How do I create a logout with Spring Security?

How do I create a logout with Spring Security? - java

I am trying to find a way just to setup a URL that will logout my user from the system. this is only for testing. Right now we are using the default login page in spring secuirty
here is my spring-secuirty.xml
<?xml version="1.0" encoding="UTF-8"?>
<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-3.1.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-3.1.xsd">
<global-method-security pre-post-annotations="enabled" />
<http use-expressions="true">
<intercept-url access="hasRole('ROLE_VERIFIED_MEMBER')" pattern="/ask-union**" />
<intercept-url access="hasRole('ROLE_VERIFIED_MEMBER')" pattern="/ask-welfare**" />
<intercept-url pattern='/*' access='permitAll' />
<form-login default-target-url="/ask-union" />
<logout logout-success-url="/" />
<session-management session-fixation-protection="newSession">
<concurrency-control max-sessions="1"/>
</session-management>
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="xxxxx#aol.com.dev" password="testing" authorities="ROLE_VERIFIED_MEMBER" />
ser-service>
</authentication-provider>
</authentication-manager>
</beans:beans>

Add this line to your config
<logout logout-url="/sign-out"/>
Then if you have a link to that URL, then it will sign you out
(Add it just below your logout success config)

I am a bit late on this. But answer may help others.
Use following code. logout-success-url is the URL you want take user after logging out.
<http auto-config="true" use-expressions="true" access-denied-page="/denied">
<logout invalidate-session="true" logout-success-url="/landing" delete-cookies="JSESSIONID" />
</http>

Double-check the URL you're using -- the absolute path should be your-domain/projectPath/sign-out, per SJS's example. If the relevant portion of your spring-security.xml file looks like the following, it should work:
<http use-expressions="true">
. . .
<logout
logout-success-url="/"
logout-url="/sign-out"/>
If you're able to authenticate, then simply browse to that path, and it should log you out. If it still doesn't try experimenting with the intermediary subdirectories specified in the URL, i.e. your-domain/projectPath/some-subdirectory/log-out.
Are you able to authenticate? It may not just be the logout aspect that's failing...

Late to the party, but for future reference -- if you are interested in seeing how the security filters are instantiated and configured from XML, take a look at the following package:
org.springframework.security.config.annotation.web.configurers
For the logout filter configuration, this will be the LogoutConfigurer class. If you review the LogoutConfigurer.createLogoutFilter() method, you'll see how the default Logout filter is created. This implies that you can do the following in an #Configuration class:
#Bean
public LogoutFilter logoutFilter() {
// NOTE: See org.springframework.security.config.annotation.web.configurers.LogoutConfigurer
// for details on setting up a LogoutFilter
SecurityContextLogoutHandler securityContextLogoutHandler = new SecurityContextLogoutHandler();
securityContextLogoutHandler.setInvalidateHttpSession(true);
LogoutFilter logoutFilter = new LogoutFilter("/", securityContextLogoutHandler);
logoutFilter.setLogoutRequestMatcher(new AntPathRequestMatcher("/logout"));
return logoutFilter;
}
If you have that, you can then either configure further by beans or have that bean picked up automatically since its method name is logoutFilter()

Related

How to relate angular 6 login with spring login

it's my first question here in stackoverflow.
I'll try to explain my problem as much comprensible as possible.
I'm trying to develop a web-app with Java 8 using spring(not spring-boot), hibernate and angular 6. I already wrote back-end side (spring and hibernate), my entities work perfectly with my db (MySQL). I divided the back-end side into 3 maven modules (persistance (entities-repositories), service, web(controllers)) and added a front-end maven module with Angular 6.
And here there is my problem.
I am not able to connect angular with spring, in particular with spring security. I cut and paste the dist folder of angular inside src>main>webapp package and i cant and/or i dont know how to reach the angular's login page using spring security. I mean, i dont know how to relate angular login with spring login.
In the spring security config xml i wrote
<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.xsd">
<http>
<intercept-url pattern="/home" access="permitAll" />
<intercept-url pattern="/login" access="permitAll" />
<intercept-url pattern="/signup" access="permitAll" />
<intercept-url pattern="/logout" access="isAuthenticated()" />
<intercept-url pattern="/**" access="hasRole('USER')" />
<form-login />
<logout />
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="admin" password="{noop}admin" authorities="ROLE_USER, ROLE_ADMIN" />
<user name="user" password="{noop}user" authorities="ROLE_USER" />
</user-service>
</authentication-provider>
</authentication-manager>
i don't know how to relate the spring login page with my angular's login page, because the angular login page is not inside into dist folder.
Here there is the structure of my project:
myproject
----myproject-ng (angulars folder)
----myproject-persistence
--------entities
--------repositories
----myproject-service
--------serivices
----myproject-web
--------controllers
--------webapp
------------myproject-ng(angulars dist folder)

Ideally AngularJs 6 Project should run independent of the backend microservice. You will have to enable Cross Origin request in controller by #CrossOrigin annotation.

Where does Spring Security get its default http configuration from?

I'm using Spring Security 3.2.4 and trying to determine where it gets its default configuration from. For example, when using the following configuration:
<http use-expressions="true">
<intercept-url pattern="/secure/login" access="permitAll" />
<intercept-url pattern="/secure/logout" access="permitAll" />
<intercept-url pattern="/secure/denied" access="permitAll" />
<session-management session-fixation-protection="migrateSession" session-authentication-error-url="/login.jsp?authFailed=true">
<concurrency-control max-sessions="10" error-if-maximum-exceeded="true" expired-url="/login.html" session-registry-alias="sessionRegistry"/>
</session-management>
<intercept-url pattern="/**" access="isAuthenticated()" />
<form-login login-page="/secure/login" default-target-url="/" authentication-failure-url="/secure/denied" />
<logout logout-url="/secure/logout" logout-success-url="/" />
<expression-handler ref="defaultWebSecurityExpressionHandler" />
</http>
Something in a Spring configuration somewhere, is telling Spring to instantiate the LoginUrlAuthenticationEntryPoint and set the login-page to "/secure/login", etc. I realize that it is the form-login tag that is doing this magic for me, but where within Spring is the form-login tag translated to mean the LoginUrlAuthenticationEntryPoint, etc?
Similarly, by default, Spring will instantiate a filter chain of name org.springframework.security.filterChains, but can't find where that bean is defined. I presume it is in an xml configuration file within one of the Spring Security jars, but I can't find it anywhere.
Where are all these defaults configured?

Check out SecurityNamespaceHandler and work your way down from there. As you can see, it is in the spring-security-config artifact, in package org.springframework.security.config.

Spring Security 3.2.0.RC1 - <http> element and deprecated method

After upgrading to Spring Security 3.2.0.RC1 I'm getting the warning "Method 'setFilterProcessesUrl' is marked deprecated" for <http auto-config="true"> in my xml config. I get this warning even for a very simple configuration:
<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.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security.xsd">
<http auto-config="true">
<intercept-url pattern="/myurl*" access="ROLE_USER" />
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="user1" password="12345" authorities="ROLE_USER" />
</user-service>
</authentication-provider>
</authentication-manager>
</beans:beans>
According to Spring Security 3.2 API documentation setFilterProcessesUrl is deprecated and setRequiresAuthenticationRequestMatcher(RequestMatcher) should be used instead. How can I change this basic XML configuration, so it doesn't use deprecated methods? I'm using Eclipse Kepler with Spring Tool Suite plugin.
UPDATE:
If I remove <http auto-config="true"> and add <form-login /> to the http element
<http>
<intercept-url pattern="/myurl*" access="ROLE_USER" />
<form-login />
</http>
I also get the "Method 'setFilterProcessesUrl' is marked deprecated" warning and if I add <logout /> I get the same warning the second time.
On the other hand, if I replace <form-login /> and <logout /> with <http-basic /> the warnings go away.

If you are using the namespace then an IDE error like this doesn't really matter, since you can guarantee that Spring Security will support the feature. You aren't actually using the method yourself.
auto-config is a bad idea generally. Someone looking at that configuration won't easily know what it actually does. Do you really want basic authentication, for example? You are best to remove auto-config and explicitly add the features you want.

Fixed in Spring Security 3.2.1. The warnings were caused by XML namespace using a deprecated method.
https://jira.springsource.org/browse/SEC-2455

Double authentication in GWT / EXT + Spring Security application

I have problems with double authentication. I have implemented the authentication form through pop-up window which is always on top. But I have problem probably with interceptors that cause the authentication request by Tomcat even before the start of application:
A username and password are being requested by http://127.0.0.1:8888.
The site says: "Spring Security Application"
If I disable interceptors, I see in log that SecurityContextHolder treats user as Anonymous.
So my question is:
Can I somehow disable that first Tomcat login screen?
My Spring-security configuration XML is:
<?xml version="1.0" encoding="UTF-8"?>
<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.xsd">
<authentication-manager alias="authenticationManager">
<authentication-provider ref="customAuthenticationProvider"/>
</authentication-manager>
<beans:bean id="customAuthenticationProvider" class="com.myCompany.model.security.CustomAuthenticationProvider" >
<beans:property name="databaseId" value="${configuration.databaseId}" />
<beans:property name="applicationId" value="${configuration.applicationId}" />
</beans:bean>
<http auto-config="true" >
<intercept-url pattern="/myApp/**" access="IS_AUTHENTICATED_ANONYMOUSLY"/>
<intercept-url pattern="/MyApp.html*" access="IS_AUTHENTICATED_ANONYMOUSLY"/>
<intercept-url pattern="/gwt/**" access="ROLE_USER"/>
<intercept-url pattern="/**/*.html" access="ROLE_USER"/>
<intercept-url pattern="/css/**" filters="none"/>
<intercept-url pattern="/**" access="ROLE_USER" />
<http-basic />
</http>
<global-method-security secured-annotations="enabled" />
</beans:beans>

In case I understand your question in a correct way you have a problem with double authentication e.g. a Tomcat authentication or an Apache Basic Auth and the Spring authentication mechanism.
While the last project I had related problems with an Apache Basic Auth and the Spring security mechanism. Before launch I had the task to "protect" the access to the site by an simple Apache Basic Auth. By enabling this in the Apache configuration Spring started to do the same: "Spring Security Application" has been shown all the time
The solution for this behaviour was to disable the auto-config:
<security:http auto-config="false" ...>
...
</security:http>

Your question is not too clear. You mention a Tomcat login screen, which I assume is the first screen of your web application, to allow a user to sign in.
If this is correct, and your login page is named, say login.html, all you have to do is configure the interceptors to allow anonymous access to this page-
<intercept-url pattern="/**/login.*" access="IS_AUTHENTICATED_ANONYMOUSLY"/>
<intercept-url pattern="/gwt/**" access="ROLE_USER"/>
<intercept-url pattern="/**/*.html" access="ROLE_USER"/>
<intercept-url pattern="/css/**" filters="none"/>
<intercept-url pattern="/**" access="ROLE_USER" />

Spring 3 Security: AccessDeniedHandler is not being invoked

I have a spring 3 application with the configurations given below. When any user tries to access a page and he/she isn't logged in, I get an Access is Denied exception with an ugly stack trace. How do I handle this exception and not let it dump out a stack trace. I implemented my own access-denied-handler but that doesn't get invoked.
Based on the type of the requested resource, I would like to show custom error messages or pages. Here is my spring configuration.
How do I get Spring to invoke my access-denied-handler . Here is my spring configuration
<security:http auto-config='true'>
<security:intercept-url pattern="/static/**" filters="none"/>
<security:intercept-url pattern="/login" filters="none"/>
<security:intercept-url pattern="/**" access="ROLE_USER" />
<security:form-login login-page="/index"
default-target-url="/home" always-use-default-target="true"
authentication-success-handler-ref="AuthenticationSuccessHandler"
login-processing-url="/j_spring_security_check"
authentication-failure-url="/index?error=true"/>
<security:remember-me key="myLongSecretCookieKey" token-validity-seconds="1296000"
data-source-ref="jdbcDataSource" user-service-ref="AppUserDetailsService" />
<security:access-denied-handler ref="myAccessDeniedHandler" />
</security:http>
<bean id="myAccessDeniedHandler"
class="web.exceptions.handlers.AccessDeniedExceptionHandler">
<property name="errorPage" value="/public/403.htm" />
</bean>
The custom class for handling this exception is given below
public class AccessDeniedExceptionHandler implements AccessDeniedHandler
{
private String errorPage;
#Override
public void handle(HttpServletRequest request, HttpServletResponse response,
AccessDeniedException arg2) throws IOException, ServletException {
response.sendRedirect(errorPage);
}
public void setErrorPage(String errorPage) {
if ((errorPage != null) && !errorPage.startsWith("/")) {
throw new IllegalArgumentException("errorPage must begin with '/'");
}
this.errorPage = errorPage;
}
}
When I run this application, this is the error that I get. I am only pasting a part of the stacktrace and the Spring Debug logs.
20:39:46,173 DEBUG AffirmativeBased:53 - Voter: org.springframework.security.access.vote.RoleVoter#5b7da0d1, returned: -1
20:39:46,173 DEBUG AffirmativeBased:53 - Voter: org.springframework.security.access.vote.AuthenticatedVoter#14c92844, returned: 0
20:39:46,178 DEBUG ExceptionTranslationFilter:154 - Access is denied (user is anonymous); redirecting to authentication entry point
org.springframework.security.access.AccessDeniedException: Access is denied
at org.springframework.security.access.vote.AffirmativeBased.decide(AffirmativeBased.java:71)
at org.springframework.security.access.intercept.AbstractSecurityInterceptor.beforeInvocation(AbstractSecurityInterceptor.java:204)
How do I fix this problem? Firstly, I want to stop spring from Throwing that exception. If it still throws it, I want to handle it and not raise any flags.
Update: I have attached a part of my web.xml as well.
<!-- Hibernate filter configuration -->
<filter>
<filter-name>HibernateFilter</filter-name>
<filter-class>org.springframework.orm.hibernate3.support.OpenSessionInViewFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>HibernateFilter</filter-name>
<url-pattern>/*</url-pattern>
<dispatcher>FORWARD</dispatcher>
<dispatcher>REQUEST</dispatcher>
</filter-mapping>
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<!--Dispatcher Servlet -->
<servlet>
<servlet-name>rowz</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>

In your configuration You require the user to be always authenticated when entering any URL on Your site:
<security:intercept-url pattern="/**" access="ROLE_USER" />
I think You should allow the user to be unauthenticated when entering the login page:
<security:intercept-url pattern="/your-login-page-url" access="ROLE_ANONYMOUS" />
<security:intercept-url pattern="/your-login-process-url" access="ROLE_ANONYMOUS" />
<security:intercept-url pattern="/your-login-failure-url" access="ROLE_ANONYMOUS" />
<security:intercept-url pattern="/**" access="ROLE_USER" />
If You use URL's like: /login/start, /login/error and /login/failure You can have:
<security:intercept-url pattern="/login/**" access="ROLE_ANONYMOUS" />
<security:intercept-url pattern="/**" access="ROLE_USER" />
Update:
Having this configuration should make the framework to redirect all unauthenticated (anonymous) users to login page, and all authenticated to AccessDeniedHandler. The AccessDeniedException is one of the core parts of the framework and ignoring it is not a good idea. It's hard to help more if You only provide parts of Your Spring Security configuration.
Be sure to read the JavaDoc for ExceptionTranslationFilter for detailed explanation of what exceptions are thrown by the framework, why and how are the handled by default.
If possible, try removing as many custom parts You added, like AuthenticationSuccessHandler, RememberMeAuthenticationFilter and AccessDeniedHandler and see if the problem pesist? Try to get the minimal congiuration and add new features step by step to see where the error comes from.
One important thing that You don't mention in Your question is what is the result of this error message? Do You get HTTP 500? Or HTTP 403? Or do You get redirected to login page?
If, as You mentioned in the question, the user is unauthenticated and he/she gets redirected to login page, than that's how it's intended to work. It looks like You get the error message logged by ExceptionTranslationFilter:172 only because You have DEBUG level set to Spring Security classes. If so, than that's also how it's intended to work, and if You don't want the error logged, than simply rise the logging level for Spring Secyruty classes.
Update 2:
The patterns with filters="none" must match the login-page, login-processing-url and authentication-failure-ur attributes set in <security:form-login /> to skip all SpringSecurity checks on pages that display the login page and process the logging in.
<security:http auto-config='true'>
<security:intercept-url pattern="/static/**" filters="none"/>
<security:intercept-url pattern="/index" filters="none"/>
<security:intercept-url pattern="/j_spring_security_check" filters="none"/>
<security:intercept-url pattern="/**" access="ROLE_USER" />
<security:form-login login-page="/index"
default-target-url="/home" always-use-default-target="true"
authentication-success-handler-ref="AuthenticationSuccessHandler"
login-processing-url="/j_spring_security_check"
authentication-failure-url="/index?error=true"/>
<security:remember-me key="myLongSecretCookieKey" token-validity-seconds="1296000"
data-source-ref="jdbcDataSource" user-service-ref="AppUserDetailsService" />
<security:access-denied-handler ref="myAccessDeniedHandler" />
</security:http>

AccessDeniedHandler is invoked when user is logged in and there is no permissions to resource (source here). If you want to handle request for login page when user is not logged in, just configure in security-context:
<http ... entry-point-ref="customAuthenticationEntryPoint">
And define customAuthenticationEntryPoint:
<beans:bean id="customAuthenticationEntryPoint" class="pl.wsiadamy.webapp.controller.util.CustomAuthenticationEntryPoint">
</beans:bean>
TIP, don't try to fight with ExceptionTranslationFilter.
I have tried to override org.springframework.security.web.access.ExceptionTranslationFilter, without effects:
<beans:bean id="exceptionTranslationFilter" class="org.springframework.security.web.access.ExceptionTranslationFilter">
<beans:property name="authenticationEntryPoint" ref="customAuthenticationEntryPoint"/>
<beans:property name="accessDeniedHandler" ref="accessDeniedHandler"/>
</beans:bean>
<beans:bean id="accessDeniedHandler"
class="org.springframework.security.web.access.AccessDeniedHandlerImpl">
<beans:property name="errorPage" value="/accessDenied.htm"/>
</beans:bean>
The ref="customAuthenticationEntryPoint" just didn't invoked.

I have added Spring Access denied page in follwing way:
Spring Frame Work: 3.1
Spring Security: 3.1, Java 1.5+
Entry in *-security.xml:
<security:access-denied-handler error-page="/<My Any error page controller name>" />
Example:
<security:access-denied-handler error-page="/accessDeniedPage.htm" />
Error page will always start with "/"
Entry for controller:
#Controller
public class RedirectAccessDenied {
#RequestMapping(value = "/accessDeniedPage.htm", method = RequestMethod.GET)
public String redirectAccessDenied(Model model) throws IOException, ServletException {
System.out.println("############### Redirect Access Denied Handler!");
return "403";
}
}
Here 403 is my JSP name.

Spring Security uses an AuthenticationEntryPoint object to decide what to do when a user requires authentication. You can create your own AuthenticationEntryPoint bean ( see javadoc ), and then set the entryPoint attribute in the http element:
<http entry-point-ref="entryPointBean" .... />
However, by default, the form-login element creates a LoginUrlAuthenticationEntryPoint which redirects all of your unauthenticated users to the login page, so you shouldn't have to do this yourself. In fact, the log you posted claims it is forwarding the user to the authentication entry point: "Access is denied (user is anonymous); redirecting to authentication entry point".
I wonder if the problem is that you turned off the filter chain for the login url. Instead of setting filters to none, which means spring security is bypassed entirely, try keeping the filters on but allowing unrestricted access like this:
<security:intercept-url pattern="/login" access="permitAll" />
If that still doesn't help, please post the rest of the log so we can see what happens after the request is transferred to the entry point.

Programmatically solution:
#Order(1)
#Configuration
#EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {
//
// ...
//
#Override
protected void configure(HttpSecurity http) throws Exception {
http.exceptionHandling().accessDeniedHandler(new AccessDeniedHandlerImpl() {
#Override
public void handle(HttpServletRequest request, HttpServletResponse response, AccessDeniedException accessDeniedException) throws IOException, ServletException {
super.handle(request, response, accessDeniedException);
accessDeniedException.printStackTrace();
}
});
//
// ...
//
}
}

Can you check your web.xml is supporting forward request?
errorPage is a FORWARD request and mostly in web.xml we support REDIRECTS only. Just a thought else your code looks ok to me.
Edited
A different point of view and This is been taken from working code only.
Have a look at Authenticated Voter class
Disable the annotations
<global-method-security pre-post-annotations="disabled"
secured-annotations="disabled" access-decision-manager-ref="accessDecisionManager">
</global-method-security>
bypassing filters
<http auto-config="true" use-expressions="true"
access-decision-manager-ref="accessDecisionManager"
access-denied-page="/accessDenied">
<intercept-url pattern="/appsecurity/login.jsp" filters="none" />
<intercept-url pattern="/changePassword" filters="none" />
<intercept-url pattern="/pageNotFound" filters="none" />
<intercept-url pattern="/accessDenied" filters="none" />
<intercept-url pattern="/forgotPassword" filters="none" />
<intercept-url pattern="/**" filters="none" />
<form-login login-processing-url="/j_spring_security_check"
default-target-url="/home" login-page="/loginDetails"
authentication-failure-handler-ref="authenticationExceptionHandler"
authentication-failure-url="/?login_error=t" />
<logout logout-url="/j_spring_security_logout"
invalidate-session="true" logout-success-url="/" />
<remember-me />
<!-- Uncomment to limit the number of sessions a user can have -->
<session-management invalid-session-url="/">
<concurrency-control max-sessions="1"
error-if-maximum-exceeded="true" />
</session-management>
</http>
custom Decision Voter
<bean id="customVoter" class="xyz.appsecurity.helper.CustomDecisionVoter" />
Access Decision Manager
<!-- Define AccessDesisionManager as UnanimousBased -->
<bean id="accessDecisionManager" class="org.springframework.security.access.vote.UnanimousBased">
<property name="decisionVoters">
<list>
<ref bean="customVoter" />
<!-- <bean class="org.springframework.security.access.vote.RoleVoter"
/> -->
<bean class="org.springframework.security.access.vote.AuthenticatedVoter" />
</list>
</property>
</bean>
Authentiation Exception Handler
<bean id="authenticationExceptionHandler"
class="org.springframework.security.web.authentication.ExceptionMappingAuthenticationFailureHandler">
<property name="exceptionMappings">
<props>
<!-- /error.jsp -->
<prop
key="org.springframework.security.authentication.BadCredentialsException">/?login_error=t</prop>
<!-- /getnewpassword.jsp -->
<prop
key="org.springframework.security.authentication.CredentialsExpiredException">/changePassword</prop>
<!-- /lockedoutpage.jsp -->
<prop key="org.springframework.security.authentication.LockedException">/?login_error=t</prop>
<!-- /unauthorizeduser.jsp -->
<prop
key="org.springframework.security.authentication.DisabledException">/?login_error=t</prop>
</props>
</property>
</bean>

It looks like spring tries to redirect users who have not logged in to the login page, which is "/index", but that itself is a protected url.
The other possibility is, it tries to display /public/403.html, but that is again protected by security configuration.
Can you add the following entries and try?
<security:intercept-url pattern="/login" filters="none" />
<security:intercept-url pattern="/public/**" filters="none" />

Develop Reference

Java is a programming language and computing platform first released by Sun Microsystems in 1995.

How do I create a logout with Spring Security? - java

Add this line to your config <logout logout-url="/sign-out"/> Then if you have a link to that URL, then it will sign you out (Add it just below your logout success config)

Related

How to relate angular 6 login with spring login

Where does Spring Security get its default http configuration from?

Spring Security 3.2.0.RC1 - <http> element and deprecated method

Double authentication in GWT / EXT + Spring Security application

Spring 3 Security: AccessDeniedHandler is not being invoked

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