Are there subfolders in the zipfile? I would think you would want to use
the full name (including path) to extract/open the file, not just the
filename:
for name in zf.namelist():
print zf.open(name)
Anthony
On Thursday, June 14, 2012 7:37:36 AM UTC-4, praveen krishna wrote:
>
> Hi,
> I have found the mistake ,the problem is in opening the file.I have
> tried in python.
> if my code is
> zipfile.ZipFile('/home/praveen/job_files/seq_data/PCGENE_Format.zip')
> for name in zf.namelist():
> filename = name.split('/')[-1]
> print filename
>
> o7p
> zipfolder()
> N77CDR1K.ATD
> NGFP_TRK.1
> NP7C1TRK.1D
> NPKGTS1
>
>
> import zipfile
> def zipfolder():
> zf =
> zipfile.ZipFile('/home/praveen/job_files/seq_data/PCGENE_Format.zip')
> for name in zf.namelist():
> filename = name.split('/')[-1]
> raw_seq = zf.open(filename)
> print raw_seq
> error:
>
> KeyError: "There is no item named 'N77CDR1K.ATD' in the archive"
> import zipfile
> def zipfolder():
> zf =
> zipfile.ZipFile('/home/praveen/job_files/seq_data/PCGENE_Format.zip')
> for name in zf.namelist():
> filename = name.split('/')[-1]
> raw_seq = zf.open(name)
> print raw_seq
>
> o/p
> zipfolder()
> <zipfile.ZipExtFile object at 0x2401d10>
> <zipfile.ZipExtFile object at 0x2401d50>
> <zipfile.ZipExtFile object at 0x2401d10>
> <zipfile.ZipExtFile object at 0x2401d50>
> <zipfile.ZipExtFile object at 0x232f350>
> I am unable to extract the entity in a file with the module zipfile.Is
> there any possibility with this module if not can you suggest me any other
> python module.I am attaching one of my sample file to this mail.
>
>
>
> On Tue, Jun 12, 2012 at 10:53 PM, pbreit <pbreitenb...@gmail.com> wrote:
>
>> Pretty hard to follow.
>>
>> One thing: you might need a db.commit() after rec.update_record(raw_seq=*
>> *file_raw,processed_seq=rev) since it's looping.
>>
>> Can you browse your DB to see what exactly is in it? If it's SQLite, you
>> can use a Firefox addon:
>> https://addons.mozilla.org/en-US/firefox/addon/sqlite-manager/
>>
>> Or add a print statement which will output in a console somewhere:
>>
>> filename = name.split('/')[-1]
>> print filename
>>
>>
>>
>
