我有一个Django应用程序，允许用户下载他们购买这些MP3文件都托管在Amazon S3中的MP3文件。我怎么能强制下载，当用户点击下载按钮，而不让他们看到（亚马逊）原始链接？ 我有一个观点，即下载该文件，但该文件已损坏。 下面是它的样子：

 高清下载（要求）：
    文件名='https://s3-eu-west-1.amazonaws.com/skempi/Ihsahn/04-emancipation-qtxmp3.mp3
    响应= Htt的presponse（MIMETYPE ='应用/武力下载）
    响应['内容处置'] ='附件;文件名=％S％文件名
    响应[X-SENDFILE] =文件名
    返回响应
 

解决方案

我不容易找到指定如何做到这一点，最终回到这个问题一次又一次，当我搜索的任何地方。因此，

通过使用博托后端Django的仓库，要做到这一点的方式是一样的东西。

 文件路径= settings.MEDIA_DIRECTORY + file.name
response_headers = {
    响应内容类型：应用/武力下载，
    响应内容处置：附件;文件名=％S'％file.name
    }
URL = s3.generate_url（6​​0，GET，
                斗= settings.AWS_STORAG​​E_BUCKET_NAME，
                键=文件路径，
                response_headers = response_headers，
                force_http =真）
返回http.Htt presponseRedirect（URL）
 

I have a Django app that allows users to download MP3 files that they purchased and these MP3 files are hosted in Amazon S3. How can I force a download when users click a "download" button without allowing them to see the original link (to Amazon)? I have a view that downloads the file but the file is corrupt. Here is how it looks like:

def download(request):
    filename = 'https://s3-eu-west-1.amazonaws.com/skempi/Ihsahn/04-emancipation-qtxmp3.mp3'
    response = HttpResponse(mimetype='application/force-download')
    response['Content-Disposition']='attachment;filename="%s"'%filename
    response["X-Sendfile"] = filename
    return response

I couldn't easily find anywhere that specified how to do this and ended up back at this question time and again when I searched. So

With django-storages using the boto backend, the way to do this is something like

filepath = settings.MEDIA_DIRECTORY + file.name
response_headers = {
    'response-content-type': 'application/force-download',
    'response-content-disposition':'attachment;filename="%s"'%file.name
    }
url = s3.generate_url(60, 'GET',
                bucket=settings.AWS_STORAGE_BUCKET_NAME,
                key=filepath,
                response_headers=response_headers,
                force_http=True)
return http.HttpResponseRedirect(url)