A Snippet that will extract the file name from a given file path
This is a simple workflow to extract the file name from a given file path with any file type. It requires dot and file extension as well.
Input arguments
In_filePath (String) ex:Data\Config.xlsx
In_dotExtention (String) ex: .xlsx this is an Excel file
Output arguments
Out_filename (String)
You can get File name as Config.
Code Snippet : https://go.uipath.com/component/extra...
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