Django-创建一个包含多个文件的Zip并将其下载ab

Question

Possible Duplicate:
Serving dynamically generated ZIP archives in Django

（如果我错过了任何潜在的复制品，请随时指点我）

我看过这个片段： http://djangosnippets.org/snippets/365/

还有这个answer:

但我想知道如何调整它们以满足我的需要：我希望通过一个链接（或通过一个视图动态生成）压缩多个文件并将其作为下载文件提供。我对Python和Django还不熟悉，所以我不知道该怎么做。

提前谢谢！

Answer 1

所以据我所知，你的问题不是如何动态生成这个文件，而是创建一个链接供人们下载。。。

我建议如下：

0）为文件创建一个模型，如果要动态生成它，则不使用FileField，而只使用生成此文件所需的信息：

class ZipStored(models.Model):
    zip = FileField(upload_to="/choose/a/path/")

1）创建并存储您的拉链。这一步很重要，您可以在内存中创建zip，然后将其转换为分配给FileField：

function create_my_zip(request, [...]):
    [...]
    # This is a in-memory file
    file_like = StringIO.StringIO()
    # Create your zip, do all your stuff
    zf = zipfile.ZipFile(file_like, mode='w')
    [...]
    # Your zip is saved in this "file"
    zf.close()
    file_like.seek(0)
    # To store it we can use a InMemoryUploadedFile
    inMemory = InMemoryUploadedFile(file_like, None, "my_zip_%s" % filename, 'application/zip', file_like.len, None)
    zip = ZipStored(zip=inMemory)
    # Your zip will be stored!
    zip.save()
    # Notify the user the zip was created or whatever
    [...]

2）创建一个url，例如获取一个与id匹配的数字，也可以使用slugfield（this）

url(r'^get_my_zip/(\d+)$', "zippyApp.views.get_zip")

3）现在这个视图，这个视图将返回与url中传递的id匹配的文件，您还可以使用slug发送文本而不是id，并通过slug字段进行get过滤。

function get_zip(request, id):
    myzip = ZipStored.object.get(pk = id)
    filename = myzip.zip.name.split('/')[-1]
    # You got the zip! Now, return it!
    response = HttpResponse(myzip.file, content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename=%s' % filename

Answer 2

我已经在威利链接的duplicate question上发布了这个，但是由于悬赏的问题不能作为副本关闭，所以最好也复制到这里：

import os
import zipfile
import StringIO

from django.http import HttpResponse


def getfiles(request):
    # Files (local path) to put in the .zip
    # FIXME: Change this (get paths from DB etc)
    filenames = ["/tmp/file1.txt", "/tmp/file2.txt"]

    # Folder name in ZIP archive which contains the above files
    # E.g [thearchive.zip]/somefiles/file2.txt
    # FIXME: Set this to something better
    zip_subdir = "somefiles"
    zip_filename = "%s.zip" % zip_subdir

    # Open StringIO to grab in-memory ZIP contents
    s = StringIO.StringIO()

    # The zip compressor
    zf = zipfile.ZipFile(s, "w")

    for fpath in filenames:
        # Calculate path for file in zip
        fdir, fname = os.path.split(fpath)
        zip_path = os.path.join(zip_subdir, fname)

        # Add file, at correct path
        zf.write(fpath, zip_path)

    # Must close zip for all contents to be written
    zf.close()

    # Grab ZIP file from in-memory, make response with correct MIME-type
    resp = HttpResponse(s.getvalue(), mimetype = "application/x-zip-compressed")
    # ..and correct content-disposition
    resp['Content-Disposition'] = 'attachment; filename=%s' % zip_filename

    return resp