我想在正确登录系统后获得,以便服务/用户始终以用户登录的登录名的形式返回响应。
但是我遇到了问题,因为服务返回状态404。
弹簧控制器:
@RestController
public class UserController {
@RequestMapping(value = "/user", method = RequestMethod.GET)
public UserView home(@CurrentUser User principal) {
return principal != null ? new UserView(principal) : null;
}
}
自己的注释:
@Target({ElementType.PARAMETER, ElementType.TYPE})
@Retention(RetentionPolicy.RUNTIME)
@Documented
@AuthenticationPrincipal
public @interface CurrentUser {
}
服务:
@Service("userDetailsService")
public class MyUserDetailsService implements UserDetailsService {
private Logger logger = LoggerFactory.getLogger(getClass());
@Autowired
DataSource dataSource;
@Override
public UserDetails loadUserByUsername(String userName) throws UsernameNotFoundException {
JdbcTemplate jdbcTemplateObject = new JdbcTemplate(dataSource);
String SQL = "select username, enabled from admin.ebpp_user where username = ?";
User user = (User)jdbcTemplateObject.queryForObject(SQL, new Object[]{userName}, new JdbcUserMapper());
logger.info("User: "+ user);
return user;
}
}
映射器:
public class JdbcUserMapper implements RowMapper {
@Override
public Object mapRow(ResultSet resultSet, int rowNum) throws SQLException {
return User.Builder.anUser()
.withUsername(resultSet.getString("username"))
.build();
}
}
最后是主要的配置类:
@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
auth.userDetailsService(userDetailsService);
auth.jdbcAuthentication()
.dataSource(dataSource)
.usersByUsernameQuery("select username, password, enabled from admin.ebpp_user where username=?")
.authoritiesByUsernameQuery("select username, authority from admin.authorities where username = ?");
}
使用userDetailsService登录:
2018-02-19 22:58:42,714 INFO[com.MyUserDetailsService] user: User{username='test'}
当它调用:localhost:8080 / rest / home然后收到“404 Not Found” 有谁知道为什么我不能通过该服务下载用户?
edit.1。
我的web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation=" http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/context/applicationContext.xml</param-value>
</context-param>
</web-app>
和applicationContext.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:jee="http://www.springframework.org/schema/jee" xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee.xsd">
<bean id="userDetailsService"
class="com.MyUserDetailsService" />
<jee:jndi-lookup id="dbDataSource"
jndi-name="java:/TestDS"
expected-type="javax.sql.DataSource" />
</beans>
edit.2
当我添加代码时:
@RequestMapping(value = "/user", method = RequestMethod.GET)
public UserView home(@CurrentUser User principal) {
Authentication auth = SecurityContextHolder.getContext().getAuthentication();
UserDetails userDetail = (UserDetails) auth.getPrincipal();
// userDetail contains data
}
然后userDetail包含数据(在调试器模式下),但“@CurrentUser User principal”jest仍为null
答案 0 :(得分:2)
当您想要接收Principal时,您已将其定义为Method参数,然后按照包含在主体中的名称加载User。
@RestController
public class UserController {
@Autowired
MyUserDetailsService userdetailservice;
@RequestMapping(value = "/user", method = RequestMethod.GET)
public UserView home(Principal principal) {
User user = userdetailservice.loadUserByUsername(principal.getName());
return principal != null ? new UserView(user) : null;
}
}