将目录中的文件名导出到表格中，所以写了这个小程序，请各位大神指点。谢谢

1.遍历指定目录及子目录中的文件，将文件名输出至表格。

2.表格第1列为序号，第2列为一级目录，第3列为二级目录，第4列为文件名，第5列为文件地址。

3.点击“选择目录”，选择要遍历的目录，点击“输出到表格”，可以遍历目录及文件名到表格，并打开表格。

代码如下：

import os
import tkinter as tk
from tkinter import filedialog
from openpyxl import Workbook
 
 
def browse_folder():
    folder_selected = filedialog.askdirectory()
    if folder_selected:
        entry_folder.delete(0, tk.END)
        entry_folder.insert(0, folder_selected)
 
 
def save_to_excel():
    folder_path = entry_folder.get()
    if not folder_path:
        return
 
    workbook = Workbook()
    sheet = workbook.active
    sheet.title = "Files"
 
    sheet.append(["序号", "一级目录", "二级目录", "文件名", "文件地址"])
 
    index = 1
    for root, dirs, files in os.walk(folder_path):
        for file in files:
            file_path = os.path.join(root, file)
            directory = os.path.relpath(root, folder_path)
            parts = directory.split(os.sep)
            first_level = parts[0] if len(parts) > 0 else ""
            second_level = parts[1] if len(parts) > 1 else ""
            sheet.append([index, first_level, second_level, file, file_path])
            index += 1
 
    file_path = filedialog.asksaveasfilename(defaultextension=".xlsx", filetypes=[("Excel files", "*.xlsx")])
    if file_path:
        workbook.save(file_path)
        open_excel(file_path)
 
 
def open_excel(file_path):
    os.startfile(file_path)
 
 
# 创建主窗口
root = tk.Tk()
root.title("目录遍历到表格")
 
# 创建输入框和按钮
entry_folder = tk.Entry(root, width=50)
entry_folder.pack(pady=10)
 
button_browse = tk.Button(root, text="选择目录", command=browse_folder)
button_browse.pack(pady=5)
 
button_save = tk.Button(root, text="输出到表格", command=save_to_excel)
button_save.pack(pady=5)
 
root.mainloop()