Question

除了他的名字外，我找不到其他用户，但是id或完整用户并不需要唯一的名字我有

@Component
public class WebSocketEventListener {
    private static final Logger logger = 
    LoggerFactory.getLogger(WebSocketEventListener.class);

    @EventListener
    public void handleWebSocketConnectListener(SessionConnectedEvent event) {
        logger.info("Received a new web socket connection");

        System.out.println("User connection : " + event.getUser());
    }
}

有一个方法getUser（）getName，但名称不是唯一的，和方法event.getUser（）提供了一些可怕的东西））-

User connection : 
org.springframework.security.authentication.UsernamePasswordAuthenticationToken@fe260c00: Principal: 
User(id=1, username=admin, password=$2a$08$VDogbqVQY23gNnLFty/6ReGLecW/bk3oCkUHrsly4HgjBIGRNBSEC, 
email=kiy9@gmail.com); Credentials: [PROTECTED]; Authenticated: true; Details: 
org.springframework.security.web.authentication.WebAuthenticationDetails@0: RemoteIpAddress: 
0:0:0:0:0:0:0:1; SessionId: 3566D00689DCC01041367983F3132937; Granted Authorities: USER

有什么方法可以从那里获得用户或id吗？

如果我尝试User user = event.getUser（）或（（User）event.getUser（））。getId（）我得到

 java.lang.ClassCastException:
 org.springframework.security.authentication.
 UsernamePasswordAuthenticationToken 
 cannot be cast to com.newcode.meeting.domain.User

Answer 1

我发现只有一种方法可以吸引用户从“ event.getUser（）”开始，我们得到“ Principal” =>

  User connection : 
  org.springframework.security.authentication.Usernam 
  ePasswordAuthenticationToken@fe260c00: Principal: 
  User(id=1, username=admin, 
  password=$2a$08$VDogbqVQY23gNnLFty/6ReGLecW/bk3oCkUHrsly4HgjBIGRNBSEC, 
  email=kiy9@gmail.com); Credentials: [PROTECTED]; Authenticated: true; Details: 
  org.springframework.security.web.authentication.WebAuthenticationDetails@0: 
  RemoteIpAddress: 
  0:0:0:0:0:0:0:1; SessionId: 3566D00689DCC01041367983F3132937; Granted Authorities: USER

只有一种方法“ getName”，但它不适合我们，因为我们的名称不是唯一的，并且不可能将“ Principal”转换为“ User”，并且我们使用String来工作）

@EventListener
public void handleWebSocketConnectListener(SessionConnectedEvent event) {
    logger.info("Received a new web socket connection:=> "event.getUser().getName());
    User user = userRepo.findUserById(getUserId(event.getUser()));
   // code
}
private Long getUserId(Principal principal) {
    String userString = principal.toString();
    int startIndex = userString.indexOf("id=");
    int endIndex = userString.indexOf(",", startIndex);
    return Long.valueOf(userString.substring(startIndex + 3, endIndex));
}

不是很优雅，但是可以用））

委托人获取用户ID

1 个答案: