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java - 带有 Spring Security 和自定义数据库的 Spring Boot

转载 作者：搜寻专家 更新时间：2023-10-30 22:08:27 24 4
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提前抱歉我的英语不好..

由于我更改了我的数据库配置，所以我没有成功登录我的应用程序。我正在使用 Spring 安全性。在进行更改之前，一切正常。

我有两个实体:

  • 用户.java
  • 用户角色.java

用户.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();
}
}



我认为问题出在我的 User 实体字段的名称或我在 SecurityConfig.java 中的两个查询，但我知道如何解决我的问题。
我必须保留我的数据库配置(字段名称)。



预先感谢您的帮助! :)




编辑

通过两次更改一切正常，但它不是好的数据库。我将发布两个数据库之间的差异以及 SecurityConfig.java

中的差异

一垒(它是有效的)
User table
User Role table

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。我无法发布链接，但您在 User.java 和 UserRole.java

中有上面的完美描述

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=?");
}

最佳答案

请更改 SQL 以按顺序返回列名而不是 *;

<!-- 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=?

更多详细信息请参阅 here ，好的，这是代码，我认为更容易理解为什么你应该按顺序返回列。

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);
                    }
                });
    }

关于java - 带有 Spring Security 和自定义数据库的 Spring Boot，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/41441621/

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