Hopefully a simple one for you (or at least seemingly)!

I import a .txt file, from which i make a ListLinePlot. I simply want to read in the name of the file, store it in a variable so I can use it to tag my plots later.

    Data = Import["C:\\Users\\Name\\Folder\\test2.txt", "CSV"];
ListLinePlot[Transpose[{Data[[All, 1]], Data[[All, 2]]}], 
 Frame -> True, 
 PlotRange -> {{Data[[All, 1]][[
      First@Flatten[Position[Data, {_, Max[Data[[All, 2]]]}]]]]
      - 0.0003, 
    Data[[All, 1]][[
      First@Flatten[Position[Data, {_, Max[Data[[All, 2]]]}]]]]
     + 0.0003}}, FrameTicks -> Automatic, Axes -> False, 
 PlotStyle -> Directive[Black, Thin], 
 FrameLabel -> {"SIMULATION TIME STEP [\[Mu]s]", 

Here is my code I'm using to plot, the filename is "test2", and it is that I would like to extract.

Thanks in advance,



2 Answers 2


I don't know if I understand you correctly. but here is what I got for you:


If you want without extension, then:


Given you already have the filename as a text string you could specify the file name and path separately when you first run, and create the full path & filename using FileNameJoin. FileNameJoin is OS agnostic so puts in forward or backslashes depending on whether you are running Windows, OSX or Linux.

Alternatively if you want something a bit more dynamic you can use Algohi's solution after first getting the filename with either

file = SystemDialogInput["FileOpen"]
(* Brings up a standard FileOpen dialog box specific to your OS *)

FileNameSetter[Dynamic[file2], "Open"]

(*  Creates a browse button in your notebook which opens a 
file open dialog box when clicked and file2 is set when 
you click on a file *)

For both options you can limit the shown files to a particular filename extension such as CSV.


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