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Need help creating a batch file for running an audit program on a single PC

Posted on 2008-06-09
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Last Modified: 2010-04-21
We use OpenAudit to audit our PCs.  I need a batch file that does the following.  I'm struggling to get my version to work, and I'm not sure why.  I would script it, but I'm not well-versed at scripts yet.

1.  Change the directory to C:\wamp\www\openaudit\scripts
- I know that I can do that with cd c:\wamp\www\openaudit\scripts

2.  Ask the user for the PC name
- I have been using 'set /P choice=Name of PC?' as my command.

3.  Plug name of PC into line of code as follows: 'cscript audit.vbs [pcname]'
- Been using if commands with no luck.  Program just quits instead of running.

Also,I would like it If the PC name is left empty, generate an error, otherwise run the program.

Thoughts?  Thanks ahead of time!
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Question by:flames1100
2 Comments
 
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Accepted Solution

by:
JesterToo earned 500 total points
ID: 21747139
try this...

@echo off
setlocal

:begin

   C:
   CD \wamp\www\openaudit\scripts
   set /P choice=Name of PC?

   if "%choice%A"=="A" goto :error

   cscript audit.vbs %choice%
   goto :done

:error
   echo ERROR -- no computer name specified.

:done
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LVL 2

Author Closing Comment

by:flames1100
ID: 31465805
Thanks so much!  That really helped.  I'll put the code I finally used below, which I modded a bit to do a little more.  I see my mistakes that I was making now.  Thanks again!

@echo off
setlocal

:begin

   C:
   CD \wamp\www\openaudit\scripts

:program

   set /P choice=Name of PC?

   if "%choice%A"=="A" goto :error

   cscript audit.vbs %choice%
   goto :done

:error
   echo ERROR -- no computer name specified.
   goto program

:error2
   echo ERROR -- invalid entry.
   goto :done

:done
set /P choice=Audit another?
if "%choice%"=="y" goto :program
if "%choice%"=="n" goto :exit
goto :error2

:exit
pause && exit
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