表单提交自动刷新？

callback函数中，没有validator的错误，函数会执行，但会把页面刷新。例如下面的代码，无法显示弹窗（modal1和modal2都不行），并且会把表单的内容重置，类似于页面刷新，请问如何解决这个问题，组织刷新？
`var validator = new Validator('form1',[
{
name:"temperature",
display:"请输入正确的体温值",
rules: 'is_money'
},{
name:"BMI",
display:"请输入正确的BML值",
rules: 'is_money'
}
],function(obj,evt){

var errors_elm = document.getElementById('error_msg');
errors_elm.style.display = 'none';
console.log(evt);
console.log(obj);
if(obj.errors.length>0){
// 判断是否错误
var error_str = '';
for (var i = 0; i < obj.errors.length; i++) {
error_str += i+1 + ':' + obj.errors[i].message + '；
';
}
errors_elm.style.display = "block";
errors_elm.innerHTML = error_str;

$("#tipModal2").modal('show');
}else{
var CRE = $("input[name='CRE']").val();

var flag = true;
if($.trim(CRE) < 64 || $.trim(CRE) > 104){
flag = false;
$('#CRE_tip').show();
}

if(flag){
$.ajax({
type: "POST", //提交的方法
url: "/get_result/", //提交的地址
data: $('#form1').serialize(),// 序列化表单值
async: false,
error: function (request) { //失败的话
//alert("Connection error");
$('#modal1').modal('show');
},
success: function (result) { //成功
console.log(result); //就将返回的数据显示出来
$('#modal2').modal('show');
}
});
return false;
}else{
// alert("hhh");
$("#tipModal").modal('show');
}
}
});`

Start with the inline Validator callback for form1 and trace how the form submission event reaches it. Reproduce the reported case with a valid form, then inspect the callback's return path and the modal/ajax branches. Done means the callback displays the intended modal without refreshing or resetting the form.

Written by the indexing model from the issue text.