How to launch any application in uipath

HI team,

How to launch any application in uipath using its location/ how to achieve double clicking on App icon on desktop.

ex: if the location of the application is “C:\ProgramData\Microsoft\Windows\Start Menu\Service Delivery Tools\APP NAME.lnk”


we can use start process activity and pass the above path as input

or we can use open application activity and pass the above path as input to the filename property

Cheers @KavithaManohar

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Create a variable of type process in the assign and in the right side, process.start(“your application location or exe file”)


, which window should i indicate without launching that app

mention the above path in the filename property in the property panel (right side of the screen)
or mention the path with in start process activity
both will be fine

Cheers @KavithaManohar

so were we able to open the application now
Cheers @KavithaManohar

Uipath have two activities to start/open application
1.Start Process
2. Open application

you can refer,



Is it just the path that we have to give? Or do we also need to include the application executable, e.g. demo.exe? Appreciate quick reply.



I got the fact that the application name should also be mentioned.
If the path has to come from Orchestrator Asset, can I just save this path in an asset in Orchestrator? Or do I have to do anything else? Because when I save this in Orchestrator and call it from InitAllApplications, it throws an error as below, any quick help pls?

The system cannot find the file specified. (Exception from HRESULT: 0x80070002)


I got the issue resolved by myself, it is a typo kinda, thank you