zipstream-new 1.1.5

Zipfile generator that takes input files as well as streams

python-zipstream

zipstream.py is a zip archive generator based on python 3.3's zipfile.py. It was created to generate a zip file generator for streaming (ie web apps). This is beneficial for when you want to provide a downloadable archive of a large collection of regular files, which would be infeasible to generate the archive prior to downloading or of a very large file that you do not want to store entirely on disk or on memory.

The archive is generated as an iterator of strings, which, when joined, form the zip archive. For example, the following code snippet would write a zip archive containing files from 'path' to a normal file:

import zipstream

z = zipstream.ZipFile()
z.write('path/to/files')

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)

zipstream also allows to take as input a byte string iterable and to generate the archive as an iterator. This avoids storing large files on disk or in memory. To do so you could use something like this snippet:

def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)

Of course both approach can be combined:

def iterable():
    for _ in xrange(10):
        yield b'this is a byte string\x01\n'

z = zipstream.ZipFile()
z.write('path/to/files', 'my_archive_files')
z.write_iter('my_archive_iter', iterable())

with open('zipfile.zip', 'wb') as f:
    for data in z:
        f.write(data)

Since recent versions of web.py support returning iterators of strings to be sent to the browser, to download a dynamically generated archive, you could use something like this snippet:

def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)

If the zlib module is available, zipstream.ZipFile can generate compressed zip archives.

Installation

pip install zipstream-new

Requirements

• Python 2.6+, 3.2+, pypy

Examples

flask

from flask import Response

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    def generator():
        z = zipstream.ZipFile(mode='w', compression=ZIP_DEFLATED)

        z.write('/path/to/file')

        for chunk in z:
            yield chunk

    response = Response(generator(), mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

# or

@app.route('/package.zip', methods=['GET'], endpoint='zipball')
def zipball():
    z = zipstream.ZipFile(mode='w', compression=ZIP_DEFLATED)
    z.write('/path/to/file')

    response = Response(z, mimetype='application/zip')
    response.headers['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

django 1.5+

from django.http import StreamingHttpResponse

def zipball(request):
    z = zipstream.ZipFile(mode='w', compression=ZIP_DEFLATED)
    z.write('/path/to/file')

    response = StreamingHttpResponse(z, content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename={}'.format('files.zip')
    return response

webpy

def GET(self):
    path = '/path/to/dir/of/files'
    zip_filename = 'files.zip'
    web.header('Content-type' , 'application/zip')
    web.header('Content-Disposition', 'attachment; filename="%s"' % (
        zip_filename,))
    return zipstream.ZipFile(path)

Running tests

With python version > 2.6, just run the following command: python -m unittest discover

Alternatively, you can use nose.