Python中的ZipFile模块出现错误的幻数错误

Question

我在Windows 7(64位)上使用Python 2.7.当我尝试用ZipFile模块解压缩zip文件时,我收到以下错误: -

Traceback (most recent call last):
  File "unzip.py", line 8, in <module>
    z.extract(name)
  File "C:\Python27\lib\zipfile.py", line 950, in extract
    return self._extract_member(member, path, pwd)
  File "C:\Python27\lib\zipfile.py", line 993, in _extract_member
    source = self.open(member, pwd=pwd)
  File "C:\Python27\lib\zipfile.py", line 897, in open
    raise BadZipfile, "Bad magic number for file header"
zipfile.BadZipfile: Bad magic number for file header

WinRAR可以提取我试图提取的文件.这是我用来从中提取文件的代码myzip.zip

from zipfile import ZipFile
z = ZipFile('myzip.zip')   //myzip.zip contains just one file, a password protected pdf        
for name in z.namelist():
    z.extract(name)

这段代码适用于我使用WinRAR创建的许多其他zip文件 myzip.zip

我尝试评论以下几行Python27\Lib\zipfile.py: -

if fheader[0:4] != stringFileHeader:
   raise BadZipfile, "Bad magic number for file header"

但这并没有真正帮助.运行我的代码实际上,我在我的shell上得到一些转储.

Answer 1

正确的ZIP文件在开头总是有"\ x50\x4B\x03\x04".您可以使用以下代码测试文件是否真的是ZIP文件:

with open('/path/to/file', 'rb') as MyZip:
  print(MyZip.read(4))

它将打印文件头,以便您检查.

UPDATE Strange,testzip()和所有其他函数都运行良好.你试过这样的代码吗？

with zipfile.GzipFile('/path/to/file') as Zip:
  for ZipMember in Zip.infolist():
    Zip.extract(ZipMember, path='/dir/where/to/extract', pwd='your-password')