问题是，当我进入登录页面时，输入用户名/密码登录我即使我正确地写了它也会收到错误（我检查了我的数据库并且条目存在）。 我知道我应该使用Logger但我仍在学习使用Hibernate的Spring Framework。
编辑：
1.测试的第一部分我输入了一个空的用户名/密码（它适用于空/空）。对于第二部分，我使用 工作的那个
2.其他DAO与Hibernate一起正常工作（例如，他们正确检索数据，我对它们没有任何问题）
控制台输出如下所示：
-----------------------------------
User Service INVOKED
User Service-- searching for User:
DAO-- Searching for:
Hibernate: select this_.id as id9_0_, this_.accountName as accountN2_9_0_, this_.password as password9_0_, this_.secGrade as secGrade9_0_, this_.userEmail as userEmail9_0_, this_.userName as userName9_0_ from USER this_ where this_.accountName=?
DAO-- End search
DAO--Not found
User Service-- UserProxyImpl instantiated
User Service-- NOT FOUND,
null
true //<-- Error returned to the controller
-----------------------------------
User Service INVOKED
User Service-- searching for User:admin
DAO-- Searching for:admin
Hibernate: select this_.id as id9_0_, this_.accountName as accountN2_9_0_, this_.password as password9_0_, this_.secGrade as secGrade9_0_, this_.userEmail as userEmail9_0_, this_.userName as userName9_0_ from USER this_ where this_.accountName=?
UserService-- Error in retrieving user
// It stops here and i don't understand why
true // still, returns error to the controller.
这是会话超时问题吗？
登录控制器中的方法：
@RequestMapping(value = "/login", method = RequestMethod.GET)
public String Login(@RequestParam(value="error", required=false) boolean error,ModelMap model) {
if (error == true) {
model.put("error", "You have entered an invalid username or password!");
} else {
model.put("error", "");
}
System.out.println(error);
return "login";
}
用户实体：
@Entity
@Table(name = "USER")
public class User implements Serializable {
/**
*
*/
private static final long serialVersionUID = -1963505165125499005L;
private long id;
private int secGrade;
private String userName;
private String accountName;
private String password;
private String userEmail;
public User(String name,
String user_name,
String password,
String email,
int secGrade){
this.userName = name;
this.accountName = user_name;
this.password = password;
this.userEmail = email;
this.secGrade = secGrade;
}
@Id
@GeneratedValue
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public int getSecGrade() {
return secGrade;
}
public void setSecGrade(int secGrade) {
this.secGrade = secGrade;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getAccountName() {
return accountName;
}
public void setAccountName(String accountName) {
this.accountName = accountName;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getUserEmail() {
return userEmail;
}
public void setUserEmail(String userEmail) {
this.userEmail = userEmail;
}
}
用于处理数据库实体的代理类：
public class UserProxyImpl implements UserProxy {
private int secGrade;
private String name;
private String user_name;
private String password;
private String email;
public UserProxyImpl() { }
public UserProxyImpl(User usr){
if( usr != null){
System.out.println("USER PROXY--- constru from "+usr.getAccountName());
this.secGrade = usr.getSecGrade();
this.name = usr.getUserName();
this.user_name = usr.getAccountName();
this.password = usr.getPassword();
this.email = usr.getUserEmail();
}
}
+ GETTERS/SETTERS
}
来自UserServiceImpl的方法，它创建用户obj并添加权限：
@Override
public UserDetails loadUserByUsername(String username)
throws UsernameNotFoundException,DataAccessException {
// Declare a null Spring User
UserDetails user = null;
System.out.println("-----------------------------------");
System.out.println("User Service INVOKED");
try {
System.out.println("User Service-- searching for User:"+username);
// Search database for a user that matches the specified username
UserProxyImpl dbUser = new UserProxyImpl(userDAO.searchDB(username));
System.out.println("User Service-- UserProxyImpl instantiated");
if(dbUser.getName() != null){
System.out.println("User Service-- FOUND,"+username);
}
else{
System.out.println("User Service-- NOT FOUND,"+username);
}
// Populate the Spring User object with details from the dbUser
// getAuthorities() will translate the access level to the correct role type
System.out.println(dbUser.getName());
user = new User(
dbUser.getUser_name(),
dbUser.getPassword().toLowerCase(),
true,
true,
true,
true,
getAuthorities(new Integer(dbUser.getSecGrade()))
);
System.out.println(user.toString());
} catch (Exception e) {
System.out.println("UserService-- Error in retrieving user");
throw new UsernameNotFoundException("Error in retrieving user");
}
// Return user to Spring for processing.
return user;
}
@Override
public Collection<GrantedAuthority> getAuthorities(Integer access) {
// Create a list of grants for this user
List<GrantedAuthority> authList = new ArrayList<GrantedAuthority>(2);
// All users are granted with ROLE_USER access
authList.add(new GrantedAuthorityImpl("ROLE_USER"));
// Check if this user has admin access
// We interpret Integer(3) as an admin user
if ( access.compareTo(3) == 0) {
authList.add(new GrantedAuthorityImpl("ROLE_ADMIN"));
}
else if ( access.compareTo(2) == 0) {
authList.add(new GrantedAuthorityImpl("ROLE_MOD"));
}
// Return list of granted authorities
return authList;
}
用于从DB中检索对象的DAO类方法：
public User searchDB(String username){
User u = (User)this.getSessionFactory().getCurrentSession()
.createCriteria(User.class)
.add(Restrictions.eq("accountName",username))
.uniqueResult();
System.out.println("DAO-- End search");
if(u != null){
System.out.println("DAO-- Found:"+u.getUserName());
return u;
}
else{
System.out.println("DAO--Not found");
return null;
}
}
Spring Security配置。 XML：
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans
xmlns="http://www.springframework.org/schema/security"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.1.xsd">
<http auto-config = 'true' use-expressions="true" access-denied-page="/denied" >
<intercept-url pattern = "/home/" access="permitAll"/>
<intercept-url pattern = "/home/login" access="permitAll"/>
<intercept-url pattern = "/home/jobs" access="permitAll"/>
<intercept-url pattern = "/home/info" access="permitAll"/>
<intercept-url pattern = "/home/common" access="hasRole('ROLE_USER')"/>
<intercept-url pattern = "/home/desk" access="hasRole('ROLE_MOD')"/>
<intercept-url pattern = "/home/admin" access="hasRole('ROLE_ADMIN')"/>
<form-login login-page="/login"
default-target-url="/home"
authentication-failure-url="/home/login?error=true"/>
<logout logout-success-url="/home" />
</http>
<authentication-manager alias="authenticationManager">
<authentication-provider ref="authenticationProvider"/>
</authentication-manager>
<beans:bean id="authenticationProvider" class="org.springframework.security.authentication.dao.DaoAuthenticationProvider">
<beans:property name="userDetailsService" ref="UserServiceImpl"/>
</beans:bean>
<!-- Use a Md5 encoder -->
<beans:bean class="org.springframework.security.authentication.encoding.Md5PasswordEncoder" id="passwordEncoder"/>
<!-- A custom service where Spring will retrieve users and their corresponding access levels -->
<beans:bean id="UserServiceImpl" class="com.x.interview_management.service.impl.UserServiceImpl"/>
</beans:beans>
的Login.jsp：
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ page session="false" %>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form"%>
<%@ taglib prefix="sec" uri="http://www.springframework.org/security/tags" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<h1>${error}</h1>
<sec:authorize access="!isAuthenticated()">
<div id = "login" style="text-align:center;">
<h3 style="text-align:center">Login with Username and Password</h3>
<form action='/InterviewManagement/j_spring_security_check' method='POST'>
<table>
<tr>
<td>Username:</td>
<td><input type='text' name='j_username' value=''><td/>
<tr/>
<tr>
<td>Password:</td>
<td><input type='password' name='j_password'/><td/>
<tr/>
<tr>
<td><input name="submit" type="submit"/></td>
<td><input name="reset" type="reset"/></td>
</tr>
</table>
</form>
</div>
</sec:authorize>
<a href="/InterviewManagement/home/">return home</a>
</body>
</html>
对于代码布局感到抱歉，这是我第一次在StackExchange上发帖。
答案 0 :(得分：1)
问题是我没有User类的默认构造函数。