form表单文件上传提交且接口回调显示提交成功

前端：
<form method="post" enctype="multipart/form-data" id="formSubmit">
      <div class="row">
           <div class="col-lg-12" style="padding-left:25px;padding-top: 5px">
                <label>请选择文件</label>
                <input type="file" name="file" title="点击选择文件" multiple="" accept="*/*" class="form-control">
           </div>
      </div>
      <div class="row">
           <div style="padding-left:25px;padding-top:10px">
                <input type="submit" class="btn btn-primary">
           </div>
      </div>
</form>

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JS： $('#formSubmit').submit(function (event) {         //首先验证文件格式         var fileName = $(this).find("input[name=file]").val();         if (fileName === '') {             alert('请选择文件');             return;         }         // mulitipart form,如文件上传类         var formData = new FormData(this);         $.ajax({             async: false,             type: "POST",             url: "/upload",             data: formData,             dataType: "JSON",             mimeType: "multipart/form-data",             contentType: false,             cache: false,             processData: false,             success: function (data) {                 if (data.success) {                     layer.alert("上传成功")                 } else {                     layer.alert(data.error)                 }             }         });         return false;     });

　　js代码的最后需要return false 防止表单重复提交，刷新页面。

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java:  @RequestMapping(value = "/upload", method = RequestMethod.POST, produces = "application/json;charset=utf-8")     @ResponseBody     public SimpleAjaxResponse uploadSingleFile(HttpServletRequest request, Model model) {         try {             DefaultMultipartHttpServletRequest httpServletRequest = (DefaultMultipartHttpServletRequest) request;             MultipartFile multipartFile = httpServletRequest.getFile("file");             return new SimpleAjaxResponse(true);         } catch (Exception e) {             LOGGER.error(e.getMessage(), e);             e.printStackTrace();             return new SimpleAjaxResponse("上传出错了");         }       }