Sep 26 '18 at 09:59

yivo

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I think it should catch `01`. – Nouman Sep 26 '18 at 10:01
@BlackThunder It doesn't. In JS console `/^(?:-?[1-9]\d*$)|(?:^0)$/.test('01')` returns false. – yivo Sep 26 '18 at 10:04
Maybe he meant 01. In any case it doesn't catch the trailing period as long as there isn't a number proceeding it. – Devin B. Jul 06 '20 at 22:12

Daniele D. · Answer 9 · 2017-02-13T09:38:23.637

This worked in my case where I needed positive and negative integers that should NOT include zero-starting numbers like 01258 but should of course include 0

^(-?[1-9]+\d*)$|^0$

Example of valid values: "3", "-3", "0", "-555", "945465464654"

Example of not valid values: "0.0", "1.0", "0.7", "690.7", "0.0001", "a", "", " ", ".", "-", "001", "00.2", "000.5", ".3", "3.", " -1", "+100", "--1", "-.1", "-0", "00099", "099"

score 0 · Answer 10 · answered Sep 15 '16 at 09:54

0

^(-+)?[1-9][0-9]*$ starts with a - or + for 0 or 1 times, then you want a non zero number (because there is not such a thing -0 or +0) and then it continues with any number from 0 to 9

answered Sep 15 '16 at 09:54

Evangelos

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Regex that matches integers in between whitespace or start/end of string only

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