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";
}
}
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();
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。
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();
}
