ZipInfo filename is mangled when os.sep is not '/'

Aperta
#94,529 1 commento 0 reazioni 0 assegnatari Vedi su GitHub

Nessuno ha ancora preso questa issue.

Valutazione

Difficoltà
3/5
Tempo stimato
1-2 giorni
Idoneità per principianti
45/100
Tipo di issue
Bug
Chiarezza
Abbastanza chiara
Stato di attività
Ferma
Stack tecnologico
python
Ambito
backend

Direzione di ricerca

Inizia in Lib/zipfile.py, alla traduzione del nome file di ZipInfo indicata intorno alle righe 378-382, e riproduci l'esempio segnalato os.sep='.'. Il lavoro è completato quando ZipInfo conserva i nomi file dell'archivio con separatori '/', senza sostituire '.' sui sistemi il cui separatore non è '/' né '\'; il contenuto non indica alcun file di test.

Scritto dal modello di indicizzazione a partire dal testo della issue.

Descrizione

OS-unsupported stdlib type-bug

Bug report

The ZipInfo object within ZipFile performs an explicit translation of the filename.

https://github.com/python/cpython/blob/7db1d2eaf367a1073191c80c7baeee41ae1f2f21/Lib/zipfile.py#L378-L382

I believe this is intended to make it easy to use on Windows where you might pass an explicit pathname to the ZipInfo object creation. On Windows the filesystem separator is commonly \ (although it supports / in many cases), so that this foces the the filename attribute to contain a filename in the unix form.
This logic is used whether the ZipInfo object is created manually (usually to add a new file), or when the filename has been taken from an archive that is being extracted.

However, the logic is broken on systems where the os.sep is anything else other than \ or /. On systems where the os.sep is . this means that if you try to create an archive with a file containing a . extension the filename in the archive will be mangled. On such a system, extracting an archive will also mangle the filename.

To demonstrate this, it is possible to do a very simple command line example:

>>> import zipfile
>>> import os
>>> os.sep = '.'
>>> zipfile.ZipInfo('hello.txt')
<ZipInfo filename='hello/txt' file_size=0>

In the real world, this breaks any possibility of using this module on RISC OS where the filesystem separator in os.sep is .. In the current Python 3 on RISC OS, the ZipFile module will always mangle filenames that have standard extensions.

I believe that the intention of the object is that:

  • the filename initialiser on the object and attribute is in unicode format (this has been enforced since Python 3 by the explicit decodes in the archive member reading code).
  • the filename attribute is formed as would be stored in the archive, using / as a directory separator (stated by documentation filename should be the full name of the archive member).
  • the filename initialiser on the object is allowed to be supplied a path name on unix and windows systems, as a convenience (the referenced code will have been relied on by existing software).

As such, I believe the referenced code is broken, and to retain the above assumptions and to allow the handling of zip archives on systems where os.sep is not / or \, the code should instead read:

        if os.sep == "\\" and os.sep in filename:
            filename = filename.replace(os.sep, "/")

This removes the overzealous replacement of os.sep in the creation of the ZipInfo object.

Further problems exist with the from_file method which I shall raise separately.

There are some issues which might be related to this (but this change does not preclude them): https://github.com/python/cpython/issues/90139 and https://github.com/python/cpython/issues/92184.

Your environment

  • CPython versions tested on: Python 3.9, 3.10
  • Operating system and architecture: On OS X, simulating the problem seen on RISC OS.
Lingua principale
Python
Stelle
77.2k
Fork
36k
Merge medio
1g 9h
PR unite (30g)
558

Guida per i contributori

Apri la guida per i contributori

Come iniziare

  1. Leggi tutta la issue e poi la guida ai contributi del progetto.
  2. Commenta sulla issue per dire che te ne occupi tu — evita che due persone facciano lo stesso lavoro.
  3. Fai un fork del repository e lavora su un branch.
  4. Apri una pull request che faccia riferimento al numero della issue.

Altre issue di python/cpython

Tutte le issue di python/cpython

Issue simili

Altre issue su Python

Ricevi le nuove issue nella tua casella

Un breve riepilogo di issue GitHub adatte ai principianti.