关于文件上传解析结果List为空的问题：List<FileItem> list = upload.parseRequest(request);list=null;
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悬赏园豆：10
[已关闭问题]
关闭于 2016-01-29 21:18
List<FileItem> list = upload.parseRequest(request);list=null; why???
Servlet里面的代码：
1 public String doUploadPicture(HttpServletRequest request){ 2 String result = "error"; 3 try{ 4 String filepath = getServletContext().getInitParameter("photopath");//配置文件里面配置了照片路径？ 5 6 DiskFileItemFactory factory = new DiskFileItemFactory();// 创建文件项目工厂对象 7 factory.setRepository(new File(TEMP_FOLDER)); //设置临时文件夹为TEMP_FOLDER 8 factory.setSizeThreshold(1024 * 1024); // 设置缓冲区大小为 1M 9 ServletFileUpload upload = new ServletFileUpload(factory);//用工厂实例化上传组件,ServletFileUpload用来解析文件上传请求 10 11 List<FileItem> list = upload.parseRequest(request);// 解析结果放在List中 12 13 FileItem item = getUploadFileItem(list); // 获取上传的文件 14 String filename = getUploadFileName(item); // 获取文件名 15 filename = filename.substring(filename.length() - 3, filename.length()); //取得原文件名的后缀名 16 String photo_prex = "." + filename; 17 Date date = new Date(); 18 SimpleDateFormat df = new SimpleDateFormat("yyyyMMddHHmmss"); 19 String time = df.format(date); 20 filename = time + photo_prex; //修改原来的文件名 21 // 真正写到磁盘上 22 item.write(new File(filepath, filename)); // 第三方提供的 23 //调用服务层插入一条数据 24 result = filename; 25 }catch(Exception e){ 26 result = "error"; 27 }finally{ 28 return result; 29 } 30 }
JAVASCRIPT代码：
1 $("#ajaxUpload").click(function(){ 2 var title = $.trim($("#pic_title").val()); 3 if(title == ""){ 4 alert("请输入文件名！"); 5 return false;} 6 7 var type = $(":input[type='file']").val().match(/^(.*)(\.)(.{1,8})$/)[3]; 8 type = type.toUpperCase(); 9 if (type == "JPG" || type == "PNG" ) { 10 } else { alert("上传类型有误,请上传jpg或png格式图片文件"); return false; } 11 12 $.ajaxFileUpload({ 13 url :'<%= server_path%>/UploadAction?action=uploadPicture', 14 secureuri : false, 15 fileElementId : 'fileToUpload', 16 dataType : 'json', 17 success : function(dataResult, status) { 18 if(dataResult.result == "ok" && status == "success" && dataResult.returnValue != ""){ 19 alert("success"); 20 var path = dataResult.returnValue; 21 var url = "<%= server_path%>/UserAction?action=doAddPicture"; 22 var json = { 23 "userid" : userid, 24 "path" : path, 25 "title" : title, 26 } 27 var dataResult = postAjax(url, dataResult); 28 if(dataResult.result == "ok"){ 29 alert("上传成功！"); 30 showUserPicture(path,title); 31 }else{alert("失败！");} 32 33 } 34 }, 35 error : function(data,status,e){ 36 alert("上传出错！") 37 } 38 }) 39 return false;//? 40 })
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