8

我有一个小脚本，可以提取一个 .zip 文件。这很好用，但仅适用于文件名中不包含带有“ä”、“ö”、“ü”（等等）等字母的文件的 .zip 文件。否则我会收到此错误：

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)

我该如何解决这个问题？

4

3 回答 3

5

一个建议：

当我这样做时，我得到了错误：

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

Traceback (most recent call last):
  File "<pyshell#21>", line 1, in <module>
    c + u'2'
UnicodeDecodeError: 'ascii' codec can't decode byte 0x81 in position 0: ordinal not in range(128)

有一个 unicode 字符串传递给某处加入。

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
于 2013-01-30T13:01:53.337 回答
2

出于便携的原因，也许你从 Windows 压缩文件并在 Linux 中解压它们，你可以将压缩文件中的所有文件路径转换为 ​​unicode，当从 zip 解压时，不要使用ZipFile.extractall，这个默认解压文件到磁盘并且不支持 unicode 路径在压缩文件中，试试这个：

import zipfile, sys, os,
zf = zipfile.ZipFile(sys.argv[1], 'r')
for m in zf.infolist():
    data = zf.read(m) # extract zipped data into memory
    # convert unicode file path to utf8
    disk_file_name = m.filename.encode('utf8')
    dir_name = os.path.dirname(disk_file_name)
    try:
        os.makedirs(dir_name)
    except OSError as e:
        if e.errno == os.errno.EEXIST:
            pass
        else:
            raise
    except Exception as e:
        raise

    with open(disk_file_name, 'wb') as fd:
        fd.write(data)
zf.close()
于 2015-10-11T00:36:59.087 回答
-1

清澈如水：消息指出ASCII解码器无法处理非 ASCII字符。您必须选择其他一些字符编码。

于 2013-01-30T12:56:03.677 回答