我有一个使用Spring Security的Spring MVC web应用程序。我想知道当前登录用户的用户名。我正在使用下面给出的代码片段。这是公认的方式吗?

我不喜欢在这个控制器中调用静态方法——恕我直言,这违背了Spring的全部目的。有没有一种方法来配置应用程序有当前的SecurityContext,或当前的认证,注入代替?

  @RequestMapping(method = RequestMethod.GET)
  public ModelAndView showResults(final HttpServletRequest request...) {
    final String currentUser = SecurityContextHolder.getContext().getAuthentication().getName();
    ...
  }

当前回答

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();
            }
            ...

    }

其他回答

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";
    }
}

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。

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。

我会这样做:

request.getRemoteUser();

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.