[issue43326] About Zipfile

2021-02-25 Thread Fcant 

New submission from Fcant :

When I Unzip a package using the zipfile module, the package’s filename has a 
newline character, which causes the Unzip to fail, so the UNZIP filename needs 
to be processed

--
components: Library (Lib)
files: Fcant_2021-02-26_11-26-37.jpg
messages: 387705
nosy: Fcscanf
priority: normal
severity: normal
status: open
title: About Zipfile
type: behavior
versions: Python 3.9
Added file: https://bugs.python.org/file49836/Fcant_2021-02-26_11-26-37.jpg

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[issue43326] About Zipfile

2021-02-26 Thread Fcant 

Fcant  added the comment:

Because these special symbols are not handled under the Window, he will report 
an error and will not work properly

From：樊乘乘
Email：fcsc...@outlook.com<mailto:fcsc...@outlook.com>
Tel：17826260016

发件人: Ronald Oussoren<mailto:rep...@bugs.python.org>
发送时间: 2021年2月26日 樊乘 17:15
收件人: fcsc...@outlook.com<mailto:fcsc...@outlook.com>
主题: [issue43326] About Zipfile

Ronald Oussoren  added the comment:

I'm not sure what you are asking here.

Looking at the screenshot I'd say that the zipfile you are trying to extract 
contains files with a newline in their name. That's a perfectly valid (although 
annoying) name on unix-y platforms.

The zipfile module does not have an API for rewriting the file name when 
extracting. It is far from clear to me that adding such a API would be 
generally useful.

Emulating this is fairly easy though, something like:

zf = zipfile.ZipFile(...)
for name in zf.namelist():
stream = zf.open(name)
data = stream.read()
stream.close()
with open(name.replace(...), "w") as stream:
stream.write(data)

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nosy: +ronaldoussoren

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[issue43326] About Zipfile

2021-02-26 Thread Fcant 


Change by Fcant :


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stage:  -> resolved
status: open -> closed

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