Reading the first line of a file (in a zipfile)

mike.aldrich at gmail.com mike.aldrich at gmail.com
Wed Apr 11 22:15:48 CEST 2007


On Apr 11, 4:10 pm, "Gabriel Genellina" <gagsl-... at yahoo.com.ar>
wrote:
> En Wed, 11 Apr 2007 16:13:42 -0300, <mike.aldr... at gmail.com> escribió:
>
>
>
>
>
> > Hi folks,
> > I am trying to read the first occurence of non-whitespace in a file,
> > within a zipfile. Here is my code:
>
> > zipnames = glob.glob("<search_dir>*")
> > for zipname in zipnames:
> >   z = zipfile.ZipFile(zipname, "r")
> >   for filename in z.namelist():
> >     count = len(z.read(filename).split('\n'))
> >     if fnmatch.fnmatch(filename, "*AUDIT*"):
> >       test = filename.split(' ')
> >       print 'File:', test[0],
> >       bytes = z.read(filename)
> >       print 'has', len(bytes), 'bytes'
> >       print 'and', count, 'lines'
>
> > The first line in the file I am examining will be a number followed by
> > more whitespace. Looks like I cannot split by whitespace?
>
> Your code does nothing with the first line on the file; you only split the  
> *filename* on whitespace. And you extract the file twice.
> You don't even try to find "the first occurence of non-whitespace". Surely  
> an example of file contents and what output you really expect from it  
> would be adequate.
>
> --
> Gabriel Genellina- Hide quoted text -
>
> - Show quoted text -

The file contents have leading whitespace, then a number:
     123456       \n
I expect to return '123456'




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