Question

I am creating a GUI with a browse button which I only want to return the path. I’ve been looking at solutions using code like below.

Tkinter.Button(subframe, text = "Browse", command = self.loadtemplate, width = 10).pack()

   def loadtemplate(self): 
        filename = tkFileDialog.askopenfilename(filetypes = (("Template files", "*.tplate")
                                                             ,("HTML files", "*.html;*.htm")
                                                             ,("All files", "*.*") ))
        if filename: 
            try: 
                self.settings["template"].set(filename)
            except: 
                tkMessageBox.showerror("Open Source File", "Failed to read file \n'%s'"%filename)

However I know Tkinter has a built in askopenfilename which is a super easy one line of code for opening files. Is there some way to modify this to return the directory instead of a file? Is there a smaller option than the larger chunk of code I posted?

Answer 1

It appears that tkFileDialog.askdirectory should work. documentation