如何在python中应用选择文件按钮

Question

我学习了如何创建选择文件按钮，但我不知道如何将其应用到代码中，如下所示。

from openpyxl import Workbook
# import copy

wb = Workbook()

with open('chat_20220222152420.txt', encoding='utf-8') as sherr:
    row = 1
    column = 1
    ws = wb.active
    for line in sherr:
        if column == 1:
            ## split the line and rejoin
            value = " ".join(line.strip().split(' ')[2:])
        else:
            value = line.strip()
            
        ws.cell(row=row, column=column, value=value)
        
        if (column := column + 1) > 3:
            row += 1
            column = 1
            
    wb.save('Chatchatchat.xlsx')

我不想使用 with open() ，而是想使用按钮来选择我想要打开的文件。下面是我尝试选择文件的代码。我只是不知道如何在上面的代码中应用它:'(

from ipywidgets import Button
from tkinter import Tk, filedialog
from IPython.display import clear_output, display

def select_files(b):
    
    root = Tk()
    root.withdraw()                                       # Hide the main window.
    root.call('wm', 'attributes', '.', '-topmost', True)  # Raise the root to the top of all windows.
    b.files = filedialog.askopenfilename(multiple=False)  # List of selected files will be set button's file attribute.
    print(b.files)                                        # Print the list of files selected.
    
fileselect = Button(description="File select")
fileselect.on_click(select_files)

display(fileselect)

Answer 1

实现此目的的一种方法是将第一个代码块移动到需要读取文件名的函数中：

from openpyxl import Workbook

def write_spreadsheet(filename):
    wb = Workbook()
    with open(filename, encoding='utf-8') as sherr:
        row = 1
        # ...

然后从 select_files 调用它（请注意，filedialog.askopenfilename 返回一个单个文件名，而不是列表）：

def select_files(b):
    root = Tk()
    root.withdraw()                                       # Hide the main window.
    root.call('wm', 'attributes', '.', '-topmost', True)  # Raise the root to the top of all windows.
    file = filedialog.askopenfilename(multiple=False)     # Ask the user to select a file.
    print(file)                                           # Print the file selected.
    write_spreadsheet(file)                               # Process the file.