我会这样做:

request.getRemoteUser();

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 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();
            }
            ...

    }

The only problem is that even after authenticating with Spring Security, the user/principal bean doesn't exist in the container, so dependency-injecting it will be difficult. Before we used Spring Security we would create a session-scoped bean that had the current Principal, inject that into an "AuthService" and then inject that Service into most of the other services in the Application. So those Services would simply call authService.getCurrentUser() to get the object. If you have a place in your code where you get a reference to the same Principal in the session, you can simply set it as a property on your session-scoped bean.

我喜欢在freemarker页面上分享我支持用户详细信息的方法。 一切都很简单，工作完美!

你只需要在default-target-url上放置身份验证重新请求(表单登录后的页面) 这是我的那个页面的controller方法:

@RequestMapping(value = "/monitoring", method = RequestMethod.GET)
public ModelAndView getMonitoringPage(Model model, final HttpServletRequest request) {
    showRequestLog("monitoring");


    Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
    String userName = authentication.getName();
    //create a new session
    HttpSession session = request.getSession(true);
    session.setAttribute("username", userName);

    return new ModelAndView(catalogPath + "monitoring");
}

这是我的超光速代码:

<@security.authorize ifAnyGranted="ROLE_ADMIN, ROLE_USER">
<p style="padding-right: 20px;">Logged in as ${username!"Anonymous" }</p>
</@security.authorize> 

就是这样，用户名将出现在授权后的每一页。