On Feb 1, 9:39 pm, Larry Bates <[EMAIL PROTECTED]> wrote:
> Jandre wrote:
> > Hi
>
> > I am a python novice and I am trying to write a python script (most of
> > the code is borrowed) to Zip a directory containing some other
> > directories and files. The script zips all the files fine but when it
> > tries to zip one of the directories it fails with the following
> > error:
> > "IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'"
>
> > The script I am using is:
>
> > import zipfile, os
>
> > def toZip( directory, zipFile ):
> > """Sample for storing directory to a ZipFile"""
> > z = zipfile.ZipFile(
> > zipFile, 'w', compression=zipfile.ZIP_DEFLATED
> > )
> > def walker( zip, directory, files, root=directory ):
> > for file in files:
> > file = os.path.join( directory, file )
> > # yes, the +1 is hacky...
> > archiveName = file[len(os.path.commonprefix( (root,
> > file) ))+1:]
> > zip.write( file, archiveName, zipfile.ZIP_DEFLATED )
> > print file
> > os.path.walk( directory, walker, z )
> > z.close()
> > return zipFile
>
> > if __name__ == "__main__":
> > toZip( 'c:\\aaa', 'c:\\aaa\\test.zip' )
>
> > I have tried to set the permissions on the folder, but when I check
> > the directory permissions it is set back to "Read Only"
>
> > Any suggestions?
>
> > Thanks
> > Johan Balt
>
> Couple of quick suggestions that may help:
>
> 1) don't use 'file' as a variable name. It will mask
> the builtin file function. If it hasn't bitten you before
> it will if you keep doing that.
>
> 2) If you put the target .zip file in the directory you are
> backing what do you expect the program to do when it comes
> to the file you are creating as you walk the directory? You
> haven't done anything to 'skip' it.
>
> 3) Your commonprefix and +1 appears to result in same
> information that the easier to use os.path.basename()
> would give you. Double check me on that.
>
> I don't see anything that references C:\\aaa\temp in your
> code. Does it exist on your hard drive? If so does it
> maybe contain temp files that are open? zipfile module
> can't handle open files. You must use try/except to
> catch these errors.
>
> Hope info helps.
>
> -Larry
Thank you Larry.
I've changed the code as epr your advice. The code is now:
import zipfile, os
def toZip( directory, zipFile ):
"""Sample for storing directory to a ZipFile"""
z = zipfile.ZipFile(
zipFile, 'w', compression=zipfile.ZIP_DEFLATED
)
def walker( zip, directory, files, root=directory ):
for f in files:
f = os.path.join( directory, f )
archiveName = os.path.basename(f)
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
print f
os.path.walk( directory, walker, z )
z.close()
return zipFile
if __name__ == "__main__":
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
I still get the same error:
Traceback (most recent call last):
File "C:\Python24\Lib\site-packages\pythonwin\pywin\framework
\scriptutils.py", line 310, in RunScript
exec codeObject in __main__.__dict__
File "C:\Python24\Scripts\dirZip.py", line 20, in ?
toZip( 'c:\\aaa\\', 'c:\\bbb\\test.zip' )
File "C:\Python24\Scripts\dirZip.py", line 14, in toZip
os.path.walk( directory, walker, z )
File "C:\Python24\lib\ntpath.py", line 329, in walk
func(arg, top, names)
File "C:\Python24\Scripts\dirZip.py", line 12, in walker
zip.write( f, archiveName, zipfile.ZIP_DEFLATED )
File "C:\Python24\lib\zipfile.py", line 405, in write
fp = open(filename, "rb")
IOError: [Errno 13] Permission denied: 'c:\\aaa\\temp'
c:\\aaa\\temp is a directory in the directory I an trying to zip. I
want to use this script to back up my work once a day and would like
to
keep the directory structure as is. I can zip the files in c:\aaa\tem
fine so I guess that there aren't any open files in the directory.
Any more ideas?
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