python+plus.uploader.createUpload

Question

js：

var task=plus.uploader.createUpload(server,
{method:"POST"},
function(t,status){ //上传完成
if(status==200){

                    //资源
                    var responseText = t.responseText;

                    //转换成json
                    var json = eval('(' + responseText + ')');

                    //上传文件的信息
                    var files = json.files;

                    //上传成功以后的保存路径
                    var img_url = files.uploadkey.url;            

                    //ajax 写入数据库

                    var urls = 'http://tongle.dongyixueyuan.com'+img_url;
                    switch(num){
                        case '0':
                            img_element['0'].src = urls;
                            img_element['0'].setAttribute('img',img_url);
                        break;
                        case '2':
                            img_element['1'].src = urls;
                            img_element['1'].setAttribute('img',img_url);
                        break;
                        case '3':
                            img_element['2'].src = urls;
                            img_element['2'].setAttribute('img',img_url);
                        break;
                    }

                    //关闭转圈等待框
                    wt.close();
                }else{
                    console.log("上传失败："+status);
                    var json = eval('(' + responseText + ')');
                    console.log(json);
                    //关闭原生的转圈等待框
                    wt.close();
                }
            });

        task.addData("client","");
        task.addData("uid",getUid());
        for(var i=0;i<files.length;i++){
            var f=files[i];
            console.info(f.path+"______"+f.name);
            task.addFile(f.path,{key:f.name});
        }
        task.start();



python的后台如何接收值，并进行图片保存，不熟悉python，请教