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
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。
其他回答
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.
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();
}
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。
我会这样做:
request.getRemoteUser();
