StringRegExpReplace with @IPAddress1

[m1975michael]

#include <INet.au3>
Local $sIPAddress = @IPAddress1
;Local $sIPAddress = "100.20.30.40"
Local $sIPAddress3 = StringRegExpReplace($sIPAddress, "^(\d{2,3}\.)(\d{2,3}\.)(\d{2,3}\.)(\d{2,3})", "$1$2$3")
MsgBox(0, "", "'" & $sIPAddress3 & "'")

I am having issues getting the expected results from StringRegEx when passing @IPAddress1 into the function.  The expected result would be the first 3 octets.  When I just use Local $sIPAddress = "100.20.30.40" I receive the expected results, 100.20.30  @IPAddress is a string based on the IsString() function.  Any suggestions would be very grateful. 

[mikell]

Your regex will fail on 127.0.0.1, try this

StringRegExpReplace($sIPAddress, "(\d+\.)(\d+\.)(\d+)\.\d+", "$1$2$3")

[Melba23]

m1975michael,

I would do it this way:

#include <INet.au3>
Local $sIPAddress = @IPAddress1
;Local $sIPAddress = "100.20.30.40"
Local $aRet = StringRegExp($sIPAddress, "(\d{1,3})", 3)  ; Note {1,3} - IPs can have single digits
Local $sIPAddress3 = $aRet[0] & "." & $aRet[1] & "." & $aRet[2]
MsgBox(0, "", $sIPAddress & " - " & $sIPAddress3)

Any better?

[m1975michael]

Much better thank you.  I was really stuck.

m1975michael,

I would do it this way:

#include <INet.au3>
Local $sIPAddress = @IPAddress1
;Local $sIPAddress = "100.20.30.40"
Local $aRet = StringRegExp($sIPAddress, "(\d{1,3})", 3)  ; Note {1,3} - IPs can have single digits
Local $sIPAddress3 = $aRet[0] & "." & $aRet[1] & "." & $aRet[2]
MsgBox(0, "", $sIPAddress & " - " & $sIPAddress3)

Any better?

 

[BrewManNH]

A very simple stringsplit would have been easier to code and you wouldn't have to figure out the pattern.

[jguinch]

StringRegExpReplace($sIPAddress, "\.\d+$", "")

[mikell]

BrewManNH,

Obviously right