Получается ошибка. Что делать?

Question

[yk_vova]

Что мне делать?
вот код:

import openpyxl
from openpyxl import Workbook
date = '333'
prih = int(input())
toilet = int(input())
van = int(input())
dataFile = open('log.txt', 'r')
last_line = dataFile.readlines()[-1]
dataFile.close()
wb = openpyxl.reader.excel.load_workbook(filename='pok.xlsx', data_only=True)
wb.active = 0
sheet = wb.active
sheet['A' + last_line] = date
sheet['B' + last_line] = prih
sheet['C' + last_line] = toilet
sheet['D' + last_line] = van
wb.save('pok.xlsx')
startfile("pok.xlsx")
writeFile = open('log.txt', 'a')
aa = int(last_line) + 1
writeFile.write('\n' + str(aa))
writeFile.close()

Вылезает данная ошибка:

Traceback (most recent call last):
  File "C:/Users/vayak/PycharmProjects/pythonProject6/main.py", line 17, in <module>
    wb.save('pok.xlsx')
  File "C:\Users\vayak\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\openpyxl\workbook\workbook.py", line 392, in save
    save_workbook(self, filename)
  File "C:\Users\vayak\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\openpyxl\writer\excel.py", line 291, in save_workbook
    archive = ZipFile(filename, 'w', ZIP_DEFLATED, allowZip64=True)
  File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.8_3.8.1520.0_x64__qbz5n2kfra8p0\lib\zipfile.py", line 1251, in __init__
    self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: 'pok.xlsx'

Answer 1

[kirillinyakin]

Исходя из ошибки, запрещен доступ к файлу, Вы закрыли файл перед запуском скрипта?