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Using .bat file to open file in diff location

Last answer on May 5, 2009 9:14:54 pm BST trueparadox, on Apr 14, 2009 5:13:08 pm BST 
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Hello,

I'm working on developing a menu for a USB flash drive.

I have .bat file that opens a spreadsheet with a program that is installed on the flash drive.
The code looks something like this:

Start /d \Folder1\folder2\folderofprogram Name of program.exe "Name of file.ods"

This code works well but the problem is I need to have the file in the same directory as the program.
Is there a way to make the .bat file open the file from a different directory on the USB flash drive?


I think this is a simple problem but was not able to find any information about it.

Thanks for any help!

Configuration: Windows XP
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1

 Neomagus, on May 5, 2009 9:14:54 pm BST
  • +3

Use UNC path name. so something like this: \\computername\folder\filename or \\mypc\c:\windows\testfile.exe

Hope this helps.

   
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