Ajax提交表單頁面刷新很快的解決方法

<form>
   <p class="col-md-9 col-sm-9 col-xs-12 col-md-offset-3">
             <input type="button" class="btn btn-info" value="重置" onclick="return resetaa()">
             <input type="button" class="btn btn-success" value="提交" onclick="return formCheck()">
</form>
 <script type="text/JavaScript">
   function formCheck(){
    $.ajax({
       type: "post",
       url:&#39;/fudaMes/orderInfo/insertOrderInfo&#39;,
       data:$(&#39;#formId&#39;).serialize(),// 你的formid
       async: true,
       error: function(request) {
       new PNotify({
           title: &#39;提交失败&#39;,
           text: &#39;信息录入失败&#39;,
           type: &#39;error&#39;,
           styling: &#39;bootstrap3&#39;
         });
       },
       success: function(data) {
      if(data=="success"){
          new PNotify({
             title: &#39;提交成功&#39;,
             text: &#39;订单信息已录入&#39;,
             type: &#39;success&#39;,
             styling: &#39;bootstrap3&#39;
           });  
       }else{
      new PNotify({
             title: &#39;提交失败&#39;,
             text: &#39;信息录入失败&#39;,
             type: &#39;error&#39;,
             styling: &#39;bootstrap3&#39;
           });
      } 
       }
     });
   }
</script>