ASterisk or digit in regex using javascript

Question

My intention is to get the following validated using regex

10.10.*.1
10.10.0.1
10.10.255.1
10.10.10.*
10.10.10.0
10.10.10.255

In simple terms, a star can appear in the last two octets and the IP address would still be valid

My code is as follows:

function ValidateIPaddress(ipaddress) {  
  if (/^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]|\*?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]|\*?)$/.test(ipaddress)) {  
    return (true)  
  }
  return (false)  
}

I am unable to set an or condition in the last two octets. Please help

@anubhava: That is considerably less restrictive than what the OP is aiming for. — Matt Burland
– Matt Burland, Commented Mar 20, 2018 at 16:23
There is very little reason of keeping it so restrictive unless a user is entering this IP :) — anubhava
– anubhava, Commented Mar 20, 2018 at 16:25

ctwheels · Accepted Answer · 2018-03-20 17:04:04Z

Method 1 - regex simplified

No need to overcomplicate the regex. Just ensure that the string conforms with the general format (the following regex) and then test that each octet is < 256 as the code snippet below shows.

^\d{1,3}\.\d{1,3}(?:\.(?:\d{1,3}|\*)){2}$

^ Assert position at the start of the line
\d{1,3} Match a digit 1 to 3 times
\. Match . literally
\d{1,3} Match a digit 1 to 3 times
(?:\.(?:\d{1,3}|\*)){2} Match the following twice
- \. Match . literally
- (?:\d{1,3}|\*) Match either a digit 1 to 3 times or * literally\
$ Assert position at the end of the line

var a = [
  // valid
  "10.10.*.1",
  "10.10.0.1",
  "10.10.255.1",
  "10.10.10.*",
  "10.10.10.0",
  "10.10.10.255",
  // invalid
  "256.1.1.1",
  "*.1.1.1",
  "1.1.1",
  "1.1.1.1.1"
]

var r = /^\d{1,3}\.\d{1,3}(?:\.(?:\d{1,3}|\*)){2}$/

a.forEach(function(ip) {
    console.log(`${ip}: ${r.test(ip) && ip.split('.').every(function(x) { return Number(x) < 256 || x === '*' })}`)
})

Method 2 - no regex

Alternatively, without even using regex:

var a = [
  // valid
  "10.10.*.1",
  "10.10.0.1",
  "10.10.255.1",
  "10.10.10.*",
  "10.10.10.0",
  "10.10.10.255",
  // invalid
  "256.1.1.1",
  "*.1.1.1",
  "1.1.1",
  "1.1.1.1.1"
]

a.forEach(function(ip) {
  var octets = ip.split('.'),
      valid = false
  if(octets.length === 4) {
    if(Number(octets[0]) < 256
      && Number(octets[1]) < 256
      && (Number(octets[2]) < 256 || octets[2] === '*')
      && (Number(octets[3]) < 256 || octets[3] === '*')
    )
      valid = true
  }
  console.log(`${ip}: ${valid}`)
})

Method 3 - single regex

This is the most bloated method, but it seems that's what you're looking for.

^(?:25[0-5]|2[0-4]\d|[01]?\d{1,2})\.(?:25[0-5]|2[0-4]\d|[01]?\d{1,2})(?:\.(?:25[0-5]|2[0-4]\d|[01]?\d{1,2}|\*)){2}$

var a = [
  // valid
  "10.10.*.1",
  "10.10.0.1",
  "10.10.255.1",
  "10.10.10.*",
  "10.10.10.0",
  "10.10.10.255",
  // invalid
  "256.1.1.1",
  "*.1.1.1",
  "1.1.1",
  "1.1.1.1.1"
]

var r = /^(?:25[0-5]|2[0-4]\d|[01]?\d{1,2})\.(?:25[0-5]|2[0-4]\d|[01]?\d{1,2})(?:\.(?:25[0-5]|2[0-4]\d|[01]?\d{1,2}|\*)){2}$/

a.forEach(function(ip) {
  console.log(`${ip}: ${r.test(ip)}`)
})

@anubhava it wasn't enough to detour the OP unfortunately haha

Matt Burland · Accepted Answer · 2018-03-20 16:27:19Z

1

You are missing on the last two octets anything to capture the single digit possibility. Try:

^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]|[0-9]|\*?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]|[0-9]|\*?)$

answered Mar 20, 2018 at 16:27

Matt Burland

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Comments

Scott Sauyet · Accepted Answer · 2018-03-20 16:54:46Z

I think this should do it.

let rx = /^(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9]|\*)\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9]|\*)$/;

const test = str => rx.test(str)

console.log("these should pass")
console.log(test("10.10.*.1"))
console.log(test("10.10.0.1"))
console.log(test("10.10.255.1"))
console.log(test("10.10.10.*"))
console.log(test("10.10.10.0"))
console.log(test("10.10.10.255"))

console.log("------------------------------")
console.log("these should fail")
console.log(test("10.267.0.1"))
console.log(test("10.10.0.1.0.1"))
console.log(test("10.*.0.1"))
console.log(test("*.42.0.1"))

But, as @ctwheels points out, this might be better with less regex.

Collectives™ on Stack Overflow

ASterisk or digit in regex using javascript

3 Answers 3

Method 1 - regex simplified

Method 2 - no regex

Method 3 - single regex

2 Comments

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Collectives™ on Stack Overflow

3 Answers 3

Method 1 - regex simplified

Method 2 - no regex

Method 3 - single regex

2 Comments

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