ajax提交表单页面刷新很快

注：使用ajax 提交表单时 type类型最好不用submit 用button合适

<form>
      <div 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()">
  </div>
</form>

<script type="text/javascript">
    function formCheck() {
        $.ajax({
            type: "post",url: '/fudames/orderInfo/insertOrderInfo',data: $('#formId').serialize(),// 你的formid
            async: true,error: function(request) {
                new PNotify({
                    title: '提交失败',text: '信息录入失败',type: 'error',styling: 'bootstrap3'
                });
            },success: function(data) {

                if (data == "success") {
                    new PNotify({
                        title: '提交成功',text: '订单信息已录入',type: 'success',styling: 'bootstrap3'
                    });
                } else {
                    new PNotify({
                        title: '提交失败',styling: 'bootstrap3'
                    });
                }

            }
        });
    }
</script>