If Computer Name Contains

Question

I need to create a batch file that moves computer name object in Active Directory only if computer name contains some string, like:

If %computername% contains "LAP" 
( dsmove "CN=%computername%,CN=Computers,DC=domain,DC=local" -newparent"OU=**Laptops**,OU=Computers,OU=Company,DC=domain,DC=local" )

    If %computername% contains "DESK" 
        (dsmove "CN=%computername%,CN=Computers,DC=domain,DC=local" -newparent "OU=**Desktops**,OU=Computers,OU=Company,DC=domain,DC=local" )

What is the correct command please?

npocmaka · Answer 1 · 2013-06-25T14:29:19.173

6

set check_computername=%computername:LAP=%
if "%check_computername%" EQU "%computername%" (
  echo computer name contains "LAP"
) else (
  echo computer name does not contain "LAP"
)

You can put your stuff in if and else blocks.

Case Insensitive solution:

echo %check_computername%| Find /I "LAP" >nul 2>&1 || echo does not contain LAP
echo %check_computername%| Find /I "LAP" >nul 2>&1 && echo does not contain LAP

edited Jun 25 '13 at 14:29

answered Jun 25 '13 at 08:08

npocmaka

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1

there is a logical error: change EQU to NEQ or adapt the ECHOs – Stephan Jun 25 '13 at 08:27
No. My computer name contains "LAP" in it. I tried to run this script and it says computer name does not contain "LAP". – Nick Jun 25 '13 at 09:18
@npocmaka solution should work beware that the substitution is case-sensitive ! – Loïc MICHEL Jun 25 '13 at 09:35
So how to do it incase-sensitive? – Nick Jun 25 '13 at 10:39
Convert computer name to uppercase before compareison see http://www.robvanderwoude.com/battech_convertcase.php – Loïc MICHEL Jun 25 '13 at 11:14

score 1 · Answer 2 · answered Jun 25 '13 at 17:34

1

@ECHO %COMPUTERNAME% | find /I "LAP"
IF NOTERRORLEVEL 1 ( dsmove ... OU=laptop ... )
GOTO :EOF
@ECHO %COMPUTERNAME% | find /I "DESK"
IF NOTERRORLEVEL 1 ( dsmove ... OU=desktop... )
GOTO :EOF

answered Jun 25 '13 at 17:34

mojo

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score 0 · Answer 3 · answered Apr 25 '19 at 19:17

The logic has to be reversed. Here's a case-insensitive solution:

setlocal enabledelayedexpansion    
set nameSearch=Lap
set checkComputerName=!computername:%nameSearch%=!
if "%checkComputerName%" NEQ "%computername%" (
  echo %nameSearch% found in %computername%
) else (
  echo %nameSearch% not found in %computername%
)

score 0 · Answer 4 · answered Dec 05 '19 at 14:13

Came across this issue today, this is how I solved it..

Say you have different names for desktops, laptops etc ie (DESKTOP0001, LAPTOP0001) etc, then this method will work nicely.

What you want to do is grab the first few characters of the name, you can use :x,y within a variable to do this.

Example

echo %compuername:~0,6%

The output for this would be DESKTO (first 6 characters starting from position 0)

(echo %computername:~1,6% would give you ESKTOP )

Quick proof test..

if %computername:~0,6% == DESKTO echo yes-Desktop

So for my use, I used

if %computername:~0,6% == DESKTO goto Desktop
if %computername:~0,6% == LAPTOP goto Laptop

goto end

:Desktop
enter Desktop commands here
goto end

:Laptop
enter Laptop commands here
goto end

:end

You solved the problem `string starts with`, but the question was about `string contains` — jeb, Dec 05 '19 at 14:30

score 0 · Answer 5 · answered Sep 07 '21 at 19:15

All the solutions work, but instead of static positioning, %computername:~0,6%, and instead of calling to run an external program "find /i", I would just use the string compare with string replacement method

If Not "%computername% == "%computername:LAP=%" (
   dsmove "CN=%computername%,CN=Computers,DC=domain,DC=local" -newparent"OU=**Laptops**,OU=Computers,OU=Company,DC=domain,DC=local"
)

If Not "%computername% == "%computername:DESK=%" (
   dsmove "CN=%computername%,CN=Computers,DC=domain,DC=local" -newparent "OU=**Desktops**,OU=Computers,OU=Company,DC=domain,DC=local"
)

If Computer Name Contains

5 Answers5