OpenApplication Working Directory

OpenRPA
1

First post, here. I am attempting to write my first OpenRPA script, and I have an application that won’t open properly unless started from a specific folder/directory. Running on Windows 11, there is the “Start in” option for Windows application shortcuts. Is there any way to specify the “Start in” folder option when using OpenApplication?

Ex.
“filename”: “C:\Applications\Automation\bin\Runme.exe”
something like:
“start_in”: “C:\Applications\Automation\bin\”

Or any other suggestions that I should use?

Thanks so much!

2

You can use the Start process activity and set WorkingDirectory property.
edit: wow, I really are bad at spelling when replying from my ipad.

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