CODEWITHSUNDEEP
pythontkinter
user2888676
user2888676

Reputation: 3

python file dialog accessing file name

Im trying to use tkinter to open a file dialog, once this file dialogue is open how do i get the file object that is returned by the function. As in how do i access it in main?

basically how do i handle return values by functions that are invoked by command

import sys
import Tkinter
from tkFileDialog import askopenfilename
#import tkMessageBox

def quit_handler():
    print "program is quitting!"
    sys.exit(0)

def open_file_handler():
    file= askopenfilename()
    print file
    return file


main_window = Tkinter.Tk()


open_file = Tkinter.Button(main_window, command=open_file_handler, padx=100, text="OPEN FILE")
open_file.pack()


quit_button = Tkinter.Button(main_window, command=quit_handler, padx=100, text="QUIT")
quit_button.pack()


main_window.mainloop()

Upvotes: 0

Views: 6790

Answers (2)

Mark Mikofski
Mark Mikofski

Reputation: 20178

the easiest way I can think of is to make a StringVar

file_var = Tkinter.StringVar(main_window, name='file_var')

change your callback command using lambda to pass the StringVar to your callback

command = lambda: open_file_handler(file_var)

then in your callback, set the StringVar to file

def open_file_handler(file_var):
    file_name = askopenfilename()
    print file_name
    #return file_name
    file_var.set(file_name)

Then in your button use command instead of open_file_handler

open_file = Tkinter.Button(main_window, command=command,
                           padx=100, text="OPEN FILE")
open_file.pack()

Then you can retrieve the file using

file_name = file_var.get()

Upvotes: 1

Rushy Panchal
Rushy Panchal

Reputation: 17532

Instead of returning the file variable, just handle it there (I also renamed the file variable so you do not override the built-in class):

def open_file_handler():
    filePath= askopenfilename() # don't override the built-in file class
    print filePath
    # do whatever with the file here

Alternatively, you can simply link the button to another function, and handle it there:

def open_file_handler():
    filePath = askopenfilename()
    print filePath
    return filePath

def handle_file():
    filePath = open_file_handler()
    # handle the file

Then, in the button:

open_file = Tkinter.Button(main_window, command=handle_file, padx=100, text="OPEN FILE")
open_file.pack()

Upvotes: 3

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