如何使用askopenfile获取有效的文件名

Question

I'm trying to create a program with Tkinter and tkFileDialog that opens a file for reading and then packs it into a text widget but, whenever I run this:

from Tkinter import *
from tkFileDialog import askopenfile
import time
m = Tk()

def filefind():
   file = askopenfile()
   f = open(str(file), "r+")
   x = f.read()
   t = Text(m)
   t.insert(INSERT, x)
   t.pack()


b = Button(m, text='File Picker', command=filefind)
b.pack()
m.mainloop()

I get this:

Exception in Tkinter callback
Traceback (most recent call last):
File "C:\Python27\lib\lib-tk\Tkinter.py", line 1536, in __call__
return self.func(*args)
File "C:\Users\super\PycharmProjects\untitled1\File Picker.py", line in    filefind
f = open(str(file), "r+")
IOError: [Errno 22] invalid mode ('r+') or filename: "<open file u'C:/Users/super/PycharmProjects/untitled1/util.h', mode 'r' at 0x00000000026E0390>"

Answer 1

Here is the issue; askopenfile() is returning an object, not just the name. If you print file , you will get <_io.TextIOWrapper name='/File/Path/To/File.txt' mode='r' encoding='UTF-8'> . You want the name= from the object. To get that, all you need to do is replace f = open(str(file), "r+") with f = open(file.name, "r+") .

Here is how it will look in your code:

from Tkinter import *
from tkFileDialog import askopenfile
import time
m = Tk()

def filefind():
    file = askopenfile()
    f = open(file.name, "r+")  # This will fix the issue.
    x = f.read()
    t = Text(m)
    t.insert(INSERT, x)
    t.pack()


b = Button(m, text='File Picker', command=filefind)
b.pack()
m.mainloop()

Edit

A cleaner way of doing this is by letting askopenfile() do the work of opening a file instead of 're-opening' it again with open() . Here is the cleaner version:

file = askopenfile()
x = file.read()