I have the following batch file: (name: u.bat)
@echo off start "" "C:\Program Files (x86)\IDM Computer Solutions\UltraEdit\Uedit32.exe" %1 rem exit I run the batch file in the temp folder to open an existing "readme.txt"
c:\temp> u readme.txt What is happening is the app is trying to open readme.txt in the C:\Program Files ....\UltraEdit directory and not the C:\temp directory.
How do I tell the batch file %1 is in the current directory?
you need to specify the PATH,
Syntax
Starts a separate window to run a specified program or command.
START ["title"] [/Dpath] [/I] [/MIN] [/MAX] [/SEPARATE | /SHARED] [/LOW | /NORMAL | /HIGH | /REALTIME | /ABOVENORMAL | /BELOWNORMAL] [/WAIT] [/B] [command/program] [parameters]
path Starting directory
Might be related to UltraEdit, but what you can do is include %~dp0 to get the current drive and path, like so
@echo off start "" "C:\Program Files (x86)\IDM Computer Solutions\UltraEdit\Uedit32.exe" .\%1 rem exit or
@echo off start "" /D . "C:\Program Files (x86)\IDM Computer Solutions\UltraEdit\Uedit32.exe" %1 rem exit 
This should also work and allow you to open more than one file:
@echo off start "" "C:\Program Files (x86)\IDM Computer Solutions\UltraEdit\Uedit32.exe" "%~f1" "%~f2" "%~f3" "%~f4" "%~f5" But you will probably find that this also works and the cmd window will close:
@echo off "C:\Program Files (x86)\IDM Computer Solutions\UltraEdit\Uedit32.exe" %* 