Change 3rd octet of IP in string format using PowerShell - regex

Question

Think I've found the worst way to do this:
$ip = "192.168.13.1"
$a,$b,$c,$d = $ip.Split(".")
[int]$c = $c
$c = $c+1
[string]$c = $c
$newIP = $a+"."+$b+"."+$c+"."+$d
$newIP
But what is the best way? Has to be string when completed. Not bothered about validating its a legit IP.

Answer 1

Using your example for how you want to modify the third octet, I'd do it pretty much the same way, but I'd compress some of the steps together:
$IP = "192.168.13.1"
$octets = $IP.Split(".") # or $octets = $IP -split "\."
$octets[2] = [string]([int]$octets[2] + 1) # or other manipulation of the third octet
$newIP = $octets -join "."
$newIP

Answer 2

You can simply use the -replace operator of PowerShell and a look ahead pattern. Look at this script below
Set-StrictMode -Version "2.0"
$ErrorActionPreference="Stop"
cls
$ip1 = "192.168.13.123"
$tests=#("192.168.13.123" , "192.168.13.1" , "192.168.13.12")
foreach($test in $tests)
{
$patternRegex="\d{1,3}(?=\.\d{1,3}$)"
$newOctet="420"
$ipNew=$test -replace $patternRegex,$newOctet
$msg="OLD ip={0} NEW ip={1}" -f $test,$ipNew
Write-Host $msg
}
This will produce the following:
OLD ip=192.168.13.123 NEW ip=192.168.420.123
OLD ip=192.168.13.1 NEW ip=192.168.420.1
OLD ip=192.168.13.12 NEW ip=192.168.420.12
How to use the -replace operator?
https://powershell.org/2013/08/regular-expressions-are-a-replaces-best-friend/
Understanding the pattern that I have used
The (?=) in \d{1,3}(?=.\d{1,3}$) means look behind.
The (?=.\d{1,3}$ in \d{1,3}(?=.\d{1,3}$) means anything behind a DOT and 1-3 digits.
The leading \d{1,3} is an instruction to specifically match 1-3 digits
All combined in plain english "Give me 1-3 digits which is behind a period and 1-3 digits located towards the right side boundary of the string"
Look ahead regex
https://learn.microsoft.com/en-us/dotnet/standard/base-types/regular-expression-language-quick-reference
CORRECTION
The regex pattern is a look ahead and not look behind.

Answer 3

If you have PowerShell Core (v6.1 or higher), you can combine -replace with a script block-based replacement:
PS> '192.168.13.1' -replace '(?<=^(\d+\.){2})\d+', { 1 + $_.Value }
192.168.14.1
Negative look-behind assertion (?<=^(\d+\.){2}) matches everything up to, but not including, the 3rd octet - without considering it part of the overall match to replace.
(?<=...) is the look-behind assertion, \d+ matches one or more (+) digits (\d), \. a literal ., and {2} matches the preceding subexpression ((...)) 2 times.
\d+ then matches just the 3rd octet; since nothing more is matched, the remainder of the string (. and the 4th octet) is left in place.
Inside the replacement script block ({ ... }), $_ refers to the results of the match, in the form of a [MatchInfo] instance; its .Value is the matched string, i.e. the 3rd octet, to which 1 can be added.
Data type note: by using 1, an implicit [int], as the LHS, the RHS (the .Value string) is implicitly coerced to [int] (you may choose to use an explicit cast).
On output, whatever the script block returns is automatically coerced back to a string.
If you must remain compatible with Windows PowerShell, consider Jeff Zeitlin's helpful answer.

Answer 4

For complete your method but shortly :
$a,$b,$c,$d = "192.168.13.1".Split(".")
$IP="$a.$b.$([int]$c+1).$d"

Answer 5

function Replace-3rdOctet {
Param(
[string]$GivenIP,
[string]$New3rdOctet
)
$GivenIP -match '(\d{1,3}).(\d{1,3}).(\d{1,3}).(\d{1,3})' | Out-Null
$Output = "$($matches[1]).$($matches[2]).$New3rdOctet.$($matches[4])"
Return $Output
}
Copy to a ps1 file and dot source it from command line, then type
Replace-3rdOctet -GivenIP '100.201.190.150' -New3rdOctet '42'
Output: 100.201.42.150
From there you could add extra error handling etc for random input etc.

Answer 6

here's a slightly different method. [grin] i managed to not notice the answer by JeffZeitlin until after i finished this.
[edit - thanks to JeffZeitlin for reminding me that the OP wants the final result as a string. oops! [*blush*]]
what it does ...
splits the string on the dots
puts that into an [int] array & coerces the items into that type
increments the item in the targeted slot
joins the items back into a string with a dot for the delimiter
converts that to an IP address type
adds a line to convert the IP address to a string
here's the code ...
$OriginalIPv4 = '1.1.1.1'
$TargetOctet = 3
$OctetList = [int[]]$OriginalIPv4.Split('.')
$OctetList[$TargetOctet - 1]++
$NewIPv4 = [ipaddress]($OctetList -join '.')
$NewIPv4
'=' * 30
$NewIPv4.IPAddressToString
output ...
Address : 16908545
AddressFamily : InterNetwork
ScopeId :
IsIPv6Multicast : False
IsIPv6LinkLocal : False
IsIPv6SiteLocal : False
IsIPv6Teredo : False
IsIPv4MappedToIPv6 : False
IPAddressToString : 1.1.2.1
==============================
1.1.2.1