使用 Python 文件对话框 askopenfilename() 方法打开文件时如何获取文件类型? [英] How do you get the file type when a file is opened using the Python filedialog askopenfilename() method?

问题描述

我用 python tkinter 编写了这个程序，它使用 tkinter 文件对话框 askopenfilename() 方法打开一个文件.它将获取该文件的内容并将其插入到 TextBox 中.

I have this program written in python tkinter which opens a file using the tkinter filedialog askopenfilename() method. It will get the contents of that file and will insert that into a TextBox.

我希望能够获取该文件的文件类型(即，如果文件是 Python 文件，那么我希望控制台打印.py"，如果是 HTML 则我希望它打印.html"; 等等).

I want to be able to get the file type of that file (i.e if the file is a Python file then I want the console to print ".py" if its HTML then I want it to print ".html" and so forth).

所以这就是我尝试过的:

So this is what I tried:

FilePath = filedialog.askopenfilename(initialdir="C:/gui/", title="Open a File", filetypes=(("All Files", "*.*"), ("Text Files", "*.txt"), ("HTML Files", "*.html"), ("CSS Files", "*.css"),("JavaScript Files", "*.js"), ("Python Files", "*.py")))

print(FilePath)

然而，这给了我:/home/Test/RandomFolder/Test.py

However, this gives me: /home/Test/RandomFolder/Test.py

是否可以只获取.py"?最后?

Is it possible to get just the ".py" at the end?

推荐答案

open_File = easygui.fileopenbox(msg="Choose a file", default=r'C:\Users\user\*.txt')

extension = open_File.split('.', 1)

print(extension)

所以我在这里所做的是获取用户路径，因为通常在路径中你不会有 . 任何地方，除非它在文件的末尾，我已经拆分字符串变成2，如果你想要.之前的所有内容，只需使用extension = open_File.split('.', 0)，然后使用上面的代码.

So what I've done here is get the users path, and because typically in a path you won't have a . anywhere unless its at the end of a file, I've split the string into 2, if you want the everything before the . just use extension = open_File.split('.', 0), and everything after use the above code.

另外，虽然我在这种情况下使用的是 Easygui，但它适用于任何文件对话框方法，原则是关于您如何从字符串中获取您想要的内容.

Also, although I am using Easygui in this case, it applies to any file dialog method, the principle is in regards to how you're getting what you want out of the string.

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