请大神帮忙编写代码 跪谢
@Controller
@RequestMapping("/*")
public class UserAction {
@RequestMapping("login.action")
public String login(HttpSession session,String username,String password,RedirectAttributes ra,String yzm){
Map parmas=new HashMap();
String sessionYzm=(String) session.getAttribute("verCode");
if(yzm.toLowerCase().equals(sessionYzm)){
}else{
return "redirect:/jsp/login.jsp";
}
parmas.put("logname", username);
parmas.put("logpwd", password);
User user=userService.slectUserByLognameAndLogpwd(parmas);
if(user==null){
return "redirect:/jsp/login.jsp";
}
String name=user.getLogname();
String pass=user.getLogpwd();
if(username.equals(name)&& password.equals(pass)){
return "main";
}else{
return "redirect:/jsp/login.jsp";
}
}
}
jsp页面:
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
登陆页面
用户登录
function refresh(){
var refreshUrl="creatYzm.action?"+new Date();
$("#yzm").attr("src",refreshUrl);
}
$().ready(function() {
$("#myForm").validate();
});
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代码就不帮你编写了。只是简单说一下,你完全可以采用异步的方式登陆,请求的时候做好加密等安全性措施的处理。后台处理玩后再在异步的回调里处理,成功跳转到你想去的画面,失败给出提示。
另外springmvc的@RequestMapping 写的是请求的url,你这么写【@RequestMapping("login.action")】.action应该交给spring去拦截。