Solved

Spring Security Log Out

Posted on 2011-03-21
1
1,738 Views
Last Modified: 2012-06-27
ello,

When I configure Spring Security 3 to log out the user I get a huge error from the ThreadPoolExecutor

Exception in thread ""http-bio-8080"-exec-7" java.lang.StackOverflowError
at com.dc.api.model.Users.getUsername(Users.java:200)

The Users.java:200 error maps to my org.springframework.security.core.userdetails.User Details implementation method

public String getUsername() {
        return this.getUsername();
    }

log out link:

 
  <a href="${facesContext.externalContext.requestContextPath}/j_spring_security_logout.html">log out</a>

Open in new window



spring security config:
   
     <logout invalidate-session="true" 
		      logout-success-url="/" 
		      logout-url="/j_spring_security_logout.html"/>

Open in new window

                                
web.xml:
<filter-mapping>
        <filter-name>springSecurityFilterChain</filter-name>
        <url-pattern>*.html</url-pattern>
        <dispatcher>FORWARD</dispatcher>
        <dispatcher>REQUEST</dispatcher>
    </filter-mapping>
    <filter-mapping>
        <filter-name>springSecurityFilterChain</filter-name>
        <url-pattern>*.xhtml</url-pattern>
        <dispatcher>FORWARD</dispatcher>
        <dispatcher>REQUEST</dispatcher>
    </filter-mapping>

Open in new window


security context file:
<?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:util="http://www.springframework.org/schema/util"
    xmlns:context="http://www.springframework.org/schema/context"
    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/util
    http://www.springframework.org/schema/util/spring-util-3.0.xsd
    http://www.springframework.org/schema/security
    http://www.springframework.org/schema/security/spring-security-3.1.xsd
    http://www.springframework.org/schema/context
    http://www.springframework.org/schema/context/spring-context-3.0.xsd">
    <context:annotation-config />
    <context:component-scan base-package="dc" />
    <global-method-security />
    <http security="none" pattern="/javax.faces.resource/**" />
    <http security="none" pattern="/services/rest-api/1.0/**" />
    <http security="none" pattern="/preregistered/*" />
 	<http access-denied-page="/auth/denied.html">
        <intercept-url
            pattern="/**/*.xhtml"
            access="ROLE_NONE_GETS_ACCESS" />
        <intercept-url
            pattern="/auth/**"
            access="ROLE_ANONYMOUS,ROLE_USER" />
         <intercept-url
            pattern="/auth/*"
            access="ROLE_ANONYMOUS" />
         <intercept-url
            pattern="/registered/*"
            access="ROLE_USER" />
          <intercept-url
            pattern="/*"
           access="ROLE_ANONYMOUS" />
        <form-login
            login-processing-url="/j_spring_security_check.html"
            login-page="/auth/login.html"
            default-target-url="/registered/home.html"
            authentication-failure-url="/auth/login.html" />
         <logout invalidate-session="true" 
		      logout-success-url="/" 
		      logout-url="/j_spring_security_logout.html"/>
        <anonymous username="guest" granted-authority="ROLE_ANONYMOUS"/>
        <remember-me user-service-ref="userManager" key="key value"/>
 	</http>
 	<!-- Configure the authentication provider -->
	<authentication-manager>
		<authentication-provider user-service-ref="userManager">
		        <password-encoder ref="passwordEncoder" />
		</authentication-provider>
    </authentication-manager>
</beans:beans>

Open in new window

0
Comment
Question by:cgray1223
1 Comment
 
LVL 92

Accepted Solution

by:
objects earned 500 total points
ID: 35179307
public String getUsername() {
        return this.getUsername();
    }

that method is calling itself
should be more like


public String getUsername() {
        return this.username;
    }
0

Featured Post

3 Use Cases for Connected Systems

Our Dev teams are like yours. They’re continually cranking out code for new features/bugs fixes, testing, deploying, testing some more, responding to production monitoring events and more. It’s complex. So, we thought you’d like to see what’s working for us.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Impossible to extract MSI from new JAVA releases 2 48
thymeleaf natural templating vs JSP 2 67
Java Loop 6 49
arguments to jar 5 14
Are you developing a Java application and want to create Excel Spreadsheets? You have come to the right place, this article will describe how you can create Excel Spreadsheets from a Java Application. For the purposes of this article, I will be u…
In this post we will learn how to connect and configure Android Device (Smartphone etc.) with Android Studio. After that we will run a simple Hello World Program.
Video by: Michael
Viewers learn about how to reduce the potential repetitiveness of coding in main by developing methods to perform specific tasks for their program. Additionally, objects are introduced for the purpose of learning how to call methods in Java. Define …
This video teaches viewers about errors in exception handling.

910 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

23 Experts available now in Live!

Get 1:1 Help Now