使用Spring引导的HTTP基本身份验证'；s基于Java的配置

我正在尝试设置一个简单的Spring引导应用程序，该应用程序使用具有硬编码密码的单个用户进行HTTP基本身份验证

到目前为止，我使用基于XML的配置实现了它

如何使用基于Java的配置实现相同的结果

• SecurityConfig.java

  @EnableWebSecurity
  @ImportResource("classpath:spring-security.xml")
  public class SecurityConfig {}
  

• spring security.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.xsd
                                   http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security.xsd">
      <http>
          <intercept-url pattern="/MyService/**" access="isAuthenticated()" />
          <http-basic />
      </http>
  
      <user-service>
          <user name="foo" password="{noop}bar" authorities="ROLE_USER" />
      </user-service>
  </beans:beans>
  

注意：我不得不使用

@EnableWebSecurity

而不是

@Configuration

来解决这个问题

---

我使用的是Spring Boot 2.3.4和Spring Security 5.3.4。

好吧，如果我理解正确，您只想设置http连接吗？ 下面是我编写的一个代码示例，它适合您的xml（我想）

然后，如果您打算真正使用它，您需要获取用户，可能是从数据库获取的，因此您还需要类似以下内容：

@Service
public class YourUserDetailsService implements UserDetailsService { //UserDetailsService is the interface we need to let Spring do its magic

    private final LoginsService LoginsService;

    public LibraryUserDetailsService(LoginsService loginsService) {
        this.loginsService = loginsService;
    }

    @Override
    public UserDetails loadUserByUsername(String password, String userName) throws UsernameNotFoundException {

 //Here you fetch, decrypt, and check that the password and username are correct
 //WARNING: This is a really simple example, do not use this in your applications code 
     
  Optional<GrantedAcces> access = 
  libraryLoginsService.findUser(userName,password);

  //I create a new user with the authorized role, this is store in the session
           return new User(access.get().getUserName,access.get().getPassword(), Collections.singleton(new SimpleGrantedAuthority("ROLE_ADMIN")));

   }

@服务
公共类YourUserDetailsService实现UserDetailsService{//UserDetailsService是让Spring发挥其魔力所需的接口
私人最终登录服务登录服务；
公共图书馆用户详细信息服务（登录服务登录服务）{
this.loginsService=loginsService；
}
@凌驾
public UserDetails loadUserByUsername（字符串密码，字符串用户名）引发UsernameNotFoundException{
//在这里，您获取、解密并检查密码和用户名是否正确
//警告：这是一个非常简单的示例，请勿在应用程序代码中使用
可选访问=
libraryLoginsService.findUser（用户名、密码）；
//我创建了一个具有授权角色的新用户，它存储在会话中
返回新用户（access.get（）.getUserName、access.get（）.getPassword（）、Collections.singleton（新的SimpleGrantedAuthority（“角色\管理员”））；
}

我希望这个能帮助你，我理解你的问题

@Service
public class YourUserDetailsService implements UserDetailsService { //UserDetailsService is the interface we need to let Spring do its magic

    private final LoginsService LoginsService;

    public LibraryUserDetailsService(LoginsService loginsService) {
        this.loginsService = loginsService;
    }

    @Override
    public UserDetails loadUserByUsername(String password, String userName) throws UsernameNotFoundException {

 //Here you fetch, decrypt, and check that the password and username are correct
 //WARNING: This is a really simple example, do not use this in your applications code 
     
  Optional<GrantedAcces> access = 
  libraryLoginsService.findUser(userName,password);

  //I create a new user with the authorized role, this is store in the session
           return new User(access.get().getUserName,access.get().getPassword(), Collections.singleton(new SimpleGrantedAuthority("ROLE_ADMIN")));

   }