python – zipfile提取时的unicode错误

我有一个小脚本,它将提取.zip文件.
这很好用,但仅适用于.zip文件,它们的文件名中不包含带有“ä”,“ö”,“ü”(等等)字母的文件.
否则我收到此错误：

Exception in thread Thread-1:
Traceback (most recent call last):
  File "threading.pyc",line 552,in __bootstrap_inner
  File "install.py",line 92,in run
  File "zipfile.pyc",line 962,in extractall
  File "zipfile.pyc",line 950,in extract
  File "zipfile.pyc",line 979,in _extract_member
  File "ntpath.pyc",line 108,in join
UnicodeDecodeError: 'ascii' codec can't decode byte 0x94 in position 32: ordinal not in range(128)

这是我的脚本的提取部分：

zip = zipfile.ZipFile(path1)
zip.extractall(path2)

我怎么解决这个问题？

最佳答案

一个建议：

我这样做时收到错误：

>>> c = chr(129)
>>> c + u'2'

Traceback (most recent call last):
  File "

有一个unicode字符串传递到某处加入.

可能是zipfile的文件路径是用unicode编码的吗？
如果你这样做怎么办：

zip = zipfile.ZipFile(str(path1))
zip.extractall(str(path2))

或这个：

zip = zipfile.ZipFile(unicode(path1))
zip.extractall(unicode(path2))

这是ntpath中的第128行：

def join(a,*p): # 63
    for b in p: # 68
                path += "\\" + b  # 128

第二个建议：

from ntpath import *

def join(a,*p):
    """Join two or more pathname components,inserting "\\" as needed.
    If any component is an absolute path,all prevIoUs path components
    will be discarded."""
    path = a
    for b in p:
        b_wins = 0  # set to 1 iff b makes path irrelevant
        if path == "":
            b_wins = 1

        elif isabs(b):
            # This probably wipes out path so far.  However,it's more
            # complicated if path begins with a drive letter:
            #     1. join('c:','/a') == 'c:/a'
            #     2. join('c:/','/a') == 'c:/a'
            # But
            #     3. join('c:/a','/b') == '/b'
            #     4. join('c:','d:/') = 'd:/'
            #     5. join('c:/','d:/') = 'd:/'
            if path[1:2] != ":" or b[1:2] == ":":
                # Path doesn't start with a drive letter,or cases 4 and 5.
                b_wins = 1

            # Else path has a drive letter,and b doesn't but is absolute.
            elif len(path) > 3 or (len(path) == 3 and
                                   path[-1] not in "/\\"):
                # case 3
                b_wins = 1

        if b_wins:
            path = b
        else:
            # Join,and ensure there's a separator.
            assert len(path) > 0
            if path[-1] in "/\\":
                if b and b[0] in "/\\":
                    path += b[1:]
                else:
                    path += b
            elif path[-1] == ":":
                path += b
            elif b:
                if b[0] in "/\\":
                    path += b
                else:
                    # !!! modify the next line so it works !!!
                    path += "\\" + b
            else:
                # path is not empty and does not end with a backslash,# but b is empty; since,e.g.,split('a/') produces
                # ('a',''),it's best if join() adds a backslash in
                # this case.
                path += '\\'

    return path

import ntpath
ntpath.join = join

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