我在django项目中有此代码。它将在服务器中Django根目录之外的路径中上传文件。现在我想要一个可以检索该文件并显示客户端的函数。它是非直接链接,类似于http://example.com/uploads/ {ID}。现在怎么可能。请注意,上传的文件是图像。def upload(request):if request.method == 'POST' and request.FILES['file1']: myfile = request.FILES['file1'] fileType = request.GET.get('type',None) if fileType==None: return json_response(65,400) if myfile.size > 5*1024*1000: return json_response(85,400) elif not (myfile.content_type.startswith('image') or myfile.content_type == 'application/pdf'): return json_response(185,400) fs = FileSystemStorage('../uploads') filename = fs.save(myfile.name, myfile) uploaded_file_url = fs.url(filename) request.user.file_set.create(name = uploaded_file_url, type = fileType, createdAt = datetime.now())
添加回答
举报
0/150
提交
取消