Java 通过对spring的任何请求获取用户详细信息
我希望在正确登录到系统之后获得,以便服务/用户始终以用户登录的形式返回响应 但是我有一个问题,因为服务返回404状态 弹簧控制器:Java 通过对spring的任何请求获取用户详细信息,java,spring,spring-security,Java,Spring,Spring Security,我希望在正确登录到系统之后获得,以便服务/用户始终以用户登录的形式返回响应 但是我有一个问题,因为服务返回404状态 弹簧控制器: @RestController public class UserController { @RequestMapping(value = "/user", method = RequestMethod.GET) public UserView home(@CurrentUser User principal) { return p
@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;
}
}
制图员:
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UserDetails登录服务外观:
2018-02-19 22:58:42,714 INFO[com.MyUserDetailsService] user: User{username='test'}
当它调用:localhost:8080/rest/home时,会收到“404notfound”
有人知道为什么我不能通过服务下载用户吗
编辑.1
my 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>
然后userDetail包含数据(在调试器模式下),但是“@CurrentUser User principal”仍然为null当您想要接收主体时,您已经将其定义为方法参数,然后按照主体中包含的名称加载用户
@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;
}
}
服务器上是否有错误?服务器上是否没有错误?是否有映射到路径/rest/home的控制器?是否可以发布web XML?
@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
}
@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;
}
}