如果您正在使用Spring 3，并且需要在控制器中使用经过身份验证的主体，那么最好的解决方案是这样做:

import org.springframework.security.authentication.UsernamePasswordAuthenticationToken;
import org.springframework.security.core.userdetails.User;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;

    @Controller
    public class KnoteController {
        @RequestMapping(method = RequestMethod.GET)
        public java.lang.String list(Model uiModel, UsernamePasswordAuthenticationToken authToken) {

            if (authToken instanceof UsernamePasswordAuthenticationToken) {
                user = (User) authToken.getPrincipal();
            }
            ...

    }

在Spring 3+中，您有以下选项。

方案一:

@RequestMapping(method = RequestMethod.GET)    
public String currentUserNameByPrincipal(Principal principal) {
    return principal.getName();
}

选择二:

@RequestMapping(method = RequestMethod.GET)
public String currentUserNameByAuthentication(Authentication authentication) {
    return authentication.getName();
}

选项3:

@RequestMapping(method = RequestMethod.GET)    
public String currentUserByHTTPRequest(HttpServletRequest request) {
    return request.getUserPrincipal().getName();

}

选项4:花哨的一个:查看更多细节

public ModelAndView someRequestHandler(@ActiveUser User activeUser) {
  ...
}

我同意必须为当前用户查询SecurityContext很糟糕，这似乎是一种非常非spring的处理这个问题的方式。

我写了一个静态的“助手”类来处理这个问题;它是脏的，因为它是一个全局和静态的方法，但我认为如果我们改变任何与安全相关的东西，至少我只需要改变一个地方的细节:

/**
* Returns the domain User object for the currently logged in user, or null
* if no User is logged in.
* 
* @return User object for the currently logged in user, or null if no User
*         is logged in.
*/
public static User getCurrentUser() {

    Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal()

    if (principal instanceof MyUserDetails) return ((MyUserDetails) principal).getUser();

    // principal object is either null or represents anonymous user -
    // neither of which our domain User object can represent - so return null
    return null;
}


/**
 * Utility method to determine if the current user is logged in /
 * authenticated.
 * <p>
 * Equivalent of calling:
 * <p>
 * <code>getCurrentUser() != null</code>
 * 
 * @return if user is logged in
 */
public static boolean isLoggedIn() {
    return getCurrentUser() != null;
}

如果您正在使用Spring Security ver >= 3.2，您可以使用@AuthenticationPrincipal注释:

@RequestMapping(method = RequestMethod.GET)
public ModelAndView showResults(@AuthenticationPrincipal CustomUser currentUser, HttpServletRequest request) {
    String currentUsername = currentUser.getUsername();
    // ...
}

这里，CustomUser是一个自定义对象，它实现了由自定义UserDetailsService返回的UserDetails。

更多信息可以在Spring Security参考文档的@AuthenticationPrincipal一章中找到。

我使用@Controller类中的@AuthenticationPrincipal注释以及@ controlleradvisor注释。例:

@ControllerAdvice
public class ControllerAdvicer
{
    private static final Logger LOGGER = LoggerFactory.getLogger(ControllerAdvicer.class);


    @ModelAttribute("userActive")
    public UserActive currentUser(@AuthenticationPrincipal UserActive currentUser)
    {
        return currentUser;
    }
}

其中UserActive是我用于已登录用户服务的类，并且扩展自org.springframework.security.core.userdetails.User。喜欢的东西:

public class UserActive extends org.springframework.security.core.userdetails.User
{

    private final User user;

    public UserActive(User user)
    {
        super(user.getUsername(), user.getPasswordHash(), user.getGrantedAuthorities());
        this.user = user;
    }

     //More functions
}

真的很容易。

我会这样做:

request.getRemoteUser();