Spring Boot Redirect to requested URL after login - spring-boot

I have a Spring Boot UI application. I am trying to redirect users to the originally requested URL after login.
When a user requests http://www.example.com/myapp/user/22, the application aptly redirects to http://www.example.com/myapp/login. Once the user logs in, the application redirects to http://www.example.com/myapp/dashboard. I would like the application to redirect to http://www.example.com/myapp/user/22.
I have gone through several links and feel I have a proper configuration, yet, redirection is not working as expected.
My Security Config is
public class SecurityConfig extends WebSecurityConfigurerAdapter {
.....
....
#Autowired
private MyAuthenticationSuccessHandler authenticationSuccessHandler;
#Override
protected void configure(HttpSecurity http) throws Exception {
http.
authorizeRequests()
.antMatchers("/user/**").authenticated()
.and().csrf().disable().formLogin()
.successHandler(authenticationSuccessHandler)
......
and My Success Handler is
#Component
public class MyAuthenticationSuccessHandler extends SimpleUrlAuthenticationSuccessHandler {
...
public MyAuthenticationSuccessHandler() {
super();
this.setDefaultTargetUrl("/myapp/dashboard");
this.setUseReferer(true);
}
#Override
public void onAuthenticationSuccess(HttpServletRequest request, HttpServletResponse response,
Authentication authentication) throws IOException, ServletException {
//Do something ..........
........
.........
super.onAuthenticationSuccess(request, response, authentication);
}
I tried using SavedRequestAwareAuthenticationSuccessHandler too.
I notice that my success handler is invoked, but the target URL is always /user/login and my login controller is invoked..
#RequestMapping("/login")
public ModelAndView login(#ModelAttribute() {
if(!userIdentified) {
//go to login page
} else {
new ModelAndView("redirect:/myapp/dashboard");
}
}
and the user is redirected to "dashboard".
What else am I missing?

Use "Referer" from session attribute to get the latest request URL. On my app, i use this one
public class CustomAuthenticationSuccessHandler extends SimpleUrlAuthenticationSuccessHandler {
public static final String REDIRECT_URL_SESSION_ATTRIBUTE_NAME = "REDIRECT_URL";
#Override
public void onAuthenticationSuccess(HttpServletRequest request, HttpServletResponse response,
Authentication authentication) throws IOException, ServletException {
Object redirectURLObject = request.getSession().getAttribute(REDIRECT_URL_SESSION_ATTRIBUTE_NAME);
if(redirectURLObject != null)
setDefaultTargetUrl(redirectURLObject.toString());
else{
setDefaultTargetUrl("/");
}
request.getSession().removeAttribute(REDIRECT_URL_SESSION_ATTRIBUTE_NAME);
super.onAuthenticationSuccess(request, response, authentication);
}
}
Edit :
Sorry i forgot to show the login controller
#RequestMapping(method = RequestMethod.GET, value = {"/login"})
String login(Model model, Principal principal, HttpServletRequest request) throws Exception{
String referer = request.getHeader("Referer"); //Get previous URL before call '/login'
//save referer URL to session, for later use on CustomAuthenticationSuccesshandler
request.getSession().setAttribute(CustomAuthenticationSuccessHandler.REDIRECT_URL_SESSION_ATTRIBUTE_NAME, referer);
return principal == null ? "login" : "redirect:/";
}

Although Singgih S answer works, BUT there is a better way as below :
Ref:
https://www.baeldung.com/spring-security-redirect-login
There is no magic in these easy to use features in Spring Security.
When a secured resource is being requested, the request will be
filtered by a chain of various filters. Authentication principals and
permissions will be checked. If the request session is not
authenticated yet, AuthenticationException will be thrown.
The AuthenticationException will be caught in the
ExceptionTranslationFilter, in which an authentication process will be
commenced, resulting in a redirection to the login page.
Therefore :
1. When redirection to the "/login" page occurs, your secured request url is saved in the session as DefaultSavedRequest object.
2. Also we know when a successful form based login occurs, one of the implementations of AuthenticationSuccessHandler is called.
so we can create a custom class and get DefaultSavedRequest in it as below :
public class CustomAuthenticationSuccessHandler extends SimpleUrlAuthenticationSuccessHandler {
#Override
public void onAuthenticationSuccess(HttpServletRequest request, HttpServletResponse response, Authentication authentication) throws IOException, ServletException {
DefaultSavedRequest defaultSavedRequest = (DefaultSavedRequest) request.getSession().getAttribute("SPRING_SECURITY_SAVED_REQUEST");
if(defaultSavedRequest != null){
getRedirectStrategy().sendRedirect(request, response, defaultSavedRequest.getRedirectUrl());
}else{
super.onAuthenticationSuccess(request, response, authentication);
}
}
}
3. We have to introduce this class in WebSecurityConfigurerAdapter :
#Override
protected void configure(HttpSecurity http) throws Exception {
http.(...).anyRequest().authenticated().and()
.formLogin()
.loginPage("/login")
.loginProcessingUrl("/login")
.successHandler(new CustomAuthenticationSuccessHandler());
So you can implement your logic in the above onAuthenticationSuccess method.
Best wishes

The Spring route, ala extending SavedRequestAwareAuthenticationSuccessHandler or SimpleUrlAuthenticationSuccessHandler can be a bit clunky to implement. In the controller (ex. a POST method that processes logins), you can do the header request yourself; ex:
HttpServletRequest request =null;
String priorUrl = request.getHeader("Referer");
You will notice that you will have the URL prior to either a manual (initiated by user) logout or a session timeout (as handled by Spring session): you'll get an https://iAmPriorUrl.com/.... Then you can do whatever you want with it.

Related

How to intercept not authorazied request to a rest controller (method level authorization) - Spring Boot

In a REST Controller I have the following method.
#GetMapping("/activate_user")
#RolesAllowed({Role.ROLE_ADMIN})
public void activateUser() {
// Some code here
}
If a user with ROLE_ADMIN calls this method, it works like it should.
If a user without ROLE_ADMIN calls this method, it return an Http-Status 403. That is also ok, but I want now to intercept this call in case the user is not authorized, run some custom code and return some JSON data back to the caller.
I don't know how it could be done with Spring?!
You can override the accessdenied exception and this way it will only be executed for 403 unauthorized.
#Override
public void handle(HttpServletRequest request, HttpServletResponse response,
AccessDeniedException accessDeniedException) throws IOException, ServletException {
response.getWriter().write("Custom Access Denied Message");
}
you can use MVC Interceptor Configuration to intercept specific URLs/APIs
#Configuration
#EnableWebMvc
public class WebConfig extends WebMvcConfigurerAdapter {
#Override
public void addInterceptors(InterceptorRegistry registry) {
registry.addInterceptor(new LocaleInterceptor());
registry.addInterceptor(new ThemeInterceptor()).addPathPatterns("/**").excludePathPatterns("/adminRole/**");
registry.addInterceptor(new SecurityInterceptor()).addPathPatterns("/auth/*", "/ui/**", "/xyz/**");
}
}
You can even exclude specific URL's.

SpringSecurity: How to Continue Forwarding Request to RestController After a Successful Authentication?

I am doing a pure backend project with REST APIs (not MVC) and would like to use SpringSecurity with JWT token to project these APIs. The implementation is good and all APIs are successfully protected with the token, and I can post a JSON string with username and password to "/login" path to get token
My problem is:
The SpringSecurity will return the response with token directly in successfulAuthentication() rather than keep forwarding to RestController (RestController's "/login" path gets no data)
And my question is:
What should I do, after a successful authentication, to allow SpringSecurity can keep forwarding the request to RestController's "/login" path so that I can do something else on the request and the newly built token beside the security in the path?
Appreciate any helps, Thank you!
My code:
#Component
public class TokenWebSecurityConfig extends WebSecurityConfigurerAdapter {
// ...
#Override
protected void configure(HttpSecurity http) throws Exception {
// ...
http.authorizeRequests()
.antMatchers("/registry").permitAll() // allow path /registry
.antMatchers("/login").permitAll() // allow path /login
.antMatchers("/verify").permitAll() // allow path /verify
.anyRequest().authenticated();
// ...
}
}
#RestController
public class EntranceEndpoint {
#RequestMapping(path = "/login", method = RequestMethod.POST)
public RestResponse<String> login(LoginMetaInfo login) {
System.out.println(login); // no output here when login
// some further operations for a successful login, and return a REST response
}
}
And this is what the SpringSecurity do on a successful login
public class TokenLoginFilter extends UsernamePasswordAuthenticationFilter {
// ...
/**
* on login success
*/
#Override
protected void successfulAuthentication(HttpServletRequest request, HttpServletResponse response, FilterChain chain, Authentication auth) throws IOException {
// here build the token and insert into response for commitment
// - the SpringSecurity soon returns the response directly, rather then keep forwarding to RestController
String token = xxxx;
response.setStatus(StatusCode.SUCCESS().getCode());
RestResponse<String> body = RestResponse.succeeded(StatusCode.SUCCESS().withMsg(LoginResponseCode.LOGIN), token);
response.setContentType(MediaType.APPLICATION_JSON);
response.setCharacterEncoding(MediaType.CHARSET);
ObjectMapper mapper = new ObjectMapper();
mapper.writeValue(response.getWriter(), body );
}
}
What about simply using HttpServletResponse's sendRedirect instead of writing to the response?
#Override
protected void successfulAuthentication(HttpServletRequest request, HttpServletResponse response, FilterChain chain, Authentication auth) throws IOException {
// do what you want here
response.sendRedirect("/login");
// response.sendRedirect("https://yoururl");
}

AuthenticationFailureHandler HttpServletResponse.sendError url

I have developed single page web application using Spring Boot and Spring MVC. I am using Spring Security and JWT to authenticate users. I have written a custom AuthenticationFailureHandler which works but I want to know how I can control the url that a user gets redirect to when an exception is thrown. My AuthenticationFailureHandler looks like this:
public class JwtAuthenticationFailureHandler implements AuthenticationFailureHandler {
#Override
public void onAuthenticationFailure(HttpServletRequest request, HttpServletResponse response, AuthenticationException exception) throws IOException, ServletException {
response.sendError(HttpStatus.UNAUTHORIZED.value(), exception.getMessage());
}
}
When the JWT expires the application throws an AccountExpiredException, the AuthenticationFailureHandler.onAuthenticationFailure method gets executed and the user gets redirected to the login page:
http://localhost:8080/login?sessionExpired=true
This is all good, but I have no idea how the sessionExpired=true query string is generated and I want to have some control over it. In the past I have used ExceptionMappingAuthenticationFailureHandlers like this:
Map<String, String> mappings = new HashMap<>();
mappings.put(BadCredentialsException.class.getCanonicalName(), BAD_CREDENTIALS_EXCEPTION_URL);
mappings.put(AccountExpiredException.class.getCanonicalName(), ACCOUNT_EXPIRED_EXCEPTION_URL);
mappings.put(CredentialsExpiredException.class.getCanonicalName(), CREDENTIALS_EXPIRED_EXCEPTION_URL);
mappings.put(DisabledException.class.getCanonicalName(), ACCOUNT_INACTIVE_EXCEPTION_URL);
mappings.put(LockedException.class.getCanonicalName(), ACCOUNT_LOCKED_EXCEPTION_URL);
mappings.put(ValidationException.class.getCanonicalName(), VALIDATION_EXCEPTION_URL);
ExceptionMappingAuthenticationFailureHandler exceptionMappingAuthenticationFailureHandler = new ExceptionMappingAuthenticationFailureHandler();
exceptionMappingAuthenticationFailureHandler.setExceptionMappings(mappings);
So based on the various exceptions above I would like to be able to redirect to the following URLs:
http://localhost:8080/login?error
http://localhost:8080/login?accountexpired
http://localhost:8080/login?credentialsexpired
http://localhost:8080/login?accountlocked
http://localhost:8080/login?accountinactive
http://localhost:8080/login?validationerror
I'm not sure who to do this with response.sendError and I don't know how the sessionExpired=true query string is being generated. I have tried throwing different exceptions but the url never changes.
I have a couple of questions. Is it possible to control the URL when using HttpServletResponse.sendError and if not is it possible ot set the HttpStatus code when using ExceptionMappingAuthenticationFailureHandler.sendRedirect?
Why don't you try to use the response.sendRedirect:
#Override
public void onAuthenticationFailure(HttpServletRequest request, HttpServletResponse response,
AuthenticationException exception) throws IOException, ServletException {
final HttpSession session = request.getSession(false);
if (session != null) {
request.getSession().setAttribute(WebAttributes.AUTHENTICATION_EXCEPTION, exception);
}
//here the logic to get the error type of the exception
String errorMessage = ????
redirectStrategy.sendRedirect(request, response,
"http://localhost:8080/login?" + errorMessage);
}

Spring Security: How can I set a RememberMe cookie url path, that differs from the context path?

How in Spring Security can I set a RememberMe cookie url path, that differs from the context path?
Supposing my website's homepage url is (url rewrite):
https://www.mysuperspecialdomain.com
And that my login page has a url like this:
https://www.mysuperspecialdomain.com/shop/account/login
After succesful login the RememberMe cookie has the path /shop (visible in the browser, e.g. Chrome). This is the project's context path.
This leads to the situation, that when I'm going to my homepage, RememberMe is not logging in. Only when I navigate to a url, that starts with https://www.myspecialdomain.com/shop it's doing it.
If you use Spring Security 4.1.0 or higher, you can configure the cookie domain, see RememberMeConfigurer#rememberMeCookieDomain:
The domain name within which the remember me cookie is visible.
but you can't change the context path.
So you have to implement your own RememberMeServices (you could create a sub class of an existing one) and add it with RememberMeConfigurer#rememberMeServices to your security configuration.
I've found a solution to my own question - manipulation of the path of the RememberMe-cookie can be done via an HttpServletResponseWrapper. This is the solution (based on this answer https://stackoverflow.com/a/7047298/7095884):
Define an HttpServletResponseWrapper:
public class RememberMeCookieResponseWrapper extends HttpServletResponseWrapper {
public RememberMeCookieResponseWrapper(HttpServletResponse response) {
super(response);
}
#Override
public void addCookie(Cookie cookie) {
if (cookie.getName().equals("shop")) {
cookie.setPath("/");
}
super.addCookie(cookie);
}
}
Define a filter, that wraps the servlet response with the just defined wrapper:
public class RememberMeCookieFilter implements Filter {
public void doFilter(ServletRequest request, ServletResponse response,
FilterChain chain) throws IOException, ServletException {
if (response instanceof HttpServletResponse) {
HttpServletResponse newResponse =
new RememberMeCookieResponseWrapper((HttpServletResponse)response);
chain.doFilter(request, newResponse);
}
}
}
Add this filter to the Spring Filter Chain in front of the authentication part:
#Configuration
#EnableWebSecurity
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {
#Override
protected void configure(HttpSecurity http) throws Exception {
http.addFilterBefore(new RememberMeCookieFilter(), UsernamePasswordAuthenticationFilter.class)
...

SSO with Spring security

I have an application, where user is pre-authorized by SSO and lands to my page, now I need to make a call to another rest api to get some data, which is running on another server, but it will be use the same authentication. So I just wanted to know, how I can provide the authentication process? Do I need to set the cookie what I am getting from the incoming request.
When the request lands on your page it should have a token or key, in the http AUTHORIZATION header, this should be used with a filter
public class AuthFilter extends OncePerRequestFilter {
private String failureUrl;
private SimpleUrlAuthenticationFailureHandler failureHandler = new SimpleUrlAuthenticationFailureHandler();
#Override
protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain chain)
throws ServletException, IOException {
try {
// check your SSO token here
chain.doFilter(request, response);
} catch (OnlineDriverEnquiryException ode) {
failureHandler.setDefaultFailureUrl(failureUrl);
failureHandler.onAuthenticationFailure(request, response, new BadCredentialsException("Captcha invalid!"));
}
}
public String getFailureUrl() {
return failureUrl;
}
public void setFailureUrl(String failureUrl) {
this.failureUrl = failureUrl;
}
}
Also read this post on how to set up the auto config. Spring security without form login

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