开发笔记:AJAX实现form表单提交

1 $(function(){
2 $("#submitbtn").click(function(){
3 login();
4 })
5 })
6 function login(){
7 var userName = $("#userName").val();//获取userName的值
8 var userPwd = $("#userPwd").val();//获取userPwd的值
9 $.ajax({
10 url:"user/judgeLogin",
11 type:"POST",
12 data:{"userName":userName,"userPwd":userPwd},
13 dataType:"json",
14 success:function(returnData){
15 alert(returnData);
16 }
17 })
18 }

1 function login(){
2 $.ajax({
3 url:"user/judgeLogin",
4 type:"POST",
5 data:$("#form").serialize(),//表单序列化
6 dataType:"json",
7 success:function(returnData){
8 alert(returnData);
9 }
10 })
11 }

1 /**
2 * 登录判断
3 * @param userDto
4 * @return
5 */
6 @ResponseBody
7 @RequestMapping("/judgeLogin")
8 public String judgeLogin(UserDto userDto) {
9 int k = userService.judgeLogin(userDto);
10 if(k > 0) {
11 return "--登录成功！--";
12 }else {
13 return "--登录失败！--";
14 }
15 }