Substring IP-Address with adding a string generates error but works

0
votes

I have following command:

$IP_start = $First.Substring(0, $First.LastIndexOf('.'))+ ".200"

$First is an IP-Address, for example 192.168.0.1

I want to change the 1 in the fourth octect into a 200.

Write-Output $IP_start

gives me the correct IP-Address 192.168.0.200, but at the same time I get the following Error:

Ausnahme beim Aufrufen von "Substring" mit 2 Argument(en): "Die Länge darf nicht kleiner als 0 (null) sein. Parametername: length" In *ps1:31 Zeichen:3 + $IP_start = $First.Substring(0, $First.LastIndexOf('.'))+ ".200" + ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + CategoryInfo : NotSpecified: (:) [], MethodInvocationException + FullyQualifiedErrorId : ArgumentOutOfRangeException

English translation

Exception when calling "substring" with 2 arguments: "The length can not be less than zero. Parameter name: length" In * ps1: 31 characters: 3...

I think everything is working fine, but that error messages bothers me.

//edit:

There's an ip.txt, where each line is like "192.168.0.1; ABCDEF"

$txt = Get-Content ip.txt
$editline = foreach ($Data in $txt) {
  $First, $Second = $Data -split ';' -replace '^\s*|\s*$'

$IP_start = $First.Substring(0, $First.LastIndexOf('.'))+ ".200"
Write-Output "modify ipaddr_first $IP_start"
}

$editline | Out-File "$output"

$first is therefore "192.168.0.1" and $second is "ABCDEF".

3
powershellsubstringip-addresslastindexof
What is the result of $First.LastIndexOf('.') on its own? I cannot reproduce your issue so something is different. LastIndex of will return -1 if it did not find anything so that would cause the issue. You tell us what $first is but we have to guess how you populated it. In either case I would say use the [ipaddress] type to help with this stackoverflow.com/questions/30061062/… – Matt
Substring is important to the question so it should be left in the title. Just makes a poor verb – Matt
What problem are you solving? Extracting a substring of an IP address string doesn't tell you a network ID (that depends on the network mask). – Bill_Stewart
I just updated my opening post. I explain how $first is filled with values – mr.kanister

3 Answers

2
votes

You can use regular expressions

$IP_start = "192.168.1.1"
$IP_start -replace "\d{1,3}$","200"

This will change any 192.168.1.xxx by 192.168.1.200

1
votes
$IP_start = "192.168.1.1"
(($IP_start -split "\.")[0..2] -join ".") + ".200"
0
votes

Creating a Synthetic Method for an IPAddress Object , a fun way of doing this... 

$Method = {
Param(
[Int]$Number,

[ValidateSet(1,2,3,4)]
[Int]$Position
)

$OctetPosition = $Position - 1
$CurrentIpAddress = New-Object -TypeName System.Collections.ArrayList
$This.IpAddressToString -Split '\.' | ForEach-Object -Process {
    $CurrentIpAddress.Add($_) | Out-Null
}

$ChangedOctet = [Int]($CurrentIpAddress[$OctetPosition]) + $Number

if( $ChangedOctet -gt 255 ){
    throw "Resulting octet is $ChangedOctet which is greater than 255"
}

$CurrentIpAddress.Item($OctetPosition) = $ChangedOctet


return ($CurrentIpAddress -join '.')

}

$IPAddress = [System.Net.IpAddress]'192.168.1.1'
$IPAddress | Add-Member -MemberType ScriptMethod -Name Add -Value $Method

#$IPaddress.Add($Number,$OctetPosition)
$IPAddress.Add(200,4)