Spring Security 3.2 - Adding Signup Form to Login Page - spring

I have integrated Spring Security login successfuly to my wb application. However, I would like to have in the login page, in addition to login form, a signup form.
In order to do validation I need to pass to that form the object SignupForm which represents my form.
How can I do it? I have tried so many approaches and nothing works.
Please help....

ApplicationContext.xml
<bean id="messageSource" class="org.springframework.context.support.ReloadableResourceBundleMessageSource">
<property name="basenames">
<list><value>classpath:com/harryfwong/config/messages</value></list>
</property>
</bean>
message.properties in com.harryfwong.config resource package
username.required=Username is required
password.required=Password is required
passwordconfirm.mustmatch=Password must match
email.required=Email is required
Spring Validator
<!-- language: java -->
public class SignUpFormValidator implements Validator {
#Override
public boolean supports(Class<?> clazz) {
return SignUpForm.class.equals(clazz);
}
#Override
public void validate(Object target, Errors errors) {
ValidationUtils.rejectIfEmptyOrWhitespace(errors, "username", "username.required");
ValidationUtils.rejectIfEmptyOrWhitespace(errors, "password", "password.required");
SignUpForm form = (SignUpForm) target;
if (!StringUtils.isBlank(form.getPassword())) {
if (!form.getPassword().equals(form.getPasswordConfirm())) {
errors.rejectValue("passwordConfirm", "passwordconfirm.mustmatch");
}
}
ValidationUtils.rejectIfEmptyOrWhitespace(errors, "email", "email.required");
}
}
Controller
<!-- language: java -->
#RequestMapping(method=RequestMethod.GET, value="signup")
public String signup(#ModelAttribute("form") SignUpForm form) {
if (form == null) {
form = new SignUpForm();
}
return "auth/signup";
}
#RequestMapping(method=RequestMethod.POST, value="signup")
public String signupSuccess(#ModelAttribute("form") SignUpForm form, BindingResult result, HttpServletRequest request) {
SignUpFormValidator suValidator = new SignUpFormValidator();
suValidator.validate(form, result);
if (result.hasErrors()) {
return signup(form);
} else {
authenticateUserAndSetSession(form, request);
return "auth/signupSuccess";
}
}
External Reference Login after signup: Auto login after successful registration

What does your SS configuration currently look like?
<form-login
login-page="/login"
login-processing-url="/loginProcess"
default-target-url="/landing"
always-use-default-target="false" />
Where /login is the request mapping
#RequestMapping(value="login")
public String login() {
// do something prior to display form below
return "auth/login";
}
auth/login.jsp
<form name='f' action="<c:url value='/loginProcess' />"
method='POST'>

Not sure what your security context config looks like - this is a response to your issue about not getting to the signup page
<http use-expressions="true">
<intercept-url pattern="/login" access="permitAll"/>
<intercept-url pattern="/loggedout" access="permitAll"/>
<intercept-url pattern="/signup" access="permitAll"/>
<intercept-url pattern="/" access="hasRole('ROLE_USER')" />
<intercept-url pattern="/**" access="hasRole('ROLE_USER')" />
...
</http>

Related

how to include error page in spring MVC 3.0?

how to include error page in spring MVC 3.0 ?
How show error page on exception occurrence .
<http auto-config="true">
<access-denied-handler ref="my403" />
<intercept-url pattern="/admin**" access="ROLE_ADMIN" />
</http>
<beans:bean id="my403" class="com.mkyong.web.exception.MyAccessDeniedHandler">
<beans:property name="errorPage" value="403" />
</beans:bean>
See either this thread on stackoverflow or work with #ControllerAdvice.
You can build your own error controller quite easy:
#ControllerAdvice
public class ExceptionHandlingController {
public static final String DEFAULT_ERROR_VIEW = "generalError";
final String NEW_LINE = System.getProperty("line.separator");
StringBuilder stackTraceString = new StringBuilder("");
#ExceptionHandler(Exception.class)
public ModelAndView handleError(HttpServletRequest req, Exception exception) {
//System.out.println("Request: " + req.getRequestURL() + " raised " + exception);
ModelAndView mav = new ModelAndView(DEFAULT_ERROR_VIEW);
mav.addObject("errorMessage", exception.getMessage());
for (StackTraceElement element : exception.getStackTrace() ){
stackTraceString.append( element );
stackTraceString.append( NEW_LINE );
}
mav.addObject("stackTrace", stackTraceString);
mav.addObject("url", req.getRequestURL());
return mav;
}
}

Getting logged in users with sessionRegistry not work when manually authenticate

I use spring security3 and spring mvc3 to build an web project. There is page called index.jsp, login user name and online user count will be displayed on
the top of this screen. There are 2 ways to login the system:
from login page, use default configuration post by 'j_spring_security_check'
ajax login with manually authentication
When I use login page to login into index page, both of count of online information and user name show correctly.
But when I use ajax login (manually authenticate), problem occurs: count of online user don't updated, it always displaying 0 while user name can show properly.
Part of the controller:
#Autowired
#Qualifier("authenticationManager")
AuthenticationManager authenticationManager;
#Autowired
SecurityContextRepository repository;
#RequestMapping(value="/ajaxLogin")
#ResponseBody
public String performLogin(
#RequestParam("j_username") String username,
#RequestParam("j_password") String password,
HttpServletRequest request, HttpServletResponse response) {
UsernamePasswordAuthenticationToken token = new UsernamePasswordAuthenticationToken(username, password);
try {
Authentication auth = authenticationManager.authenticate(token);
SecurityContextHolder.getContext().setAuthentication(auth);
repository.saveContext(SecurityContextHolder.getContext(), request, response);
logger.info("Authentication successfully! ");
return "{\"status\": true}";
} catch (BadCredentialsException ex) {
return "{\"status\": false, \"error\": \"Bad Credentials\"}";
}
}
spring-security.xml
<beans:beans xmlns="http://www.springframework.org/schema/security"
xmlns:beans="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.0.3.xsd">
<http auto-config="true" use-expressions="true">
<intercept-url pattern="/login" access="permitAll" />
<intercept-url pattern="/index" access="permitAll" />
<form-login login-page="/login" default-target-url="/index"
authentication-failure-url="/loginfailed" />
<logout logout-success-url="/logout" />
<session-management invalid-session-url="/index">
<concurrency-control max-sessions="1"
error-if-maximum-exceeded="false" />
</session-management>
</http>
<authentication-manager alias="authenticationManager">
<authentication-provider>
<jdbc-user-service data-source-ref="dataSource"
users-by-username-query="
select login_id,login_pwd, is_enabled
from t_user where login_id=?"
authorities-by-username-query="
select u.login_id, r.authority from t_user u, t_roles r
where u.u_id = r.u_id and u.login_id =? " />
</authentication-provider>
</authentication-manager>
Method I used to get online login user count:
public class BaseController {
protected Logger logger = Logger.getLogger(this.getClass());
#Autowired
SessionRegistry sessionRegistry;
#ModelAttribute("numUsers")
public int getNumberOfUsers() {
logger.info("in getNumberOfUsers() ...");
return sessionRegistry.getAllPrincipals().size();
}
}
Code used to show login user name:
<div>
<security:authorize ifAllGranted="ROLE_USER">
<p>Welcome <security:authentication property="principal.username" />!
Logout</p>
</security:authorize>
</div>
code used to show count of logged in users:
<div style="color:#3CC457">
${numUsers} user(s) are logged in!
</div>
I guess that because when I manually authenticate, spring security not create new session for the user. I validate it by write customized SessionCounterListener.
public class SessionCounterListener implements HttpSessionListener {
private Logger logger = Logger.getLogger(this.getClass());
private static int totalActiveSessions;
public static int getTotalActiveSession(){
return totalActiveSessions;
}
#Override
public void sessionCreated(HttpSessionEvent event) {
totalActiveSessions++;
logger.info("sessionCreated - add one session into counter" + event.getSession().getId());
}
#Override
public void sessionDestroyed(HttpSessionEvent event) {
totalActiveSessions--;
logger.info("sessionDestroyed - deduct one session from counter" + event.getSession().getId());
}
}
Below is key content of log file for the action sequence: normal login -> normal logout -> ajax login -> ajax logout.
sessionDestroyed - deduct one session 1spueddcmdao019udc43k3uumw
sessionCreated - add one session 14nro6bzyjy0x1jtvnqjx31v1
sessionDestroyed - deduct one session 14nro6bzyjy0x1jtvnqjx31v1
sessionCreated - add one session e6jqz5qy6412118iph66xvaa1
Actually, ajax login/logout not give any output.
So now, how can I get correct login user count? And why the different authenticate ways has different method to deal with session? Any help will be appreciated.
As you are manually adding Principal to SecurityContext, it will not add user to SessionRegistry. You need to add user session to SessionRegistry manually.
SecurityContextHolder.getContext().setAuthentication(auth);
sessionRegistry.registerNewSession(request.getSession().getId(), auth.getPrincipal());
Hope it helps!!
In your Spring spring-security.xml file, the URL for the AJAX authentication (/ajaxLogin) is not explicitly allowed. Thus the request should be blocked by Spring. I would suggest to add this:
<intercept-url pattern="/ajaxLogin" access="permitAll" />

How to obtain a referrer URL with Spring MVC 3.1

I am building a website that has 2 pages. A recipe list page, a recipe detail page, and a sign in page. A user can sign in to the website by clicking on a sign in button on the recipe list page or recipe detail page. When a user click the sign in button, the user will be brought to the sign in page. I would like to redirect the user back again to the recipe detail page if they click the sign in button from the recipe detail page, or to the recipe list page if they click the sign in button from the recipe list page.
I wrote this method in a Controller class. This method will be called whenever user sign in to the website. I stored the referer URL into the session. The intention of saving this URL into a session is to keep track the page where user click on the sign in button. And also to redirect user to that page in the authentication handler that I wrote.
#RequestMapping(value = "/login", method = RequestMethod.GET)
public String openLoginPage(Model uiModel, HttpServletRequest request) {
String referrer = request.getHeader("Referer");
request.getSession().setAttribute("url_prior_login", referrer);
return RECIPE_LOGIN_PAGE;
}
I also created an authentication handler class called SuccessHandler.
public class SuccessHandler extends SavedRequestAwareAuthenticationSuccessHandler {
/*
* (non-Javadoc)
*
* #see org.springframework.security.web.authentication.
* SavedRequestAwareAuthenticationSuccessHandler
* #onAuthenticationSuccess(javax.servlet.http.HttpServletRequest,
* javax.servlet.http.HttpServletResponse,
* org.springframework.security.core.Authentication)
*/
#Override
public void onAuthenticationSuccess(HttpServletRequest request, HttpServletResponse response,
Authentication authentication) throws ServletException, IOException {
String url = (String) request.getSession().getAttribute("url_prior_login");
Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal();
if (principal != null && principal instanceof RecipeUser) {
request.getSession().setAttribute("RecipeUser", (RecipeUser) principal);
}
getRedirectStrategy().sendRedirect(request, response, url);
}
}
This class will redirect user to either the recipe list page or the recipe detail page when they sign in to the website. I register this class inside a security-context.xml file:
<http use-expressions="true" auto-config="false" entry-point-ref="authenticationEntryPoint">
<intercept-url pattern="/login" access="permitAll" />
<form-login login-page="/login" authentication-failure-url="/loginfail"
default-target-url="/login"
authentication-success-handler-ref="successHandler" />
<logout logout-success-url="/" />
</http>
<authentication-manager alias="authManager">
<authentication-provider user-service-ref='myUserDetailsService' />
</authentication-manager>
<beans:bean id="myUserDetailsService" class="com.safe.stack.service.security.UserService">
<beans:property name="dataSource" ref="dataSource" />
</beans:bean>
<beans:bean id="successHandler"
class="com.safe.stack.service.security.SuccessHandler" />
<beans:bean id="authenticationEntryPoint"
class="org.springframework.security.web.authentication.LoginUrlAuthenticationEntryPoint">
<beans:property name="loginFormUrl" value="/login"/>
</beans:bean>
Is this a good way to do what I want to achieve ? Is there a better way to do this ? I could not find any example on how to this using Spring MVC.
Thank you
This is how I do it in Spring 3.1.4
public void onAuthenticationSuccess(HttpServletRequest request,
HttpServletResponse response, Authentication authentication)
throws IOException, ServletException {
// Supplies the default target Url that will be used if
// no saved request is found in the session
setDefaultTargetUrl("/member/dashboard");
super.onAuthenticationSuccess(request, response, authentication);
}
Extending SavedRequestAwareAuthenticationSuccessHandler is correct since the redirection is done automatically.
Just add new attribute "always-use-default-target" to "form-login" tag and set it to "false" (by default it is set to "true", even if don't specify it) like this:
<form-login
login-page="/login" authentication-failure-url="/loginfail"
default-target-url="/login"
authentication-success-handler-ref="successHandler"
always-use-default-target="false"
/>
You can read more about it in Spring Security documentation here, on end of section: 3.2.3.
If you add this, you won't need custom onAuthenticationSuccess anymore and you won't need to store anything in session on login page.

Spring Security - Redirect if already logged in

I'm new to Spring:
I do not want authenticated user from accessing the login page. What is the proper way to handle redirects for the '/login' if the user is already authenticated? Say, I want to redirect to '/index' if already logged in.
I have tried 'isAnonomous()' on login, but it redirects to access denied page.
<security:http auto-config="true" use-expressions="true" ...>
<form-login login-processing-url="/resources/j_spring_security_check"
default-target-url="/index"
login-page="/login" authentication-failure-url="/login?login_error=t" />
<logout logout-url="/resources/j_spring_security_logout" />
...
<security:intercept-url pattern="/login" access="permitAll" />
<security:intercept-url pattern="/**" access="isAuthenticated()" />
</security:http>
In the controller function of your login page:
check if a user is logged in.
then forward/redirect him to the index page in that case.
Relevant code:
Authentication auth = SecurityContextHolder.getContext().getAuthentication();
if (!(auth instanceof AnonymousAuthenticationToken)) {
/* The user is logged in :) */
return new ModelAndView("forward:/index");
}
Update
Or in another scenario where the mapping may be containing path variable like #GetMapping(path = "/user/{id}") in this case you can implement this logic as well:
#GetMapping(value = "/login")
public String getLogin() throws Exception {
Authentication auth = SecurityContextHolder.getContext().getAuthentication();
if (!(auth instanceof AnonymousAuthenticationToken)) {
User loggedInUser = userService.findByEmail(auth.getName())
.orElseThrow(Exception::new);
/* The user is logged in :) */
return "redirect:/user/" + loggedInUser.getUserId();
}
return "login";
}
To successfully redirect from login page, if user is already logged in, add the following to your login.jsp:
Add a security taglib header to the top of your jsp:
<%#taglib uri="http://www.springframework.org/security/tags" prefix="sec"%>
Then add the following tag inside your "head" tag (preferably near the top):
<sec:authorize access="isAuthenticated()">
<% response.sendRedirect("main"); %>
</sec:authorize>
This will redirect to main.html (or whatever your main .jsp is mapped to) if the user accessing the login page is already logged-in.
Doing this through a controller didn't work for me, since the valid login page practice is to let the spring security's "form-login" bean do all the redirecting work, so there was no login controller for me to modify.
login.xhtml
<h:head >
<f:metadata>
<f:event type="preRenderView" listener="#{loginBean.onPageLoad}"/>
</f:metadata>
</h:head>
loginBean
public void onPageLoad(){
Authentication auth = SecurityContextHolder.getContext().getAuthentication();
if (!(auth instanceof AnonymousAuthenticationToken)) {
try {
FacesContext.getCurrentInstance().getExternalContext().redirect(url);
} catch (IOException e) {
e.printStackTrace();
}
}
}
hey you can do that.
<h:head>
<sec:authorize access="isAuthenticated()">
<meta http-equiv="refresh" content="0;url=http://your index.xhtml url (full url)" />
</sec:authorize>
</h:head>
This method is very simple and convenient, is not it?

How to retrieve pK using spring security

I implement this method of the UserDetailService interface,
public UserDetails loadUserByUsername(final String username)
throws UsernameNotFoundException, DataAccessException {
final EmailCredential userDetails = persistentEmailCredential
.getUniqueEmailCredential(username);
if (userDetails == null) {
throw new UsernameNotFoundException(username + "is not registered");
}
final HashSet<GrantedAuthority> authorities = new HashSet<GrantedAuthority>();
authorities.add(new GrantedAuthorityImpl("ROLE_USER"));
for (UserRole role:userDetails.getAccount().getRoles()) {
authorities.add(new GrantedAuthorityImpl(role.getRole()));
}
return new User(userDetails.getEmailAddress(), userDetails
.getPassword(), true, true, true, true,
authorities);
}
In the security context I do some thing like this:
<!-- Login Info -->
<form-login default-target-url='/dashboard.htm' login-page="/login.htm"
authentication-failure-url="/login.htm?authfailed=true"
always-use-default-target='false' />
<logout logout-success-url="/login.htm" invalidate-session="true" />
<remember-me user-service-ref="emailAccountService" key="fuellingsport" />
<session-management>
<concurrency-control max-sessions="1" />
</session-management>
</http>
Now I want to pop out the Pk of the logged in user.
How can I show it in my jsp pages?
Any idea?
I guess you are looking for something like this, which can be used in jsp.
<security:authentication property="principal.username"/>
You can implement your own class extending User with required fields and return it from UserDetailsService.
Then you can access these fields as pointed out by Raghuram:
<security:authentication property="principal.pk"/>

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