我正在使用具有弹簧安全性的angularjs 我的app.js中有以下代码
function run($rootScope, $location, $cookieStore, $http) {
// keep user logged in after page refresh
$rootScope.globals = $cookieStore.get('globals') || {};
if ($rootScope.globals.currentUser) {
$http.defaults.headers.common['Authorization'] = 'Basic ' + $rootScope.globals.currentUser.authdata; // jshint ignore:line
}
}
当我按下退出时,我必须删除当前用户 $ rootScope.globals.currentUser ,以便应用程序正常工作
我该怎么做 我试图用java代码删除它
@RequestMapping(value = "/logout", method = RequestMethod.GET)
public String logout(HttpServletRequest request, HttpServletResponse response) {
HttpSession session= request.getSession(false);
SecurityContextHolder.clearContext();
session= request.getSession(false);
if(session != null) {
session.invalidate();
}
for(Cookie cookie : request.getCookies()) {
cookie.setMaxAge(0);
}
return "redirect:/j_spring_security_logout";
}
but It didn't work
答案 0 :(得分:0)
您可以使用angular $ cookies服务并将其删除,如下所示:
$cookies.remove('coockieName');