To open the specific file inside a specific folder

Hi all,

I want to open the specific file inside a specific folder. This folder name is known before as it is saved as a variable previously.

For eg: \ssiiasauto\auto\class\set. This is the file path.
Inside “set” I need to look for folder “ABCD2020” which is known before.

Inside this “ABCD2020” folder, I need to open a specific file.

Any help is greatly appreciated.
Thanks

Hi @Christin_John

We can use Start process activity for open a specific file in UiPath,

FolderPath = \ssiiasauto\auto\class\set
FileName (Variable) = ABCD2020
File exteension - PDF (lets assume its a PDF file)

in Start process activty give “ssiiasauto\auto\class\set” + FileName + “.PDF”

Thanks

Before opening the process using start process, also check whther the file is there or not too using Path Exists activity, it would be a good practice around

@prasath_S Hi Prasanth,
Thanks for the reply.
But it seems not working for me.

  1. when I give only the file path in the start process: \ssiiasauto\auto\class\set , it opens the folder until set. But when I add “+ Filename”. It doesn’t open anything.
    I have left my arguments column blank in the above case.

@Christin_John

Please check filename variable has value or is it null

Also check if the file is present in that directory or not

For debugging purpose you can give the full filepath and check if its opening without any variable.

Thanks