使用python Tkinter调用由askopenfilename命令确定的值

Question

Hi I've made a simple GUI that contains a browse button which uses askopenfilename :

browsebutton = Button(mGui,text='Browse',command=askopenfilename)
browsebutton.place(x=400,y=50)

I have been trying to print this file name to a text file as part of a larger script, I have tried many different strategies, my last attempt was this:

conf.write("receptor="'invoke(browsebutton)'"\n")

I'm having touble finding out how to call this function, I just started programming and I have tried multiple strategies but none seem to work. I am using Python 2.5, thank you for your help. -Paul

Answer 1

Create function with askopenfilename() and assign it to button

import Tkinter as tk
from tkFileDialog import askopenfilename 

def some_function():
    filename = askopenfilename()
    if filename:
        print "selected:", filename
    else:
        print "file not selected"

mGui = tk.Tk()
browsebutton = tk.Button(mGui,text='Browse',command=some_function)
browsebutton.pack()

mGui.mainloop()