How to pass the file location from "askopenfilename" and use it for a second function

This is where classes make things a little easier:

from tkinter import *from tkinter import filedialogimport osclass GUI():    def __init__(self):        self.root = Tk()        self.filename = ""        #root.filename = filedialog.askopenfilename(initialdir="/C")        theLabel = Label(self.root, text="The Editor")        theLabel.grid(row=0)        button1 = Button(self.root, text="Browse", command=self.open)        button2 = Button(self.root, text="Run", command=self.other_func)        button3 = Button(self.root, text="Quit", command=self.root.quit)        button1.grid(row=1)        button2.grid(row=2)        button3.grid(row=3)        self.root.mainloop()    def open(self):        result = filedialog.askopenfilename(initialdir="C:/")        print("Function open read:")        print(result)        self.filename = result        #print("Set class attribute, calling other function")        #self.other_func()    def other_func(self):        with open(self.filename) as f: # 'with' is preferred for it's error handling            for c in f:                print(c)GUI()

Well you can return the variable:

def open():    result = os.path.dirname(os.path.abspath(__file__))    return resultfile_path = open()

Later you can use file_path and pass it to a new function.

You could also immediately call that function from inside the open function while passing result:

def open():    result = os.path.dirname(os.path.abspath(__file__))    your_function(result)

I changed the way you get the file path. I would also advise against naming a function open(), since it already is a function in itself:

https://docs.python.org/3/library/functions.html#open