@RequestMapping(value = "/userLogin", method = RequestMethod.POST)
public String userLogin(@RequestParam("userName") String userName,
@RequestParam("password") String password,
HttpServletRequest request, Model model) {
System.out.println(userName);
System.out.println(password);
User user = userBasic.getUserByName(userName);
if (user != null) {
if (user.getPassword().equals(password)) {
HttpSession session = request.getSession();
session.setAttribute("userName", userName);
model.addAttribute("state", "success");
model.addAttribute("message", "登录成功！");
return "user_project";
} else {
model.addAttribute("failure", "用户名或密码错误！");
return "error";
}
} else {
model.addAttribute("state", "failure");
model.addAttribute("message", "该用户不存在！");
return "error";
}
}
在这个方法所在类上有@RequestMapping("/user")
我想要的是调用这个登录函数后如果密码正确就跳转到user_project.jsp这个页面
下面是我的index.jsp
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form" %>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/xhtml/DTD/xhtml-transitional.dtd">
<html lang="zh-CN">
<head>
<base href="<%=basePath%>">
<title>登录系统</title>
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache">
<meta http-equiv="expires" content="0">
<meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
<meta http-equiv="description" content="This is my page">
<meta http-equiv="Content-Type" content="text/html;charset=UTF-8">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
</head>
<body>
<form id="form_login">
用户名：<input type="text" name="userName" id="userName"/>
密码：<input type="password" name="password" id="password"/>
<button type="button" value="登录" onclick="toLogin()">登录</button>
</form>
<script src="http://libs.baidu.com/jquery/1.9.0/jquery.js" type="text/javascript"></script>
<script>
function toLogin(){
$.ajax({
url:"<%=basePath%>user/userLogin",
data:{"userName":"brianzhuang","password":"124652"},
type:"POST",
dataType:"html",
success: function(data,textstatus){
alert(textstatus);
},
error:function(jqxhr,textstatus,error){
alert(jqxhr);
alert(error);
alert(textstatus);
}
});
}
</script>
</body>
</html>
1，我想要利用jquery对ajax的封装来实现，ajax是异步的请求，是不是不支持跳转页面？
2，index页面中点击登录按钮后走的是error的处理方法，三个alert，为什么？调用不到controller的方法？？？
我快受不了这个了，求大神帮我解决！！！
