对于我写的上一个Spring MVC应用程序，我没有注入SecurityContext holder，但我确实有一个基本控制器，我有两个实用程序方法与此相关…isAuthenticated() & getUsername()。在内部，它们执行您所描述的静态方法调用。

至少在以后需要重构时，它只在一个地方。

您可以使用Spring AOP方法。 例如，如果你有一些服务，它需要知道当前的本金。您可以引入自定义注释，例如@Principal，它表明该服务应该是依赖于主体的。

public class SomeService {
    private String principal;
    @Principal
    public setPrincipal(String principal){
        this.principal=principal;
    }
}

然后在您的通知中，我认为需要扩展MethodBeforeAdvice，检查特定的服务有@Principal注释并注入Principal名称，或者将其设置为'ANONYMOUS'。

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;
}

在回答了这个问题之后，Spring世界发生了很多变化。Spring简化了在控制器中获取当前用户的过程。对于其他bean, Spring采用了作者的建议并简化了'SecurityContextHolder'的注入。更多细节在评论中。

这是我最终选择的解决方案。而不是在我的控制器中使用SecurityContextHolder，我想在底层注入一些使用SecurityContextHolder的东西，但从我的代码中抽象出那个单例类。我发现除了滚动我自己的界面之外，没有其他方法可以做到这一点，如下所示:

public interface SecurityContextFacade {

  SecurityContext getContext();

  void setContext(SecurityContext securityContext);

}

现在，我的控制器(或任何POJO)看起来是这样的:

public class FooController {

  private final SecurityContextFacade securityContextFacade;

  public FooController(SecurityContextFacade securityContextFacade) {
    this.securityContextFacade = securityContextFacade;
  }

  public void doSomething(){
    SecurityContext context = securityContextFacade.getContext();
    // do something w/ context
  }

}

而且，由于接口是一个解耦点，单元测试很简单。在这个例子中，我使用Mockito:

public class FooControllerTest {

  private FooController controller;
  private SecurityContextFacade mockSecurityContextFacade;
  private SecurityContext mockSecurityContext;

  @Before
  public void setUp() throws Exception {
    mockSecurityContextFacade = mock(SecurityContextFacade.class);
    mockSecurityContext = mock(SecurityContext.class);
    stub(mockSecurityContextFacade.getContext()).toReturn(mockSecurityContext);
    controller = new FooController(mockSecurityContextFacade);
  }

  @Test
  public void testDoSomething() {
    controller.doSomething();
    verify(mockSecurityContextFacade).getContext();
  }

}

接口的默认实现如下所示:

public class SecurityContextHolderFacade implements SecurityContextFacade {

  public SecurityContext getContext() {
    return SecurityContextHolder.getContext();
  }

  public void setContext(SecurityContext securityContext) {
    SecurityContextHolder.setContext(securityContext);
  }

}

最后，产品Spring配置如下所示:

<bean id="myController" class="com.foo.FooController">
     ...
  <constructor-arg index="1">
    <bean class="com.foo.SecurityContextHolderFacade">
  </constructor-arg>
</bean>

Spring是所有东西的依赖注入容器，它没有提供类似注入的方法，这似乎有点愚蠢。我知道SecurityContextHolder继承自acegi，但仍然。问题是，它们非常接近——只要SecurityContextHolder有一个getter来获得底层的SecurityContextHolderStrategy实例(这是一个接口)，你就可以注入它。事实上，我甚至为此开了一期Jira杂志。

One last thing - I've just substantially changed the answer I had here before. Check the history if you're curious but, as a coworker pointed out to me, my previous answer would not work in a multi-threaded environment. The underlying SecurityContextHolderStrategy used by SecurityContextHolder is, by default, an instance of ThreadLocalSecurityContextHolderStrategy, which stores SecurityContexts in a ThreadLocal. Therefore, it is not necessarily a good idea to inject the SecurityContext directly into a bean at initialization time - it may need to be retrieved from the ThreadLocal each time, in a multi-threaded environment, so the correct one is retrieved.

Yes, statics are generally bad - generally, but in this case, the static is the most secure code you can write. Since the security context associates a Principal with the currently running thread, the most secure code would access the static from the thread as directly as possible. Hiding the access behind a wrapper class that is injected provides an attacker with more points to attack. They wouldn't need access to the code (which they would have a hard time changing if the jar was signed), they just need a way to override the configuration, which can be done at runtime or slipping some XML onto the classpath. Even using annotation injection in the signed code would be overridable with external XML. Such XML could inject the running system with a rogue principal. This is probably why Spring is doing something so un-Spring-like in this case.

我会这样做:

request.getRemoteUser();

我得到认证的用户 HttpServletRequest.getUserPrincipal ();

例子:

import javax.servlet.http.HttpServletRequest;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.web.authentication.preauth.RequestHeaderAuthenticationFilter;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.support.RequestContext;

import foo.Form;

@Controller
@RequestMapping(value="/welcome")
public class IndexController {

    @RequestMapping(method=RequestMethod.GET)
    public String getCreateForm(Model model, HttpServletRequest request) {

        if(request.getUserPrincipal() != null) {
            String loginName = request.getUserPrincipal().getName();
            System.out.println("loginName : " + loginName );
        }

        model.addAttribute("form", new Form());
        return "welcome";
    }
}

如果你正在使用Spring 3，最简单的方法是:

 @RequestMapping(method = RequestMethod.GET)   
 public ModelAndView showResults(final HttpServletRequest request, Principal principal) {

     final String currentUser = principal.getName();

 }

试试这个

认证认证= .getAuthentication SecurityContextHolder.getContext () (); 字符串userName = authentication.getName();

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();
            }
            ...

    }

为了让它只显示在JSP页面中，你可以使用Spring Security标签库:

http://static.springsource.org/spring-security/site/docs/3.0.x/reference/taglibs.html

要使用任何标记，必须在JSP中声明安全标记库:

<%@ taglib prefix="security" uri="http://www.springframework.org/security/tags" %>

然后在jsp页面中执行如下操作:

<security:authorize access="isAuthenticated()">
    logged in as <security:authentication property="principal.username" /> 
</security:authorize>

<security:authorize access="! isAuthenticated()">
    not logged in
</security:authorize>

注意:正如@SBerg413在评论中提到的，您需要添加

use-expressions = " true "

到security.xml配置中的“http”标签以使其工作。

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。

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。

将Principal定义为控制器方法中的依赖项，spring将在调用时将当前已验证的用户注入到方法中。

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。