filedialog, tkinter and opening files

问题:

I'm working for the first time on coding a Browse button for a program in Python3. I've been searching the internet and this site, and even python standard library.

I have found sample code and very superficial explanations of things, but I haven't been able to find anything that addresses the problem I'm having directly, or a good enough explanation so I can customize code to my needs.

Here is the relevant snippet:

Button(self, text = "Browse", command = self.load_file, width = 10)\
        .grid(row = 1, column = 0, sticky = W) .....


 def load_file(self):

    filename = filedialog.askopenfilename(filetypes = (("Template files", "*.tplate")
                                                         ,("HTML files", "*.html;*.htm")
                                                         ,("All files", "*.*") ))
    if filename: 
        try: 
            self.settings["template"].set(filename)
        except: 
            messagebox.showerror("Open Source File", "Failed to read file \n'%s'"%filename)
            return

The method is a hybrid of some code I found along the way with my own customizations. It seems like I finally got it to work (kinda), though its not exactly how I need it.

I get this error when I activate the 'Browse' button: NameError: global name 'filedialog' is not defined.

I've found fairly similar issues along the way but all the solutions suggested I have covered. I went into the 'filedialog' help section of IDLE but didn't glean anything from there either.

Would someone mind providing a break down and a little guidance on this; none of my books address it specifically, and I've checked all the solutions provided to others—I'm lost.

回答1:

The exception you get is telling you filedialog is not in your namespace. filedialog (and btw messagebox) is a tkinter module, so it is not imported just with from tkinter import *

>>> from tkinter import *
>>> filedialog
Traceback (most recent call last):
  File "<interactive input>", line 1, in <module>
NameError: name 'filedialog' is not defined
>>> 

you should use for example:

>>> from tkinter import filedialog
>>> filedialog
<module 'tkinter.filedialog' from 'C:\Python32\lib\tkinter\filedialog.py'>
>>>

or

>>> import tkinter.filedialog as fdialog

or

>>> from tkinter.filedialog import askopenfilename

So this would do for your browse button:

from tkinter import *
from tkinter.filedialog import askopenfilename
from tkinter.messagebox import showerror

class MyFrame(Frame):
    def __init__(self):
        Frame.__init__(self)
        self.master.title("Example")
        self.master.rowconfigure(5, weight=1)
        self.master.columnconfigure(5, weight=1)
        self.grid(sticky=W+E+N+S)

        self.button = Button(self, text="Browse", command=self.load_file, width=10)
        self.button.grid(row=1, column=0, sticky=W)

    def load_file(self):
        fname = askopenfilename(filetypes=(("Template files", "*.tplate"),
                                           ("HTML files", "*.html;*.htm"),
                                           ("All files", "*.*") ))
        if fname:
            try:
                print("""here it comes: self.settings["template"].set(fname)""")
            except:                     # <- naked except is a bad idea
                showerror("Open Source File", "Failed to read file\n'%s'" % fname)
            return


if __name__ == "__main__":
    MyFrame().mainloop()

回答2:

Did you try adding the self prefix to the fileName and replacing the method above the Button ? With the self, it becomes visible between methods.

...

def load_file(self):
    self.fileName = filedialog.askopenfilename(filetypes = (("Template files", "*.tplate")
                                                     ,("HTML files", "*.html;*.htm")
                                                     ,("All files", "*.*") ))
...

回答3:

I had to specify individual commands first and then use the * to bring all in command.

from tkinter import filedialog
from tkinter import *