PowerShell 7.3 - Difficulty in Using `Start-Process` with Custom Working Directory for Executables
I need help solving Quick question that's been bugging me - I'm trying to run an executable using `Start-Process` in PowerShell 7.3, but I'm working with an scenario with specifying a custom working directory... The command I’m using is as follows: ```powershell $exePath = 'C:\Path\To\YourExecutable.exe' $workingDirectory = 'C:\Path\To\WorkingDir' Start-Process -FilePath $exePath -WorkingDirectory $workingDirectory ``` However, when I run this, the process starts but doesn’t recognize any files within the `$workingDirectory`. I’ve checked the permissions, and they seem fine. The command executes without any errors, but the executable fails to access the necessary files and throws an behavior like "File not found". I tried using the `-NoNewWindow` switch, but that didn't change the outcome. It also seems that the executable runs fine when invoked from the command line with the same working directory, which is puzzling. I attempted to launch the executable using the full path to the files it needs using `-ArgumentList`, but that didn't resolve the scenario either: ```powershell Start-Process -FilePath $exePath -WorkingDirectory $workingDirectory -ArgumentList 'C:\Path\To\RequiredFile.txt' ``` I've verified that the `Start-Process` command works correctly with other executables that do not depend on a specific working directory. Is there something about how `Start-Process` sets the working directory that could be affecting the execution of this particular program? Any insights or suggestions would be greatly appreciated! What's the best practice here? I'm open to any suggestions. I'm working in a Linux environment. Cheers for any assistance!