如何正确下载带有django请求的文件

我正在运行一个Django应用程序，在那里我可以上传文件。现在我想使用请求下载文件。我试图创建一个下载文件的视图，这样我就可以通过请求进行调用。但这不太管用

My model: 

class FileCollection(models.Model):
    name = models.CharField(max_length=120, null=True, blank=True)
    store_file = models.FileField(storage=PrivateMediaStorage(), null=True, blank=True)
    creation_date = models.DateTimeField(null=True, blank=True)

它告诉我TypeError：应该是str、bytes或os.PathLike对象，而不是FieldFile

如果我传入apiview/admin中提供的url，我会得到：FileNotFoundError:[Errno 2]没有这样的文件或目录

还尝试：

def fileview(request):
    path = FileCollection.objects.first()
    obj = path.store_file
    o = str(obj)
    file_path = os.path.join(settings.MEDIA_ROOT, o)
    print(file_path)
    if os.path.exists(file_path):
        with open(file_path, 'rb') as fh:
            response = HttpResponse(fh.read(),
                                    content_type="application/vnd.ms-excel")
            response[
                'Content-Disposition'] = 'inline; filename=' + os.path.basename(
                file_path)
            return response

但这给了我ValueError：视图文件_storage_api.api.v1.views.fileview没有返回HttpResponse对象。它没有返回任何结果

这样做对吗

我非常感谢你的帮助或提示。 非常感谢

file_path = file.store_file

不是文件路径，而是文件字段的实例 试用

file_path = file.store_file.name

然后使用第二个代码段

编辑：以下是我使用的代码： l_targetFile是实际文件的路径

    l_prjPath = os.path.realpath(os.path.dirname(__file__)).replace(<adapt the path here>)
    l_userFileName= <file field>.name.replace('<upload to sub-dir>','')
    l_targetFile = l_prjPath + '/media/' + l_fileObj.file_obj.name
    #return the file
    response = FileResponse(open(l_targetFile, 'rb'),\
                            (l_responseDisposition == 'attachment'))        

    #process the filename as stored on the local machine in case of download
    try:
        #check if it will throw
        l_tmpUserName = l_userFileName.encode('ascii')

        #no error use the non-encoded filename
        l_fileExpr = 'filename="{0}"'.format(l_userFileName)
    except UnicodeEncodeError:
        # Handle a non-ASCII filename
        l_fileExpr = "filename*=utf-8''{}".format(quote(l_userFileName))

    response['Content-Disposition'] = '{0};{1};'.format(l_responseDisposition,l_fileExpr)
    if '.pdf' in l_userFileName:
        response['Content-Type'] = 'application/pdf'
    elif l_dataSource == CMSArchiveEntry.SOURCE_TAG:
        response['Content-Type'] = 'text/html; charset=utf-8'
    return response

---

谢谢我的朋友！当我尝试这一点，我仍然得到文件未找到错误，即使该文件存在。它在AWS bucket中，但django仍然应该找到它，对吗？我通过以下方式进行管理：file\u path=file.store\u file read\u file=file\u path.readAWS不应该更改任何内容，但配置显然存在一些问题。name应返回一个相对于媒体根目录的路径。我将在下面添加用于呈现文件响应的代码

    l_prjPath = os.path.realpath(os.path.dirname(__file__)).replace(<adapt the path here>)
    l_userFileName= <file field>.name.replace('<upload to sub-dir>','')
    l_targetFile = l_prjPath + '/media/' + l_fileObj.file_obj.name
    #return the file
    response = FileResponse(open(l_targetFile, 'rb'),\
                            (l_responseDisposition == 'attachment'))        

    #process the filename as stored on the local machine in case of download
    try:
        #check if it will throw
        l_tmpUserName = l_userFileName.encode('ascii')

        #no error use the non-encoded filename
        l_fileExpr = 'filename="{0}"'.format(l_userFileName)
    except UnicodeEncodeError:
        # Handle a non-ASCII filename
        l_fileExpr = "filename*=utf-8''{}".format(quote(l_userFileName))

    response['Content-Disposition'] = '{0};{1};'.format(l_responseDisposition,l_fileExpr)
    if '.pdf' in l_userFileName:
        response['Content-Type'] = 'application/pdf'
    elif l_dataSource == CMSArchiveEntry.SOURCE_TAG:
        response['Content-Type'] = 'text/html; charset=utf-8'
    return response