Question

我需要一些帮助。我正在使用这个方法来帮助我打开一个带有对话框的文件 :

from Tkinter import *
import tkFileDialog

fileOpen = Tk()
fileOpen.withdraw() #hiding tkinter window

file_path = tkFileDialog.askopenfilename(
title="Open file", filetypes=[("txt file",".txt"),("All files",".*")])

if file_path != "":
    print "you chose file with path:", file_path

else:
    print "you didn t open anything!"

print file_path

It works fine, and the file is opened in Python, but I don t know how to make further commands on that file, for example if I wanted to read it or write on it.
I have tried referencing it as fileOpen, but that doesn t seem to work, and I have no idea what the variable would be instead.

Answer 1

完全不打开文件。要打开它, 请调用 open( file_ path) 。这给了您一个可以 read from 或 write to 的对象。请阅读文件 I/ O < / a > 上的 Python 教程 s < a href=" http://docs. python. org/tumental/ inputoutput. html# reading- and- wride- files > 。

友情链接