Python Django - 创建多个文件的 Zip 并使其可下载
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Django - Create A Zip of Multiple Files and Make It Downloadable
提问by yretuta
Possible Duplicate:
Serving dynamically generated ZIP archives in Django
可能的重复：
在 Django 中提供动态生成的 ZIP 档案
(Feel free to point me to any potential duplicates if I have missed them)
（如果我错过了任何潜在的重复项，请随时指出我）
I have looked at this snippet: http://djangosnippets.org/snippets/365/
我看过这个片段：http: //djangosnippets.org/snippets/365/
and this answer:
和这个答案：
but I wonder how I can tweak them to suit my need: I want multiple files to be zipped and the archive available as a download via a link (or dynamically generated via a view). I am new to Python and Django so I don't know how to go about it.
但我想知道如何调整它们以满足我的需要：我希望压缩多个文件，并且可以通过链接下载存档（或通过视图动态生成）。我是 Python 和 Django 的新手，所以我不知道如何去做。
Thank in advance!
预先感谢！
采纳答案by dbr
I've posted this on the duplicate questionwhich Willy linked to, but since questions with a bounty cannot be closed as a duplicate, might as well copy it here too:
我已经在威利链接到的重复问题上发布了这个，但由于悬赏问题不能作为重复关闭，也不妨在这里复制它：
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
回答by Nate Gentile
So as I understand your problem is not how to generate dynamically this file, but creating a link for people to download it...
因此，据我所知，您的问题不是如何动态生成此文件，而是创建一个供人们下载它的链接...
What I suggest is the following:
我的建议如下：
0) Create a model for your file, if you want to generate it dynamically don't use the FileField, but just the info you need for generating this file:
0) 为您的文件创建一个模型，如果您想动态生成它，请不要使用 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) Create and store your Zip. This step is important, you create your zip in memory, and then cast it to assign it to the FileField:
1) 创建并存储您的 Zip。这一步很重要，您在内存中创建 zip，然后将其转换为将其分配给 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) Create a url, for example get a number matching the id, you can also use a slugfield (this)
2）创建一个url，例如获取一个与id匹配的数字，你也可以使用一个slugfield（this）
url(r'^get_my_zip/(\d+)$', "zippyApp.views.get_zip")
3) Now the view, this view will return the file matching the id passed in the url, you can also use a slug sending the text instead of the id, and make the get filtering by your slugfield.
3) 现在是视图，该视图将返回与 url 中传递的 id 匹配的文件，您也可以使用发送文本而不是 id 的 slug，并通过您的 slugfield 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
