How to open a file in a For each?

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Hi @joesparty

In your list of files, i assume you have the full path of the file along with its name and extension.

If not, you can use Directory.GetFiles("Path") to get the list of files to an array.

In the for each loop, use the Start Process activity and in that, give the path to notepad.exe along with the path name of the file you want to open. To do this… refer the link below…

Hope it helps and if this works for you, please mark the answer as the solution so it could help others as well :slight_smile:

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