I use python's zipfile module to extract a .zip archive (Let's take this file at http://img.dafont.com/dl/?f=akvaleir for example.)
f = zipfile.ZipFile('akvaleir.zip', 'r')
for fileinfo in f.infolist():
print fileinfo.filename
f.extract(fileinfo, '.')
Its output:
Akval�ir_Normal_v2007.ttf
Akval�ir, La police - The Font - Fr - En.pdf
Both files are unaccessable after extraction because there are invalid encoded characters in their filenames. The problem is zipfile module doesn't have an option to specify output filenames.
However, "unzip akvaleir.zip" escapes the filename well:
root@host:~# unzip akvaleir.zip
Archive: akvaleir.zip
inflating: AkvalВir_Normal_v2007.ttf
inflating: AkvalВir, La police - The Font - Fr - En.pdf
I tried capturing output of "unzip -l akvaleir.zip" in my python program and these two filenames are:
Akval\xd0\x92ir_Normal_v2007.ttf
Akval\xd0\x92ir, La police - The Font - Fr - En.pdf
How can I get the correct filename like what unzip command does without capturing output of "unzip -l akvaleir.zip"?
Instead of the extract
method, use the open
method and save the resulting pseudofile to disk under whatever name you wish, for example with shutil.copyfileobj
.