浏览器上传文件+django后台处理

Posted 2022-03-14 huanhuaqingfeng

1，浏览器端端js程序

     

 

 

2，django后端处理程序

     接收来自浏览器上传的文件，并把文件存储在一指定的路径下。

product_id = request.POST.get(‘product_id‘)
sign = request.POST.get(‘sign‘)
file_size = request.POST.get(‘file_size‘)
sw_ver = request.POST.get(‘sw_ver‘)

file_dict = request.FILES.items()
if not file_dict:
      dict[‘status‘] = ‘no file upload‘
      return dict
for (k, v) in file_dict:
      logger.info("dic[%s]=%s", k, v)
      file_data = request.FILES.getlist(k)
      for fl in file_data:
            filename = fl._get_name()
            logger.info("filename=%s", filename)

           path_file=base_path+"upgrade/"+product_id+‘/‘

           logger.info("path_file=%s", path_file)
           if not os.path.exists(path_file):
                os.makedirs(path_file)
          logger.info("path_file=%s", path_file)
          path_file+=filename
          # logger.info( "path_file1=%s",path_file1)
          # path_file=path_file1.encode(‘gb2312‘)
          logger.info("path_file=%s", path_file)
          try:
              with open(path_file, "wb") as f:
              if fl.multiple_chunks():
                   logger.info("multiple_chunks")
                   for content in fl.chunks():
                        f.write(content)
             else:
                   data=fl.read() ###.decode(‘utf-8‘)
                   logger.info("not multiple_chunks=%s", data)
                   f.write(data)
            #except Exception as e:
            except:
                 #logger.info("error=%s", repr(e))
                   logger.info("file write fail")
                   dict[‘status‘] = ‘file write fail‘
                   return dict