On 29 Aug, 21:18, David Bolen <[EMAIL PROTECTED]> wrote:
> mcl <[EMAIL PROTECTED]> writes:
> > I am trying to unzip an 18mb zip containing just a single 200mb file
> > and I get a Memory Error. When I run the code on a smaller file 1mb
> > zip, 11mb file, it works fine.
> (...)
> > def unzip_file_into_dir(file, dir):
> > #os.mkdir(dir, 0777)
> > zfobj = zipfile.ZipFile(file)
> > for name in zfobj.namelist():
> > if name.endswith('/'):
> > os.mkdir(os.path.join(dir, name))
> > else:
> > outfile = open(os.path.join(dir, name), 'wb')
> > outfile.write(zfobj.read(name))
> > outfile.close()
>
> The "zfobj.read(name)" call is reading the entire file out of the zip
> into a string in memory. It sounds like it's exceeding the resources
> you have available (whether overall or because the Apache runtime
> environment has stricter limits).
>
> You may want to peek at a recent message from me in the "Unable to
> read large files from zip" thread, as the suggestion there may also be
> suitable for your purposes.
>
> http://groups.google.com/group/comp.lang.python/msg/de04105c170fc805?...
> -- David
David,
Thank you. I read your post and I basically understood the concept,
butI could not get my head around the code, I need to write for my
solution. (Newbie and a bit long in the tooth)
To solve my problem, I think my best approach would be to read my
zipped file / files from the zip archive when I need them. Max three
users, occasional use. So no big overloading of host's server.
pseudo code
zfhdl = zopen(zip,filename) # Open File in Zip Archive for
Reading
while True:
ln = zfhdl.readline() # Get nextline of file
if not ln: # if EOF file
break
dealwithline(ln) # do whatever is necessary with
file
zfhdl.close
That is probably over simplified, and probably wrong but you may get
the idea of what I am trying to achieve.
Richard
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