Question

我尝试从URL下载img，将其添加到zip存档中，然后通过Django HttpResponse响应此存档。

import os
import requests
import zipfile
from django.http import HttpResponse

url = 'http://some.link/img.jpg' 
file = requests.get(url)
data = file.content
rf = open('tmp/pic1.jpg', 'wb')
rf.write(data)
rf.close()
zipf = zipfile.ZipFile('tmp/album.zip', 'w') # This file is ok
filename = os.path.basename(os.path.normpath('tmp/pic1.jpg'))
zipf.write('tmp/pic1.jpg', filename)
zipf.close()
resp = HttpResponse(open('tmp/album.zip', 'rb'))
resp['Content-Disposition'] = 'attachment; filename=album.zip'
resp['Content-Type'] = 'application/zip'
return resp # Got corrupted zip file

当我将文件保存到tmp文件夹中时-可以，我可以将其解压缩。但是，当我响应此文件时，如果我尝试在Atom编辑器中打开（仅用于测试），则会在MacOS或意外的EOF上收到“错误1/2/21”。我还使用StringIO而不是保存zip文件，但这不会影响结果。

Answer 1

如果您使用的是Python 3，则可以这样做：

import os, io, zipfile, requests
from django.http import HttpResponse

# Get file
url = 'https://some.link/img.jpg'
response = requests.get(url)
# Get filename from url
filename = os.path.split(url)[1]
# Create zip
buffer = io.BytesIO()
zip_file = zipfile.ZipFile(buffer, 'w')
zip_file.writestr(filename, response.content)
zip_file.close()
# Return zip
response = HttpResponse(buffer.getvalue())
response['Content-Type'] = 'application/x-zip-compressed'
response['Content-Disposition'] = 'attachment; filename=album.zip'

return response

那是不保存文件的。下载的文件直接进入io。

要响应保存的文件，请使用以下语法：

response = HttpResponse(open('path/to/file', 'rb').read())

Django的响应ZIP文件

1 个答案: