RE: Extracting the File Path of a .csv File From the File Name
Hi, I've been trying to extract the file path of a csv file in order to open it. I've been using the following script:dt = Open("File.csv");
path = Arg(dt << get property("Source"), 1);
print("path: " || path);This prints a file not found error to the log. The file path returned shows the following:C:\Users\jaz\Desktop\File.csv was not found.My file is stored in a folder in the Desktop but the p...