Python: Using open file dialogue to assign file to a variable?

Question

I searched around and found many ways on how to assign a file to a variable via opening a file from a direct link, but i was wondering how to assign a file to a variable to be read by a function, when the user uses the file dialogue to open a file (e.g. File > open > user selected file)?

I'm trying to open a music file that is to be passed to pyglet.media.load(variable containing file).

It returns the error: NameError: f is not defined

from tkinter import *
from tkinter.filedialog import askopenfilename
import pyglet
from threading import Thread

app = Tk()
app.title("Music Player")
app.geometry("600x200")
have_avbin = True 

def openFile():
    song =  filedialog.askopenfilename(filetypes = (("MP3 files", "*.mp3"),("All files","*.*")))
    f = song
    return f



#Creates menu bar for opening MP3s, and closing the program
menu = Menu(app)
file = Menu(menu)
file.add_command(label='Open', command=  openFile) # replace 'print' with the name of your open function
file.add_command(label='Exit', command=app.destroy) # closes the tkinter window, ending the app
menu.add_cascade(label='File', menu=file)
app.config(menu=menu)

#Run each app library mainloop in different python thread to prevent freezing
def playMusic():
    global player_thread
    player_thread = Thread(target=real_playMusic)
    player_thread.start()

def stopMusic():
    global player_thread
    player_thread = Thread(target=real_stopMusic)
    player_thread.start()

#Play open file function attached to button
def real_playMusic():
    music = pyglet.media.load(f);
    music.play()
    pyglet.app.run()

#Stop the music function
def real_stopMusic():
     pyglet.app.exit()




#Play button creation
btnPlay = Button(app, text ="Play", command = playMusic)
btnPlay.grid()


#Pause button creation
btnPause = Button(app)
btnPause.grid()
btnPause.configure(text = "Stop", command = stopMusic)



app.mainloop() # keep at the end

Answer 1

have_avbin = True
f='' # initialize this variable

def openFile():
    global f # tell the function that we plan on modifying this global variable
    f = filedialog.askopenfilename(filetypes = (("MP3 files", "*.mp3"),("All files","*.*")))