[英]Spring boot with Spring Security and custom database
對不起,我的英語不好。
由於更改了數據庫配置,因此無法成功登錄我的應用程序。 我正在使用Spring安全性。 在進行更改之前,一切正常。
我有兩個實體:
User.java
package betizy.models;
//imports
@Entity
@Table(name = "use_user")
public class User {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name="USE_ID")
private Long id;
@NotNull
@Column(name = "USE_USERNAME")
private String username;
@NotNull
@Column(name = "USE_PASSWORD")
private String password;
@NotNull
@Column(name = "USE_EMAIL")
private String email;
//getters and setters
}
UserRole.java
package betizy.models;
//imports
@Entity
@Table(name = "usr_user_role")
public class UserRole {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name="USR_ID")
private Long id;
@ManyToOne
@JoinColumn(name = "USR_USE_ID")
private User user;
@NotNull
@Column(name = "USR_ROLE")
private String role;
//getters and setters
}
login.html
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" xmlns:th="http://www.thymeleaf.org">
<head>
<title>Spring Security Example </title>
<script src="/webjars/angularjs/1.4.9/angular.js"></script>
<script src="webjars/jquery/2.0.3/jquery.min.js"></script>
<link rel="stylesheet" href="/webjars/bootstrap/3.3.6/css/bootstrap.css">
<script src="/js/index.js"></script>
</head>
<body ng-app="Betizy">
<div header></div>
<div th:if="${param.error}">
Invalid username and password.
</div>
<div th:if="${param.logout}">
You have been logged out.
</div>
<form th:action="@{/login}" method="post">
<div><label> User Name : <input type="text" name="username" required/> </label></div>
<div><label> Password: <input type="password" name="password" required/> </label></div>
<div><input type="submit" value="Sign In"/></div>
</form>
<div>To open a new account click <a href="/register">here</a>. </div>
<div footer></div>
</body>
</html>
SecurityConfig.java
package betizy.security;
//imports
@Configuration
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {
@Autowired
DataSource dataSource;
@Autowired
public void configAuthentication(AuthenticationManagerBuilder auth) throws Exception {
auth.jdbcAuthentication().dataSource(dataSource)
//.passwordEncoder(passwordEncoder())
.usersByUsernameQuery(
"select * from use_user where use_user.use_username=?")
.authoritiesByUsernameQuery(
"select * from usr_user_role inner join use_user on use_user.use_id = usr_user_role.usr_use_id where use_user.use_username=?");
}
@Bean
public PasswordEncoder passwordEncoder(){
PasswordEncoder encoder = new BCryptPasswordEncoder();
return encoder;
}
@Override
protected void configure(HttpSecurity http) throws Exception {
http.authorizeRequests()
//.antMatchers("/hello").access("hasRole('ROLE_ADMIN')")
.antMatchers("/", "/register", "/user/create", "/webjars/**", "/js/**").permitAll()
.anyRequest().authenticated()
.and()
.formLogin().loginPage("/login").permitAll()
.usernameParameter("username").passwordParameter("password")
.and()
.logout().permitAll()
.and()
.exceptionHandling().accessDeniedPage("/403")
.and()
.csrf().disable();
}
}
我認為問題出在我的User
實體字段的名稱或我在SecurityConfig.java中的兩個查詢的名稱,但是我知道該如何解決我的問題。
我必須保留數據庫配置(字段名稱)。
預先感謝您的幫助 ! :)
編輯
經過兩次更改,所有方法都可以使用,但它不是一個好的數據庫。 我將發布兩個數據庫之間的差異以及SecurityConfig.java中的差異
使用SecurityConfig.java
@Autowired
public void configAuthentication(AuthenticationManagerBuilder auth) throws Exception {
auth.jdbcAuthentication().dataSource(dataSource)
.usersByUsernameQuery(
"select username,password, enabled from users where username=?")
.authoritiesByUsernameQuery(
"select username, role from user_roles where username=?");
}
其次,它不起作用。 我無法發布鏈接,但是您在User.java和UserRole.java中擁有上面的完美描述
使用SecurityConfig.java
@Autowired
public void configAuthentication(AuthenticationManagerBuilder auth) throws Exception {
auth.jdbcAuthentication().dataSource(dataSource)
.usersByUsernameQuery(
"select use_username, use_password, use_email from use_user where use_username=?")
.authoritiesByUsernameQuery(
"select use_username, usr_role from usr_user_role, use_user where use_id = usr_use_id and use_username=?");
}
請更改SQL以按順序而不是*
返回列名;
<!-- change to your own column name, return the columns in order-->
select <username,password,enabled> from use_user where use_user.use_username=?
和AUTHORITIES SQL
<!-- change to your own column name, the authority must be second column -->
select <username,authority> from usr_user_role inner join use_user on use_user.use_id = usr_user_role.usr_use_id where use_user.use_username=?
更詳細的信息,請參見此處 ,確定,這是代碼,我認為更容易理解為什么應按順序返回各列。
protected List<UserDetails> loadUsersByUsername(String username) {
return getJdbcTemplate().query(this.usersByUsernameQuery,
new String[] { username }, new RowMapper<UserDetails>() {
@Override
public UserDetails mapRow(ResultSet rs, int rowNum)
throws SQLException {
// get user info
String username = rs.getString(1);
String password = rs.getString(2);
boolean enabled = rs.getBoolean(3);
return new User(username, password, enabled, true, true, true,
AuthorityUtils.NO_AUTHORITIES);
}
});
}
protected List<GrantedAuthority> loadUserAuthorities(String username) {
return getJdbcTemplate().query(this.authoritiesByUsernameQuery,
new String[] { username }, new RowMapper<GrantedAuthority>() {
@Override
public GrantedAuthority mapRow(ResultSet rs, int rowNum)
throws SQLException {
//get GrantedAuthority
String roleName = JdbcDaoImpl.this.rolePrefix + rs.getString(2);
return new SimpleGrantedAuthority(roleName);
}
});
}
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