SpringMVC结合Ajaxfileupload异步多文件上传至远程服务器

项目中我们可能会使用到异步上传并且有时需要是异步的，那么Ajaxfileupload是一个不错的选择,最近刚 做了一个多文件上传到远程服务器，远程服务器提供接口，当远程服务器上传结束后返回一个文件的服务器地址。分享出来给有需要的参考 下，Ajaxfileupload默认是一次只能上传一个文件的，所以需要稍稍修改下他的源码就可以上传多个文件，这里就不多说，百度一下自然就知道。

<input?type="file"?id="playeraddress"?name="playeraddress"?/>
<input?type="file"?id="cover"?name="cover"?/>
//这里就是两个file?id自己定义

$.ajaxFileUpload({
????url?:?web_path+'upload/upload.do',????secureuri?:?false,????data?:?data,//需要传递的数据?json格式
????fileElementId?:['playeraddress','cover'],????dataType?:?'json',????success?:?function(data)?{
???????????//上传成功后的回调。
????},????error?:?function(data)?{
????}
});

/**
?????*
?????*?
?????*?@param?request
?????*?@param?response
?????*?@param?audioItem
?????*?@return
?????*/
????@RequestMapping(value?=?"/upload.do",?method?=?{?RequestMethod.POST?})
????@ResponseBody
????public?String?addMusic(HttpServletRequest?request,????????????HttpServletResponse?response,?AudioItem?audioItem)?{
????????//这里无需理会，这只是
????????String?uploadFileUrl?=?GlobalUtil?.getValue("upload_audioItem_audio_url");
????????String?uploadIconUrl?=?GlobalUtil.getValue("upload_audioItem_pic_url");

????????CommonsMultipartResolver?resolver?=?new?CommonsMultipartResolver(
????????????????request.getServletContext());
????????if?(resolver.isMultipart(request))?{
????????????MultipartHttpServletRequest?multiRequest?=?(MultipartHttpServletRequest)?request;
????????????//?取得request中的所有文件名
????????????Iterator<String>?iter?=?multiRequest.getFileNames();
????????????while?(iter.hasNext())?{
????????????????String?fileName?=?iter.next();
????????????????//?取得上传文件
????????????????MultipartFile?file?=?multiRequest.getFile(fileName);
????????????????/**
?????????????????*?获取文件名
?????????????????*/
????????????????String?originalFilename?=?file.getOriginalFilename();
????????????????
????????????????if?("playeraddress".equals(fileName)&&!"".equals(originalFilename))?{
????????????????????try?{
????????????????????????//这里的upload方法就是以http?post的方式上传文件的?稍后贴出
????????????????????????String?url?=?BaseService.upload(uploadFileUrl,????????????????????????????????file.getOriginalFilename(),????????????????????????????????file.getInputStream());
????????????????????}?catch?(IOException?e)?{
????????????????????????//?TODO?Auto-generated?catch?block
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????????if?("cover".equals(fileName)&&!"".equals(originalFilename)){
????????????????????try?{
????????????????????????String?icon?=?BaseService.upload(uploadIconUrl,????????????????????????????????file.getInputStream());
????????????????????}?catch?(IOException?e)?{
????????????????????????//?TODO?Auto-generated?catch?block
????????????????????????e.printStackTrace();
????????????????????}
????????????????}
????????????}
????????????//这里需要注意?使用Ajaxfileupload需要使用以下方式返回结果?回调不了success?方法
????????????response.setContentType("text/html");
??
????????????try?{
????????????????response.getWriter().write("ok");
????????????}?catch?(IOException?e)?{
????????????????e.printStackTrace();
????????????}

????????}
????????return?null;
????}

//最后一步了?
????public?static?String?upload(String?httpurl,?String?fileName,?InputStream?inputStream)?{
????????String?result?=?"";
????????try?{
????????????//这里看你的网络环境?按需设置代理??正常都不需要理会
????????????if?(true)?{
????????????????System.setProperty("http.proxyHost",?BaseService.PROXY_IP);
????????????????System.setProperty("http.proxyPort",?BaseService.PROXY_PORT);
????????????}
????????????String?BOUNDARY?=?"---------7d4a6d158c9";?//?定义数据分隔线
????????????URL?url?=?new?URL(httpurl);
????????????HttpURLConnection?conn?=?(HttpURLConnection)?url.openConnection();
????????????//?发送POST请求必须设置如下两行
????????????conn.setDoOutput(true);
????????????conn.setDoInput(true);
????????????conn.setUseCaches(false);
????????????conn.setRequestMethod("POST");
????????????conn.setRequestProperty("connection",?"Keep-Alive");
????????????conn.setRequestProperty("user-agent",????????????????????"Mozilla/4.0?(compatible;?MSIE?6.0;?Windows?NT?5.1;?SV1)");
????????????conn.setRequestProperty("Charsert",?"UTF-8");
????????????conn.setRequestProperty("Content-Type",????????????????????"multipart/form-data;?boundary="?+?BOUNDARY);
????????????OutputStream?out?=?new?DataOutputStream(conn.getOutputStream());
????????????byte[]?end_data?=?("rn--"?+?BOUNDARY?+?"--rn").getBytes();//?定义最后数据分隔线
????????????StringBuilder?sb?=?new?StringBuilder();
????????????sb.append("--");
????????????sb.append(BOUNDARY);
????????????sb.append("rn");
????????????sb.append("Content-Disposition:?form-data;name="file"?+?1
????????????????????+?"";filename=""?+?fileName?+?""rn");
????????????sb.append("Content-Type:application/octet-streamrnrn");
????????????byte[]?data?=?sb.toString().getBytes();
????????????out.write(data);
????????????DataInputStream?in?=?new?DataInputStream(inputStream);
????????????int?bytes?=?0;
????????????byte[]?bufferOut?=?new?byte[1024];
????????????while?((bytes?=?in.read(bufferOut))?!=?-1)?{
????????????????out.write(bufferOut,?0,?bytes);
????????????}
????????????out.write("rn".getBytes());?//?多个文件时，二个文件之间加入这个
????????????in.close();
????????????out.write(end_data);
????????????out.flush();
????????????out.close();
????????????//?定义BufferedReader输入流来读取URL的响应
????????????BufferedReader?reader?=?new?BufferedReader(new?InputStreamReader(
????????????????????conn.getInputStream()));

????????????String?line?=?null;
????????????while?((line?=?reader.readLine())?!=?null)?{
????????????????result+=line;
????????????}
????????}?catch?(Exception?e)?{
????????????System.out.println("发送POST请求出现异常！"?+?e);
????????}????
????????return?new?JSONObject(result).getString("url");
????}
//?到这就结束了！希望能帮到你。