打开文件按钮-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()   

Answer 1

rather than

from tkFileDialog import askopenfilename

you probably just want

import tkFileDialog

Answer 2

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

Simply, import tkFileDialog will do.

Answer 3

instead of using from tkFileDialog import askopenfilename

use import Tkinter.Filedialog as tkFiledialog