📜  tkinter filedialog 文件名 - Python 代码示例

📅  最后修改于: 2022-03-11 14:45:11.709000             🧑  作者: Mango

代码示例3
def open_file():
    file = askopenfile(mode='r', filetypes=[
                       ('Text files', '*.txt'), ('CSV Files', '*.csv')])
    if file is not None:
        print(file.name.split("/")[-1]) # this will print the file name

btn = Button(root, text='Open', command=lambda: open_file())
btn.pack(side=TOP, pady=10)