Open File Button - Python Tkinter

Question

I am having a problem where I cannot link my open Button to the "open" command, help please!

Error= fileName = tkFileDialog.askopenfilename()
NameError: global name 'tkFileDialog' is not defined

What I have:

from Tkinter import *
from tkFileDialog   import askopenfilename

frm = Frame(parent)
frm.pack(fill=X)
Button(frm, text=' Browse  ',   command=self.getFileName).pack(side=LEFT)

def getFileName(self):
    fileName = tkFileDialog.askopenfilename()
    iconEntry.insert(0, fileName)

SimpleEditor().mainloop()

mgilson · Accepted Answer · 2014-01-26 03:14:52Z

3

rather than

from tkFileDialog import askopenfilename

you probably just want

import tkFileDialog

answered Jan 26, 2014 at 3:14

mgilson

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Comments

nEO · Accepted Answer · 2014-10-06 11:28:45Z

0

yeah. Since you are already doing "tkFileDialog.askopenfilename()", you don't need to do this "from tkFileDialog import askopenfilename"

Simply, import tkFileDialog will do.

answered Oct 6, 2014 at 11:28

nEO

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Comments

Manpreet Singh · Accepted Answer · 2017-11-08 17:55:41Z

0

instead of usingfrom tkFileDialog import askopenfilename

useimport Tkinter.Filedialog as tkFiledialog

answered Nov 8, 2017 at 17:55

Manpreet Singh

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