Regex for ip validation

Question

Hi my regex pattern is

^((?:(?:^|\.)(?:\d|[1-9]\d|1\d{2}|2[-4]\d|25[0-5])){4})$

this allow 0.0.0.0 but i do not want to allow 0.0.0.0 please someone help me

Do you need to allow: `1.0.0.0`, `0.1.0.0`, `0.0.1.0`, `0.0.0.1`? (in other words, if there is only one number not equal to zero, can it be at any position?) — Casimir et Hippolyte, May 26 '15 at 09:23

score 2 · Answer 1 · answered May 26 '15 at 08:51

2

Just add a negative lookahead assertion at the start.

^(?!0+\.0+\.0+\.0+$)((?:(?:^|\.)(?:\d|[1-9]\d|1\d{2}|2[-4]\d|25[0-5])){4})$

DEMO

answered May 26 '15 at 08:51

Avinash Raj

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score 2 · Accepted Answer · answered May 26 '15 at 08:51

2

^(?=.*[1-9])((?:(?:^|\.)(?:\d|[1-9]\d|1\d{2}|2[-4]\d|25[0-5])){4})$

You can do this through positive lookahead.See demo.

https://regex101.com/r/yW3oJ9/8

answered May 26 '15 at 08:51

vks

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2

I know it might not be the case, but this regex is prone to catastrophic backtracking in some very [weird scenarios like this](https://regex101.com/r/yE3oO3/1). Perhaps, the string is not a user's input and will never occur. However, my suggestion works ok even in such cases. Avinash's is also performing great. In general, I'd avoid `.*` if it is not necessary. – Wiktor Stribiżew May 26 '15 at 09:19
@stribizhev hehehe.........i suspect, that ip address would be possible in year 3050 when population is billions :P – vks May 26 '15 at 09:23

score 0 · Answer 3 · answered May 26 '15 at 09:04

A look-ahead can be used to set a length limit. In this case, it can be quite concise, since the string should only contain digits and . symbol.

Thus, I suggest using (?![0.]+$) as we only need to check if we have no 0s and periods up to the end:

^(
 (?![0.]+$)                  # Here is the look-ahead.
 (?:
   (?:^|\.)
   (?:
      \d|[1-9][0-9]|1\d{2}|2[-4]\d|25[0-5]
   )
 ){4}
)$

See demo

A one-line JS version:

^((?![0.]+$)(?:(?:^|\.)(?:\d|[1-9][0-9]|1\d{2}|2[-4]\d|25[0-5])){4})$

floum · Answer 4 · 2015-05-26T17:43:13.980

0

Is this part of an assignment?

If not, I would use no regex and prefer String#scan :

ip_subnets = ip.scan(/\d+/).map(&:to_i) # you can use split('.') instead of scan
ip_subnets.all? { |i| (0..255).include?(i) } && 
  ip_subnets.any? { |i| i != 0 } && 
  ip_subnets.size == 4

edited May 26 '15 at 17:43

answered May 26 '15 at 13:56

floum

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Regex for ip validation

4 Answers4