解压缩受密码保护的文件时出现权限错误

Question

我正在尝试解压缩许多受密码保护的文件，但是我不断收到一些权限错误。我试图以管理员身份运行vscode来执行此操作，但是仍然出现相同的错误。

这里是代码：

input_file = ".\\pa-dirty-price-crawler\\folders"

import zipfile
with zipfile.ZipFile(input_file, 'r') as zip_ref:
    zip_ref.extractall(input_file, pwd=b'qpsqpwsr')

这里是错误：

Traceback (most recent call last):
  File "c:/Users/usr/workspace/pa-dirty-price-crawler/src/outlook.py", line 23, in <module>
    with zipfile.ZipFile(input_file, 'r') as zip_ref:
  File "C:\ProgramData\Anaconda3\lib\zipfile.py", line 1240, in __init__
    self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: '.\\pa-dirty-price-crawler\\folders'

我不知道可以执行相同操作的另一个库，但是如果有人对修复此问题有任何建议，我将不胜感激。

编辑：

当我尝试这样指定整个文件路径名时：

input_file = "C:\\Users\\usr\\workspace\\pa-dirty-price-crawler\\folders"

import zipfile
with zipfile.ZipFile(input_file, 'r') as zip_ref:
    zip_ref.extractall(pwd=b'qpsqpwsr')

我仍然收到此错误：

Traceback (most recent call last):
  File "c:/Users/usr/workspace/pa-dirty-price-crawler/src/outlook.py", line 23, in <module>
    with zipfile.ZipFile(input_file, 'r') as zip_ref:
  File "C:\ProgramData\Anaconda3\lib\zipfile.py", line 1240, in __init__
    self.fp = io.open(file, filemode)
PermissionError: [Errno 13] Permission denied: 'C:\\Users\\usr\\workspace\\pa-dirty-price-crawler\\folders'

Answer 1

zipfile.ZipFile使用文件而不是目录。因此，您的代码需要两个变量：输入file和output目录：

input_file = ".\\pa-dirty-price-crawler\\folders\myzipfile.zip"
output_directory = ".\\pa-dirty-price-crawler\\folders

import zipfile
with zipfile.ZipFile(input_file, 'r') as zip_ref:
    zip_ref.extractall(output_directory, pwd=b'qpsqpwsr')