How do I set an executable's working directory via the command line, prior to executing it?

Question

If I run a program on the command line whose location is resolved through the Path environment variable, the program's working directory is generally set to its installation directory.

I would like to run such a program from a console window and set its working directory to the current or other explicit directory. I was able to do this by temporarily copying the program to my working directory -- is there another way to accomplish this within the cmd.exe or powershell.exe environments?

The windows shell analogy to this task is to create a shortcut and set the "Start In" property accordingly.

Solution 1

To explicitly set the working directory, a PowerShell solution would be to use the Start-Process cmdlet with the -WorkingDirectory parameter.

Start-Process -FilePath notepad.exe -WorkingDirectory c:\temp

Using the alias start, positional parameter, and partial parameter name this could be written as:

start notepad.exe -wo c:\temp

CMD also has a START command. For this, use the /D parameter to specify the working directory:

START /D c:\temp notepad.exe

Solution 2

The below will work, make appropriate substitutions and save it with a .cmd extension.

@echo off 
C:
chdir C:\desired\directory
C:\full\path\of\command.exe

Put this batch file in a directory in your %PATH% and you should be able to invoke it from any cmd.exe instance.

Comments

[MBraedley]

Note that funkiness may occur if the working directory and executable aren't on the same partition/drive.

[Pacerier]

@LawrenceC, However this solution requires a temporary file. It's better if this could be achieved using no temporary files