How to open file from specified folder?

I am trying to open file by using ‘Start Process’ the file opened but its minimizing how to avoid that one.

Hi @MahalingPatil ,

Please find attached workflow.PrintFiles.xaml (8.5 KB)

You can change how much delay you want it.

1 Like

After open files which are different types how to send hot keys to make ‘save as’

F12 is not working please help me

The shortcut key will change for type of file you want to save it. Like

Excel File - Alt + F2
PDF - Ctrl+Shift+s
Word- F12

How to make selector dynamic.

go through the following link:

Hello,

i don’t know if you found an answer but you can just use the activity “open application” specify in the filename the path of application (powerpoint, excel, …) and in the argument, give the path of the file in you folder.

Tried that way, but it breaks apart the filepath string and throws multiple errors.
It seperated at every space as a individual filepath.

Hi @MahalingPatil

To open the file u can use the steps below

  1. use send hotkey activitiy and use Win+r command this will open the run window

  2. then in the type field , type the file path and and click on ok ,

  3. thus file u can open

Take this attached workflow for reference , here I am opening a notepad file u can see

notepad1.zip (73.8 KB)

Hope it helps

Mark it as solution if you got it

Nived N :robot:

Happy Automation :slight_smile::slight_smile:

1 Like

Thank you so much for that!

use directory.get file function and store all files in a file array.
use for each and pass array variable.now inside foreach use start process and pass array variable.at the end use kill process.this way you can open and close files in a specific folder

1 Like