本题中的获取session,判断session中是否有用户数据，请问如何获取
我已经在登录页面设置了sessionStorage变量，但是如何在Java拦截器中获取呢？
老师顺便看看我这样做的对不对？
拦截器代码
public class LoginInterceptor implements HandlerInterceptor {
Logger logger = LoggerFactory.getLogger(LoginInterceptor.class);
@Override
public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler) throws Exception {
System.out.println("preHandle拦截已开启");
return true;
}
@Override
public void afterCompletion(HttpServletRequest request, HttpServletResponse response, Object handler, Exception ex) throws Exception {
System.out.println("after拦截以开启");
StringBuffer info = new StringBuffer();
info.append(request.getRemoteAddr());
info.append("|");
info.append(request.getRequestURL());
info.append("|");
info.append(request.getHeader("user-agent"));
logger.info(info.toString());
}
}登录后的页面
<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
<head>
<title>Title</title>
</head>
<body>
当前用户：${user.username} <a href="login.jsp">退出</a>
</body>
<script>
let username = "${user.username}";
let password = "${user.password}";
sessionStorage.setItem('username', username);
sessionStorage.setItem('password', password);
</script>
</html>登录页面
<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
<head>
<title>Title</title>
</head>
<body>
<div id="show">${info}</div>
<form action="/login" method="post">
用户名：<input type="text" name="username"><br>
密 码：<input type="password" name="password"><br>
<input type="submit" value="登录">
</form>
</body>
</html>Controller
@Controller
public class LoginController {
@PostMapping("/login")
public ModelAndView view(User user) {
ModelAndView modelAndView = new ModelAndView();
if (user.getUsername().equals("admin") && user.getPassword().equals("admin")) {
modelAndView.setViewName("/main.jsp");
modelAndView.addObject("user", user);
} else {
modelAndView.setViewName("/login.jsp");
modelAndView.addObject("info", "用户名或密码错误，请重新登录！");
}
return modelAndView;
}
}35
收起
正在回答
1回答
同学你好，1、一般情况下，并不会在前端页面设置session。建议同学在LoginController中设置session，然后在main.jsp页面中进行获取。
2、在拦截器中可以获取session信息的，同学可以先通过session判断用户是否登录，如果已登录，则将用户信息写入到日志中。参考代码如下所示：
import com.imooc.restful.entity.User;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.web.servlet.HandlerInterceptor;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.http.HttpSession;
public class LoginInterceptor implements HandlerInterceptor {
//1.创建日志对象，并加载当前类
Logger logger= LoggerFactory.getLogger(LoginInterceptor.class);
//拦截器， 对登录的用户拦截判断，已登录的放行，未登录的响应回去先登录
@Override
public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler) throws Exception {
System.out.println(request.getRequestURI());
User user =(User) request.getSession().getAttribute("user");
if(user!=null){
//登录成功，将session中的账号信息存放到log日志中
logger.info(user.getName()+":"+user.getPassword());
System.out.println("成功");
return true;
}else {
response.sendRedirect("/login.jsp");
return false;
}
}
}祝学习愉快！
-
慕粉1555086949 #1
我在Controller中设置了session，但是在拦截器中获取是null
@Controller public class LoginController { @PostMapping("/login") public ModelAndView view(User user, HttpSession httpSession) { ModelAndView modelAndView = new ModelAndView(); if (user.getUsername().equals("admin") && user.getPassword().equals("admin")) { modelAndView.setViewName("/main.jsp"); httpSession.setAttribute("user", user); } else { modelAndView.setViewName("/login.jsp"); } return modelAndView; } }拦截器
public class LoginInterceptor implements HandlerInterceptor { Logger logger = LoggerFactory.getLogger(LoginInterceptor.class); @Override public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler) throws Exception { System.out.println("preHandle拦截已开启"); User user = (User) request.getSession().getAttribute("user"); if (user != null) { logger.info(user.getUsername() + "|" + user.getPassword()); System.out.println("记录成功"); return true; } else { System.out.println("信息错误"); response.sendRedirect("/login.jsp"); return false; } } @Override public void afterCompletion(HttpServletRequest request, HttpServletResponse response, Object handler, Exception ex) throws Exception { System.out.println("after拦截以开启"); StringBuffer info = new StringBuffer(); info.append(request.getRemoteAddr()); info.append("|"); info.append(request.getRequestURL()); info.append("|"); info.append(request.getHeader("user-agent")); logger.info(info.toString()); } }
2022-08-18 14:43:22 -
同学你好，1、测试同学代码是可以获取到session中的数据的。运行效果图如下所示：
2、
1）猜测同学可能是没有成功登陆导致的，同学可以先登陆，然后在访问其他路径试一下，查看是否有被拦截并查看是否有获取到session值。
2）也有可能是页面缓存导致的，建议同学清除浏览器与编译器缓存试一下。
祝学习愉快！
2022-08-18 16:23:04 -
可以登录成功，我把拦截器配置注释掉之后，会跳转到main.jsp，也可以显示当前用户名。
但是开启之后就不会跳转了，而且在拦截器方法中也没有数据
拦截器配置<mvc:interceptors> <mvc:interceptor> <mvc:mapping path="/**"/> <mvc:exclude-mapping path="/resources/**"/> <mvc:exclude-mapping path="/login.jsp"/> <bean class="pers.qy.spring.interceptor.LoginInterceptor"/> </mvc:interceptor> </mvc:interceptors>
2022-08-18 16:43:50
恭喜解决一个难题，获得1积分~
来为老师/同学的回答评分吧
0 星