在Django中提供动态生成的ZIP压缩文件

Question

如何在Django中为用户提供动态生成的ZIP存档？

我正在制作一个网站,用户可以选择任意书籍组合并将其下载为ZIP存档.我担心为每个请求生成这样的存档会使我的服务器慢慢爬行.我还听说Django目前没有一个很好的解决方案来提供动态生成的文件.

Answer 1

解决方案如下.

使用Python模块zipfile创建zip存档,但由于该文件指定了StringIO对象(ZipFile构造函数需要类似文件的对象).添加要压缩的文件.然后在您的Django应用程序中返回StringIO对象的内容,并将HttpResponsemimetype设置为application/x-zip-compressed(或至少application/octet-stream).如果需要,可以设置content-disposition标题,但这不应该是真正需要的.

但要注意,在每个请求上创建zip存档是个坏主意,这可能会导致您的服务器崩溃(如果存档很大,则不计算超时).性能方面的方法是在文件系统中的某处缓存生成的输出,并仅在源文件发生更改时重新生成它.更好的想法是提前准备档案(例如通过cron作业)并让你的网络服务器像往常一样提供服务.

Answer 2

这是一个Django视图来做到这一点:

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

Answer 3

许多答案在这里建议使用StringIO或BytesIO缓冲.但是这不是必需的,因为HttpResponse它已经是一个类似文件的对象:

response = HttpResponse(content_type='application/zip')
zip_file = zipfile.ZipFile(response, 'w')
for filename in filenames:
    zip_file.write(filename)
response['Content-Disposition'] = 'attachment; filename={}'.format(zipfile_name)
return response

Answer 4

对于python3我使用io.ByteIO因为StringIO的被弃用,以实现这一目标.希望能帮助到你.

import io

def my_downloadable_zip(request):
    zip_io = io.BytesIO()
    with zipfile.ZipFile(zip_io, mode='w', compression=zipfile.ZIP_DEFLATED) as backup_zip:
        backup_zip.write('file_name_loc_to_zip') # u can also make use of list of filename location
                                                 # and do some iteration over it
     response = HttpResponse(zip_io.getvalue(), content_type='application/x-zip-compressed')
     response['Content-Disposition'] = 'attachment; filename=%s' % 'your_zipfilename' + ".zip"
     response['Content-Length'] = zip_io.tell()
     return response

Answer 5

Django不直接处理动态内容的生成(特别是Zip文件).这项工作将由Python的标准库完成.您可以在此处了解如何在Python中动态创建Zip文件.

如果您担心它会降低服务器的速度,那么如果您希望有许多相同的请求,则可以缓存请求.您可以使用Django的缓存框架来帮助您.

总的来说,压缩文件可能是CPU密集型的,但Django不应该比另一个Python Web框架慢.

Answer 6

我使用了Django 2.0和Python 3.6.

import zipfile
import os
from io import BytesIO

def download_zip_file(request):
    filelist = ["path/to/file-11.txt", "path/to/file-22.txt"]

    byte_data = BytesIO()
    zip_file = zipfile.ZipFile(byte_data, "w")

    for file in filelist:
        filename = os.path.basename(os.path.normpath(file))
        zip_file.write(file, filename)
    zip_file.close()

    response = HttpResponse(byte_data.getvalue(), content_type='application/zip')
    response['Content-Disposition'] = 'attachment; filename=files.zip'

    # Print list files in zip_file
    zip_file.printdir()

    return response

Answer 7

无耻的插件:你可以使用django-zipview达到同样的目的.

之后pip install django-zipview:

from zipview.views import BaseZipView

from reviews import Review


class CommentsArchiveView(BaseZipView):
    """Download at once all comments for a review."""

    def get_files(self):
        document_key = self.kwargs.get('document_key')
        reviews = Review.objects \
            .filter(document__document_key=document_key) \
            .exclude(comments__isnull=True)

        return [review.comments.file for review in reviews if review.comments.name]