CODEWITHSUNDEEP
pythonzippython-zipfile
tkbx
tkbx

Reputation: 16275

Create .zip in Python?

I'm trying to create a function in my script that zips the contents of a given source directory (src) to a zip file (dst). For example, zip('/path/to/dir', '/path/to/file.zip'), where /path/to/dir is a directory, and /path/to/file.zip doesn't exist yet. I do not want to zip the directory itself, this makes all the difference in my case. I want to zip the files (and subdirs) in the directory. This is what I'm trying:

def zip(src, dst):
    zf = zipfile.ZipFile("%s.zip" % (dst), "w")
    for dirname, subdirs, files in os.walk(src):
        zf.write(dirname)
        for filename in files:
            zf.write(os.path.join(dirname, filename))
    zf.close()

This creates a zip that is essentially /. For example, if I zipped /path/to/dir, extracting the zip creates a directory with "path" in it, with "to" in that directory, etc.

Does anyone have a function that doesn't cause this problem?

I can't stress this enough, it needs to zip the files in the directory, not the directoy itself.

Upvotes: 38

Views: 55735

Answers (3)

andrewdotn
andrewdotn

Reputation: 34823

The zipfile.write() method takes an optional arcname argument that specifies what the name of the file should be inside the zipfile.

You can use this to strip off the path to src at the beginning. Here I use os.path.abspath() to make sure that both src and the filename returned by os.walk() have a common prefix.

#!/usr/bin/env python2.7

import os
import zipfile

def zip(src, dst):
    zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED)
    abs_src = os.path.abspath(src)
    for dirname, subdirs, files in os.walk(src):
        for filename in files:
            absname = os.path.abspath(os.path.join(dirname, filename))
            arcname = absname[len(abs_src) + 1:]
            print 'zipping %s as %s' % (os.path.join(dirname, filename),
                                        arcname)
            zf.write(absname, arcname)
    zf.close()

zip("src", "dst")

With a directory structure like this:

src
└── a
    ├── b
    │   └── bar
    └── foo

The script prints:

zipping src/a/foo as a/foo
zipping src/a/b/bar as a/b/bar

And the contents of the resulting zip file are:

Archive:  dst.zip
  Length     Date   Time    Name
 --------    ----   ----    ----
        0  01-28-13 11:36   a/foo
        0  01-28-13 11:36   a/b/bar
 --------                   -------
        0                   2 files

Upvotes: 57

Jon-Eric
Jon-Eric

Reputation: 17275

Use the arcname parameter to control the name/path in the zip file.

For example, for a zip file that contains only files, no directories:

zf.write(os.path.join(dirname, filename), arcname=filename)

Or to invent a new directory inside the zip file:

zf.write(os.path.join(dirname, filename), arcname=os.path.join("my_zip_dir", filename))

Upvotes: 1

agoebel
agoebel

Reputation: 401

From what I can tell you are close. You could use dirname and basename to make sure you are grabbing the right path name:

>>> os.path.dirname("/path/to/dst")
'/path/to'
>>> os.path.basename("/path/to/dst")
'dst'

Then using chdir you can make sure you are in the parent so the paths are relative.

def zip(src, dst):
    parent = os.path.dirname(dst)
    folder = os.path.basename(dst)

    os.chdir(parent):
    for dirname, subdirs, filenames in os.walk(folder):
        ...

This creates:

dst/a.txt
dst/b
dst/b/c.txt
...etc...

If do not want to include the name "dst" you can just do os.chdir(dst) and then os.walk('.').

Hope that helps.

Upvotes: 1

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