[issue45981] Get raw file name in bytes from ZipFile

[Devourer Station]

Devourer Station <accelerator0099 at gmail.com> added the comment:

In file Lib/zipfile.py:
1357>  flags = centdir[5]
1358>  if flags & 0x800:
1359>    # UTF-8 file names extension
1360>    filename = filename.decode('utf-8')
1361>  else:
1362>    # Historical ZIP filename encoding
1363>    filename = filename.decode('cp437')

ZipFile simply decodes all non-utf8 file names by encoding CP437.

In file Lib/zipfile.py:
352>  # This is used to ensure paths in generated ZIP files always use
353>  # forward slashes as the directory separator, as required by the
354>  # ZIP format specification.
355>  if os.sep != "/" and os.sep in filename:
356>    filename = filename.replace(os.sep, "/")

And it replaces every '\\' with '/' on windows.

Consider we have a file named '\x97\x5c\x92\x9b', which is '予兆' in Japanese encoded in SHIFT_JIS.
You may have noticed the problem:

  '\x5c' is '\\'(backslash) in ASCII

So you will see ZipFile decodes the bytes by CP437, and replaces all '\\' with '/'.
And the Japanese character '予' is replaced partially, it is no longer itself.

Someone says we can replace '/' with '\\' back, and decode it by CP437 to get the raw bytes.
But what if both '/'('\x2f') and '\\'('\x5c') appear in the raw filename?

Simply replacing '\\' in a bytestream without knowning the encoding is by no means a good way.
Maybe we can provide a rawname field in the ZipInfo struct?

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<https://bugs.python.org/issue45981>
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