What is the regex for "Any positive integer, excluding 0"

Question

How can ^\d+$ be improved to disallow 0?

EDIT (Make it more concrete):

Examples to allow:
1
30
111
Examples to disallow:
0
00
-22

It doesn't matter if positive numbers with a leading zero are allowed or not (e.g. 022).

This is for Java JDK Regex implementation.

Answer 1

Ugly, but match the exact range 1..2147483647

^(214748364[0-7]|21474836[0-3]\d|2147483[0-5]\d{2}|214748[0-2]\d{3}|21474[0-7]\d{4}|2147[0-3]\d{5}|214[0-6]\d{6}|21[0-3]\d{7}|20\d{8}|1\d{9}|[1-9]\d{0,8})$

Note:

• 2000000000 to 2147483647 -> 214748364[0-7]|21474836[0-3]\d|2147483[0-5]\d{2}|214748[0-2]\d{3}|21474[0-7]\d{4}|2147[0-3]\d{5}|214[0-6]\d{6}|21[0-3]\d{7}|20\d{8}
• 1000000000 to 1999999999 -> 1\d{9}
• 1 to 999999999 -> [1-9]\d{0,8}

Answer 2

Try this:

^[1-9]\d*$

...and some padding to exceed 30 character SO answer limit :-).

Here is Demo

Answer 3

Any positive integer, excluding 0: ^\+?[1-9]\d*$
Any positive integer, including 0: ^(0|\+?[1-9]\d*)$

Answer 4

My pattern is complicated, but it covers exactly "Any positive integer, excluding 0" (1 - 2147483647, not long). It's for decimal numbers and doesn't allow leading zeros.

^((1?[1-9][0-9]{0,8})|20[0-9]{8}|(21[0-3][0-9]{7})|(214[0-6][0-9]{6})
|(2147[0-3][0-9]{5})|(21474[0-7][0-9]{4})|(214748[0-2][0-9]{3})
|(2147483[0-5][0-9]{2})|(21474836[0-3][0-9])|(214748364[0-7]))$

Answer 5

This RegEx matches any Integer positive out of 0:

(?<!-)(?<!\d)[1-9][0-9]*

It works with two negative lookbehinds, which search for a minus before a number, which indicates it is a negative number. It also works for any negative number larger than -9 (e.g. -22).

Answer 6

This should only allow decimals > 0

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

Answer 7

Try this one, this one works best to suffice the requiremnt.

[1-9][0-9]*

Here is the sample output

String 0 matches regex: false
String 1 matches regex: true
String 2 matches regex: true
String 3 matches regex: true
String 4 matches regex: true
String 5 matches regex: true
String 6 matches regex: true
String 7 matches regex: true
String 8 matches regex: true
String 9 matches regex: true
String 10 matches regex: true
String 11 matches regex: true
String 12 matches regex: true
String 13 matches regex: true
String 14 matches regex: true
String 15 matches regex: true
String 16 matches regex: true
String 999 matches regex: true
String 2654 matches regex: true
String 25633 matches regex: true
String 254444 matches regex: true
String 0.1 matches regex: false
String 0.2 matches regex: false
String 0.3 matches regex: false
String -1 matches regex: false
String -2 matches regex: false
String -5 matches regex: false
String -6 matches regex: false
String -6.8 matches regex: false
String -9 matches regex: false
String -54 matches regex: false
String -29 matches regex: false
String 1000 matches regex: true
String 100000 matches regex: true

Answer 8

^\d*[1-9]\d*$

this can include all positive values, even if it is padded by Zero in the front

Allows

1

01

10

11 etc

do not allow

0

00

000 etc..

Answer 9

^[1-9]*$ is the simplest I can think of

Answer 10

You might want this (edit: allow number of the form 0123):

^\\+?[1-9]$|^\\+?\d+$

however, if it were me, I would instead do

int x = Integer.parseInt(s)
if (x > 0) {...}

Answer 11

You might try a negative lookahead assertion:

^(?!0+$)\d+$

Answer 12

Just for fun, another alternative using lookaheads:

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

As many digits as you want, but at least one must be [1-9].

Answer 13

Sorry to come in late but the OP wants to allow 076 but probably does NOT want to allow 0000000000.

So in this case we want a string of one or more digits containing at least one non-zero. That is

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

Answer 14

Got this one:

^[1-9]|[0-9]{2,}$

Someone beats it? :)