CODEWITHSUNDEEP
regex
Anand
Anand

Reputation: 111

Regex to match any number except 0 and 1 consisting of max four digits

I tried the following to match any number except 0 and 1 (say, 2 to 9999), but it does not seem to work as desired.

\d[0-9]?[0-9]?[^0-1]*

Upvotes: 4

Views: 25745

Answers (4)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626861

You can match all numbers from 2 to 9999 using

\b(?![01]\b)\d{1,4}\b

Or (if you have individual strings)

^(?![01]$)\d{1,4}$

See demo

The (?!...) is a negative lookahead that is used here to define exceptions.

More details

  • \b - word boundary (if ^ is used - start of the string)
  • (?![01]\b) - a negative lookahead that fails the match if there is 0 or 1 ([01] is a character class that matches a single char from the set defined in the class) as a whole word (or string if $ is used instead of \b)
  • \d{1,4} - 1, 2, 3 or 4 digits
  • \b - a trailing word boundary (no digit, letter or _ can appear immediately to the right, if there can be a letter or _, replace with (?!\d)).

Upvotes: 10

wyzeman
wyzeman

Reputation: 1

This should match anything between 1 and infinity and match 1 or more digit from 1 to 9 trailing by 0 or more 0.

  • 1
  • 10
  • 11
  • 100
  • 101010101
  • 99
  • even 42 works

but it's wont work for 003 004 024:

[1-9]+[0-9]{0,}

Upvotes: -1

Tushar
Tushar

Reputation: 87203

Exclude 0 and 1 from character class in regex.

[2-9]{1}\d{0,3}

This will match all the numbers which does not starts with 0 and 1.

RegEx101 Demo

EDIT

To match all numbers except 0 and 1

[2-9]{1,4}

RegEx101 Demo

Upvotes: 10

LInsoDeTeh
LInsoDeTeh

Reputation: 1038

If you want to get all numbers between 2 and 9999, you'd need to check two cases, either if it's a length-1 number and then exclude cyphers 0 and 1, or allow everything, if the length is > 1. So the solutions would be something like:

(([2-9]{1})|([1-9]{1}[0-9]{1,}))

This does not allow 0 and 1, but allows 111, 322, 20, or 24.

Upvotes: 1

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