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  • HttpsCookieFilter - IllegalStateException: getOutputStream() has already been called for this response

    - by Mat Banik
    Following exception is thrown every once in a while and it shows up in localhost log file in tomcat log directory. If anyone know how to get rid of it, all help would be appreciated. BTW the filter is working fine I just don't know why this exception is happening. Stack trace: java.lang.IllegalStateException: getOutputStream() has already been called for this response at org.apache.catalina.connector.Response.getWriter(Response.java:611) at org.apache.catalina.connector.ResponseFacade.getWriter(ResponseFacade.java:198) at javax.servlet.ServletResponseWrapper.getWriter(ServletResponseWrapper.java:112) at javax.servlet.ServletResponseWrapper.getWriter(ServletResponseWrapper.java:112) at org.springframework.web.servlet.view.freemarker.FreeMarkerView.processTemplate(FreeMarkerView.java:366) at org.springframework.web.servlet.view.freemarker.FreeMarkerView.doRender(FreeMarkerView.java:283) at org.springframework.web.servlet.view.freemarker.FreeMarkerView.renderMergedTemplateModel(FreeMarkerView.java:233) at org.springframework.web.servlet.view.AbstractTemplateView.renderMergedOutputModel(AbstractTemplateView.java:167) at org.springframework.web.servlet.view.AbstractView.render(AbstractView.java:250) at org.springframework.web.servlet.DispatcherServlet.render(DispatcherServlet.java:1047) at org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:817) at org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:719) at org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:644) at org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:549) at javax.servlet.http.HttpServlet.service(HttpServlet.java:617) at javax.servlet.http.HttpServlet.service(HttpServlet.java:717) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.springframework.web.filter.CharacterEncodingFilter.doFilterInternal(CharacterEncodingFilter.java:88) at org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:76) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at com.opensymphony.sitemesh.webapp.SiteMeshFilter.doFilter(SiteMeshFilter.java:65) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.springframework.orm.hibernate3.support.OpenSessionInViewFilter.doFilterInternal(OpenSessionInViewFilter.java:198) at org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:76) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.tuckey.web.filters.urlrewrite.RuleChain.handleRewrite(RuleChain.java:176) at org.tuckey.web.filters.urlrewrite.RuleChain.doRules(RuleChain.java:145) at org.tuckey.web.filters.urlrewrite.UrlRewriter.processRequest(UrlRewriter.java:92) at org.tuckey.web.filters.urlrewrite.UrlRewriteFilter.doFilter(UrlRewriteFilter.java:381) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:368) at org.springframework.security.web.access.intercept.FilterSecurityInterceptor.invoke(FilterSecurityInterceptor.java:109) at org.springframework.security.web.access.intercept.FilterSecurityInterceptor.doFilter(FilterSecurityInterceptor.java:83) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.access.ExceptionTranslationFilter.doFilter(ExceptionTranslationFilter.java:97) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.authentication.AnonymousAuthenticationFilter.doFilter(AnonymousAuthenticationFilter.java:78) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.authentication.rememberme.RememberMeAuthenticationFilter.doFilter(RememberMeAuthenticationFilter.java:119) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.doFilter(AbstractAuthenticationProcessingFilter.java:187) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.authentication.logout.LogoutFilter.doFilter(LogoutFilter.java:105) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter(SecurityContextPersistenceFilter.java:57) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter(SecurityContextPersistenceFilter.java:79) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.access.channel.ChannelProcessingFilter.doFilter(ChannelProcessingFilter.java:109) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.session.ConcurrentSessionFilter.doFilter(ConcurrentSessionFilter.java:109) at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:380) at org.springframework.security.web.FilterChainProxy.doFilter(FilterChainProxy.java:169) at org.springframework.web.filter.DelegatingFilterProxy.invokeDelegate(DelegatingFilterProxy.java:237) at org.springframework.web.filter.DelegatingFilterProxy.doFilter(DelegatingFilterProxy.java:167) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) //Here is the servlet I suspect is trowing the exception. at package.HttpsCookieFilter.doFilter(HttpsCookieFilter.java:38) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191) at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127) at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102) at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109) at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298) at org.apache.coyote.http11.Http11NioProcessor.process(Http11NioProcessor.java:886) at org.apache.coyote.http11.Http11NioProtocol$Http11ConnectionHandler.process(Http11NioProtocol.java:721) at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.run(NioEndpoint.java:2256) at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1110) at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:603) at java.lang.Thread.run(Thread.java:717) The HttpsCookieFilter class: public class HttpsCookieFilter implements Filter { private static Logger log = Logger.getLogger(HttpsCookieFilter.class); @Override public void destroy() { } @Override public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException { final HttpServletRequest req = (HttpServletRequest) request; final HttpServletResponse res = (HttpServletResponse) response; final HttpSession session = req.getSession(false); if (session != null) { setCookie(req, res); } try{ chain.doFilter(request, response); // <- Exception thrown from here }catch (IllegalStateException e){ log.warn("HttpsCookieFilter redirect problem! ", e); } } @Override public void init(FilterConfig arg0) throws ServletException { } private void setCookie( HttpServletRequest request, HttpServletResponse response) { Cookie cookie = new Cookie("JSESSIONID", request.getSession(false).getId()); cookie.setMaxAge(-1); cookie.setPath(getCookiePath(request)); cookie.setSecure(false); response.addCookie(cookie); } private String getCookiePath(HttpServletRequest request) { String contextPath = request.getContextPath(); return contextPath.length() > 0 ? contextPath : "/"; } } web.xml <?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee/web-app_2_5.xsd"> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <listener> <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class> </listener> <listener> <listener-class>org.springframework.security.web.session.HttpSessionEventPublisher</listener-class> </listener> <filter> <filter-name>httpsCookieFilter</filter-name> <filter-class>com.iteezy.server.web.servlet.HttpsCookieFilter</filter-class> </filter> <filter-mapping> <filter-name>httpsCookieFilter</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <filter> <filter-name>filterChainProxy</filter-name> <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class> </filter> <filter-mapping> <filter-name>filterChainProxy</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> ... The reason for integrating this filter comes from Spring security FAQs: I'm using Tomcat (or some other servlet container) and have enabled HTTPS for my login page, switching back to HTTP afterwards. It doesn't work - I just end up back at the login page after authenticating. This happens because sessions created under HTTPS, for which the session cookie is marked as “secure”, cannot subsequently be used under HTTP. The browser will not send the cookie back to the server and any session state will be lost (including the security context information). Starting a session in HTTP first should work as the session cookie won't be marked as secure.

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  • Security of logging people in automatically from another app?

    - by Simon
    I have 2 apps. They both have accounts, and each account has users. These apps are going to share the same users and accounts and they will always be in sync. I want to be able to login automatically from one app to the other. So my solution is to generate a login_key, for example: 2sa7439e-a570-ac21-a2ao-z1qia9ca6g25 once a day. And provide a automated login link to the other app... for example if the user clicks on: https://account_name.securityhole.io/login/2sa7439e-a570-ac21-a2ao-z1qia9ca6g25/user/123 They are logged in automatically, session created. So here we have 3 things that a intruder has to get right in order to gain access; account name, login key, and the user id. Bad idea? Or should I can down the path of making one app an oauth provider? Or is there a better way?

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  • Is There A Security Risk With Users That Are Also Groups?

    - by Rob P.
    I know a little about users and groups; in the past I might have had a group like 'DBAS' or 'ADMINS' and I'd add individual users to each group... But I was surprised to learn I could add users to other users - as if they were groups. For example if my /etc/group contained the following: user1:x:12501: user2:x:12502:user1 admin:x:123:user2,jim,bob Since user2 is a member of the admin group, and user1 is a member of user2 - is user1 effectively an admin? If the admin group is in the sudoers file, can user1 use it as well? I've tried to simulate this and I haven't been able to do so as user1...but I'm not sure it's impossible. EDIT: SORRY - updated error in question.

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  • Is this Java 7 security thread an issue if you have Java 7 installed but not as the default?

    - by user1361315
    I have a MBP with osx mountain lion installed, and I believe from what I read Mac's only ship with Java 6 by default. I'm not at my computer at the moment, but I am pretty sure I have installed Java 7 but it isn't my default java version (I think I installed it and I have to explicitly reference it to use it). Does this mean I am safe from this particular thread? Reference: http://www.pcworld.com/businesscenter/article/261748/researchers_find_critical_vulnerability_in_java_7_patch_hours_after_release.html

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  • Problem with Remember Me Service in Spring Security

    - by Gearóid
    Hi, I'm trying to implement a "remember me" functionality in my website using Spring. The cookie and entry in the persistent_logins table are getting created correctly. Additionally, I can see that the correct user is being restored as the username is displayed at the top of the page. However, once I try to access any information for this user when they return after they were "remembered", I get a NullPointerException. It looks as though the user isn't being set in the session again. My applicationContext-security.xml contains the following: <remember-me data-source-ref="dataSource" user-service-ref="userService"/> ... <authentication-provider user-service-ref="userService" /> <jdbc-user-service id="userService" data-source-ref="dataSource" role-prefix="ROLE_" users-by-username-query="select email as username, password, 1 as ENABLED from user where email=?" authorities-by-username-query="select user.id as id, upper(role.name) as authority from user, role, users_roles where users_roles.user_fk=id and users_roles.role_fk=role.name and user.email=?"/> I thought it may have had something to do with users-by-username query but surely login wouldn't work correctly if this query was incorrect? Any help on this would be greatly appreciated. Thanks, gearoid.

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  • Authlogic, logout, credential capture and security

    - by Paddy
    Ok this is something weird. I got authlogic-oid installed in my rails app today. Everything works perfectly fine but for one small nuisance. This is what i did: I first register with my google openid. Successful login, redirection and my email, along with my correct openid is stored in my database. I am happy that everything worked fine! Now when i logout, my rails app as usual destroys the session and redirects me back to my root url where i can login again. Now if i try to login it still remembers my last login id. Not a big issue as i can always "Sign in as a different user" but i am wondering if there is anyway to not only logout from my app but also logout from google. I noticed the same with stack overflow's openid authentication system. Why am i so bothered about this, you may ask. But is it not a bad idea if your web apps end user, who happens to be in a cyber cafe, thinks he has logged out from your app and hence from his google account only to realize later that his google account had got hacked by some unworthy loser who just happened to notice that the one before him had not logged out from google and say.. changed his password!! Should i be paranoid? Isn't this a major security lapse while implementing the openid spec? Probably today someone can give me a workaround for this issue and the question is solved for me. But what about the others who have implemented openid in their apps and not implemented a workaround?

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  • WCF Double Hop questions about Security and Binding.

    - by Ken Maglio
    Background information: .Net Website which calls a service (aka external service) facade on an app server in the DMZ. This external service then calls the internal service which is on our internal app server. From there that internal service calls a stored procedure (Linq to SQL Classes), and passes the serialized data back though to the external service, and from there back to the website. We've done this so any communication goes through an external layer (our external app server) and allows interoperability; we access our data just like our clients consuming our services. We've gotten to the point in our development where we have completed the system and it all works, the double hop acts as it should. However now we are working on securing the entire process. We are looking at using TransportWithMessageCredentials. We want to have WS2007HttpBinding for the external for interoperability, but then netTCPBinding for the bridge through the firewall for security and speed. Questions: If we choose WS2007HttpBinding as the external services binding, and netTCPBinding for the internal service is this possible? I know WS-* supports this as does netTCP, however do they play nice when passing credential information like user/pass? If we go to Kerberos, will this impact anything? We may want to do impersonation in the future. If you can when you answer post any reference links about why you're answering the way you are, that would be very helpful to us. Thanks!

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  • Cross-platform game development: ease of development vs security

    - by alcuadrado
    Hi, I'm a member and contributor of the Argentum Online (AO) community, the first MMORPG from Argentina, which is Free Software; which, although it's not 3D, it's really addictive and has some dozens of thousands of users. Really unluckily AO was developed in Visual Basic (yes, you can laugh) but the former community, so imagine, the code not only sucks, it has zero portability. I'm planning, with some friends to rewrite the client, and as a GNU/Linux frantic, want to do it cross-platform. Some other people is doing the same with the server in Java. So my biggest problem is that we would like to use a rapid development language (like Java, Ruby or Python) but the client would be pretty insecure. Ruby/Python version would have all it's code available, and the Java one would be easily decompilable (yes, we have some crackers in the community) We have consider the option to implement the security module in C/C++ as a dynamic library, but it can be replaced with a custom one, so it's not really secure. We are also considering the option of doing the core application in C++ and the GUI in Ruby/Python. But haven't analysed all it's implications yet. But we really don't want to code the entire game in C/C++ as it doesn't need that much performance (the game is played at 18fps on average) and we want to develop it as fast as possible. So what would you choose in my case? Thank you!

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  • Where are the real risks in network security?

    - by Barry Brown
    Anytime a username/password authentication is used, the common wisdom is to protect the transport of that data using encryption (SSL, HTTPS, etc). But that leaves the end points potentially vulnerable. Realistically, which is at greater risk of intrusion? Transport layer: Compromised via wireless packet sniffing, malicious wiretapping, etc. Transport devices: Risks include ISPs and Internet backbone operators sniffing data. End-user device: Vulnerable to spyware, key loggers, shoulder surfing, and so forth. Remote server: Many uncontrollable vulnerabilities including malicious operators, break-ins resulting in stolen data, physically heisting servers, backups kept in insecure places, and much more. My gut reaction is that although the transport layer is relatively easy to protect via SSL, the risks in the other areas are much, much greater, especially at the end points. For example, at home my computer connects directly to my router; from there it goes straight to my ISPs routers and onto the Internet. I would estimate the risks at the transport level (both software and hardware) at low to non-existant. But what security does the server I'm connected to have? Have they been hacked into? Is the operator collecting usernames and passwords, knowing that most people use the same information at other websites? Likewise, has my computer been compromised by malware? Those seem like much greater risks. What do you think?

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  • how to retrive pK using spring security

    - by aditya
    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 thanks in advance

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  • Problem with Spring security's logout

    - by uther-lightbringer
    Hello, I've got a problem logging out in Spring framework. First when I want j_spring_security_logout to handle it for me i get 404 j_spring_security_logout not found: sample-security.xml: <http> <intercept-url pattern="/messageList.htm*" access="ROLE_USER,ROLE_GUEST" /> <intercept-url pattern="/messagePost.htm*" access="ROLE_USER" /> <intercept-url pattern="/messageDelete.htm*" access="ROLE_ADMIN" /> <form-login login-page="/login.jsp" default-target-url="/messageList.htm" authentication-failure-url="/login.jsp?error=true" /> <logout/> </http> Sample url link to logout in JSP page: <a href="<c:url value="/j_spring_security_logout" />">Logout</a> When i try to use a custom JSP page i.e. I use login form for this purpose then I get better result at least it gets to login page, but another problem is that you dont't get logged off as you can diretcly type url that should be guarded buy you get past it anyway. Slightly modified from previous listings: <http> <intercept-url pattern="/messageList.htm*" access="ROLE_USER,ROLE_GUEST" /> <intercept-url pattern="/messagePost.htm*" access="ROLE_USER" /> <intercept-url pattern="/messageDelete.htm*" access="ROLE_ADMIN" /> <form-login login-page="/login.jsp" default-target-url="/messageList.htm" authentication-failure-url="/login.jsp?error=true" /> <logout logout-success-url="/login.jsp" /> </http> <a href="<c:url value="/login.jsp" />">Logout</a> Thank you for help

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  • security deleting a mysql row with jQuery $.post

    - by FFish
    I want to delete a row in my database and found an example on how to do this with jQuery's $.post() Now I am wondering about security though.. Can someone send a POST request to my delete-row.php script from another website? JS function deleterow(id) { // alert(typeof(id)); // number if (confirm('Are you sure want to delete?')) { $.post('delete-row.php', {album_id:+id, ajax:'true'}, function() { $("#row_"+id).fadeOut("slow"); }); } } PHP: delete-row.php <?php require_once("../db.php"); mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die("could not connect to database " . mysql_error()); mysql_select_db(DB_NAME) or die("could not select database " . mysql_error()); if (isset($_POST['album_id'])) { $query = "DELETE FROM albums WHERE album_id = " . $_POST['album_id']; $result = mysql_query($query); if (!$result) die('Invalid query: ' . mysql_error()); echo "album deleted!"; } ?>

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  • E-Commerce Security: Only Credit Card Fields Encrypted?!

    - by bizarreunprofessionalanddangerous
    I'd like your opinions on how a major bricks-and-mortar company is running the security for its shopping Web site. After a recent update, when you are logged into your shopping account, the session is now not secured. No 'https', no browser 'lock'. All the personal contact info, shopping history -- and if I'm not mistaken submit and change password -- are being sent unencrypted. There is a small frame around the credit card fields that is https. There's a little notice: "Our website is secure. Our website uses frames and because of this the secure icon will not appear in your browser" On top of this the most prominent login fields for the site are broken, and haven't gotten fixed for a week or longer (giving the distinct impression they have no clue what's going on and can't be trusted with anything). Now is it just me -- or is this simply incomprehensible for a billion dollar company, significant shopping site, in the year 2010. No lock. "We use frames" (maybe they forget "Best viewed in IE4"). Customers complaining, as you can see from their FAQ "explaining" why you aren't seeing https. I'm getting nowhere trying to convince customer service that they REALLY need to do something about this, and am about to head for the CEO. But I just want to make sure this is as BIZARRE and unprofessional and dangerous a situation as I think it is. (I'm trying to visualize what their Web technical team consists of. I'm getting A) some customer service reps who were given a 3 hour training course on Web site maintenance, B) a 14 year old boy in his bedroom masquerading as a major technical services company, C) a guy in a hut in a jungle with an e-commerce book from 1996.)

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  • What is the best prctice for using security in JAX-WS

    - by kislo_metal
    Here is scenario : I have some web services (JAX-WS) that need to be secured. Currently for authentication needs I providing addition SecurityWService that give authorized user some userid & sessionid that is need to be described in request to other services. It would be more better to use some java security. We have many of them but could not defined what is better to use. Q1 : It is understand that I should use SSL in transport layer, but what should I use for user authorization. Is there is better way to establishing session, validating user etc. ? Here is some key description : Most web services clents is php based. I am using jax-ws implementation as a Stateless session EJB. Deploying to glassfish v3. Q2: what is the best framework / technology for user authorization / authentication in case of using JSF 2.0 and ejb3.1 technologies ( Realms? WSIT? )? Thank You!

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  • Lack of security in many PHP applications?

    - by John
    Over the past year of freelancing, I inherited two web projects, both of them built in PHP, both of them with sensitive information like credit card info, bank info, etc... In one application, when I typed http://thecompany.com/admin/, and without being asked for a username and password, I saw every user's sensitive information, including credit card numbers, bank account numbers etc... In another application, I was able to bypass the login screen by simply typing http://the2ndcompany.com/customer.php?user_id=777, and again, without any prompts for username and password, i was able to see user 777's credit card info. I cycled through a few more user_ids (any integer) and saw each person's credit card info. Is something wrong here? Or is this the quality of work that the "average" programmer produces? Because if this is what the average programmer produces, does that means I'm an...gasp...elite programmer?? No..that can't be right....something doesn't make sense. So my question is, is it just coincidence that I inherited two applications both of which are dangerously lacking in security? Or are there are a lot of bad PHP programmers out there?

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  • How to manually set an authenticated user in Spring Security / SpringMVC

    - by David Parks
    After a new user submits a 'New account' form, I want to manually log that user in so they don't have to login on the subsequent page. The normal form login page going through the spring security interceptor works just fine. In the new-account-form controller I am creating a UsernamePasswordAuthenticationToken and setting it in the SecurityContext manually: SecurityContextHolder.getContext().setAuthentication(authentication); On that same page I later check that the user is logged in with: SecurityContextHolder.getContext().getAuthentication().getAuthorities(); This returns the authorities I set earlier in the authentication. All is well. But when this same code is called on the very next page I load, the authentication token is just UserAnonymous. I'm not clear why it did not keep the authentication I set on the previous request. Any thoughts? Could it have to do with session ID's not being set up correctly? Is there something that is possibly overwriting my authentication somehow? Perhaps I just need another step to save the authentication? Or is there something I need to do to declare the authentication across the whole session rather than a single request somehow? Just looking for some thoughts that might help me see what's happening here.

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  • Struts 2 security

    - by Dewfy
    Does Struts 2 has complete solution for simple login task? I have simple declaration in struts.xml: <package namespace="/protected" name="manager" extends="struts-default" > <interceptors> <interceptor-stack name="secure"> <interceptor-ref name="roles"> <param name="allowedRoles">registered</param> </interceptor-ref> </interceptor-stack> </interceptors> <default-action-ref name="pindex"/> <action name="pindex" > <interceptor-ref name="completeStack"/> <interceptor-ref name="secure"/> <result>protected/index.html</result> </action> </package> Accessing to this resource shows only (Forbidden 403). So what should I do on the next step to: Add login page (standart Tomcat declaration on web.xml with <login-config> not works) ? Provide security round trip. Do I need write my own servlet or exists struts2 solutions? Thanks in advance!

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  • Protecting my apps security from deassembling

    - by sandis
    So I recently tested deassembling one of my android apps, and to my horror I discovered that the code was quite readable. Even worse, all my variable names where intact! I thought that those would be compressed to something unreadable at compile time. The app is triggered to expire after a certain time. However, now it was trivial for me to find my function named checkIfExpired() and find the variable "expired". Is there any good way of making it harder for a potential hacker messing with my app? Before someone states the obvious: Yes, it is security through obscurity. But obviously this is my only option since the user always will have access to all my code. This is the same for all apps. The details of my deactivation-thingy is unimportant, the point is that I dont want deassembler to understand some of the things I do. side questions: Why are the variable names not compressed? Could it be the case that my program would run faster if I stopped using really long variable names, as are my habit?

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  • DWR and Spring Security - User is deauthenticated in few seconds

    - by Vojtech
    I am trying to implement user authentication via DWR as follows: public class PublicRemote { @Autowired @Qualifier("authenticationManager") private AuthenticationManager authenticationManager; public Map<String, Object> userLogin(String username, String password, boolean stay) { Map<String, Object> map = new HashMap<>(); UsernamePasswordAuthenticationToken authRequest = new UsernamePasswordAuthenticationToken(username, password); try { Authentication authentication = authenticationManager.authenticate(authRequest); SecurityContextHolder.getContext().setAuthentication(authentication); map.put("success", "true"); } catch (Exception e) { map.put("success", "false"); } return map; } public Map<String, Object> getUserState() { Map<String, Object> map = new HashMap<>(); Authentication authentication = SecurityContextHolder.getContext().getAuthentication(); boolean authenticated = authentication != null && authentication.isAuthenticated(); map.put("authenticated", authenticated); if (authenticated) { map.put("authorities", authentication.getAuthorities()); } return map; } } The authentication works correctly and by calling getUserState() I can see that the user is successfully logged in. The problem is that this state will stay only for few seconds. In probably 5 seconds, the getAuthentication() starts returning null. Is there some problem with session in DWR or is it some misconfiguration of Spring Security?

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  • Security of Flex for payment website

    - by Mario
    So, it's been about 3 years since I wrote and went live with my company's main internet facing website. Originally written in php, I've since just been making minor changes here and there to progress the site as we've needed to. I've wanted to rewrite it from the ground up in the last year or so and now, we want to add some major features so this is a perfect time. The website in question is as close to a banking website as you'd get (without being a bank; sorry for the obscurity, but the less info I can give out, the better). For the rewrite, I want to separate the presentation layer from the processing layer as much as I can. I want the end user to be stuck in a box and not be able to get out so to speak (this is all because of PCI complacency, being PEN tested every 3 months, etc...) So, being probed every 3 months has increasingly made me nervous. We haven't failed yet and there hasen't been a breach yet, but I want to make sure I continue to pass (as much as I can anyways) So, I'm considering rewriting the presentation layer in Adobe Flex and do all the processing in PHP (effectively IMO, separating presentation from processing) - I would do all my normal form validation in flex (as opposed to javascript or php) and do my reads and writes to the db via php. My questions are: I know Flash has something like 99% market penetration - do people find this to be true? Has anyone seen on their own sites being in flash that someone couldn't access it? Flash in general has come under alot of attacks about security and the like - i know this. I would use a swf encryptor - disable debugging (which i got snagged on once on a different application), continue to use https and any other means i can think of. At the end of the day, everyone knows if someone wants in to the data bad enough, their going to find a ways in; i just wanna make it as difficult for them as i can. Any thoughts are appreciated. -Mario

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  • UDP security and identifying incoming data.

    - by Charles
    I have been creating an application using UDP for transmitting and receiving information. The problem I am running into is security. Right now I am using the IP/socketid in determining what data belongs to whom. However, I have been reading about how people could simply spoof their IP, then just send data as a specific IP. So this seems to be the wrong way to do it (insecure). So how else am I suppose to identify what data belongs to what users? For instance you have 10 users connected, all have specific data. The server would need to match the user data to this data we received. The only way I can see to do this is to use some sort of client/server key system and encrypt the data. I am curious as to how other applications (or games, since that's what this application is) make sure their data is genuine. Also there is the fact that encryption takes much longer to process than unencrypted. Although I am not sure by how much it will affect performance. Any information would be appreciated. Thanks.

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  • Security strategies for storing password on disk

    - by Mike
    I am building a suite of batch jobs that require regular access to a database, running on a Solaris 10 machine. Because of (unchangable) design constraints, we are required use a certain program to connect to it. Said interface requires us to pass a plain-text password over a command line to connect to the database. This is a terrible security practice, but we are stuck with it. I am trying to make sure things are properly secured on our end. Since the processing is automated (ie, we can't prompt for a password), and I can't store anything outside the disk, I need a strategy for storing our password securely. Here are some basic rules The system has multiple users. We can assume that our permissions are properly enforced (ie, if a file with a is chmod'd to 600, it won't be publically readable) I don't mind anyone with superuser access looking at our stored password Here is what i've got so far Store password in password.txt $chmod 600 password.txt Process reads from password.txt when it's needed Buffer overwritten with zeros when it's no longer needed Although I'm sure there is a better way.

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  • How to strengthen Mysql database server Security?

    - by i need help
    If we were to use server1 for all files (file server), server2 for mysql database (database server). In order for websites in server1 to access to the database in server2, isn't it needed to connect to to ip address of second (mysql server) ? In this case, is remote mysql connection. However, I seen from some people comment on the security issue. remote access to MySQL is not very secure. When your remote computer first connects to your MySQL database, the password is encrypted before being transmitted over the Internet. But after that, all data is passed as unencrypted "plain text". If someone was able to view your connection data (such as a "hacker" capturing data from an unencrypted WiFi connection you're using), that person would be able to view part or all of your database. So I just wondering ways to secure it? Allow remote mysql access from server1 by allowing the static ip adress allow remote access from server 1 by setting port allowed to connect to 3306 change 3306 to other port? Any advice?

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  • [GEEK SCHOOL] Network Security 1: Securing User Accounts and Passwords in Windows

    - by Matt Klein
    This How-To Geek School class is intended for people who want to learn more about security when using Windows operating systems. You will learn many principles that will help you have a more secure computing experience and will get the chance to use all the important security tools and features that are bundled with Windows. Obviously, we will share everything you need to know about using them effectively. In this first lesson, we will talk about password security; the different ways of logging into Windows and how secure they are. In the proceeding lesson, we will explain where Windows stores all the user names and passwords you enter while working in this operating systems, how safe they are, and how to manage this data. Moving on in the series, we will talk about User Account Control, its role in improving the security of your system, and how to use Windows Defender in order to protect your system from malware. Then, we will talk about the Windows Firewall, how to use it in order to manage the apps that get access to the network and the Internet, and how to create your own filtering rules. After that, we will discuss the SmartScreen Filter – a security feature that gets more and more attention from Microsoft and is now widely used in its Windows 8.x operating systems. Moving on, we will discuss ways to keep your software and apps up-to-date, why this is important and which tools you can use to automate this process as much as possible. Last but not least, we will discuss the Action Center and its role in keeping you informed about what’s going on with your system and share several tips and tricks about how to stay safe when using your computer and the Internet. Let’s get started by discussing everyone’s favorite subject: passwords. The Types of Passwords Found in Windows In Windows 7, you have only local user accounts, which may or may not have a password. For example, you can easily set a blank password for any user account, even if that one is an administrator. The only exception to this rule are business networks where domain policies force all user accounts to use a non-blank password. In Windows 8.x, you have both local accounts and Microsoft accounts. If you would like to learn more about them, don’t hesitate to read the lesson on User Accounts, Groups, Permissions & Their Role in Sharing, in our Windows Networking series. Microsoft accounts are obliged to use a non-blank password due to the fact that a Microsoft account gives you access to Microsoft services. Using a blank password would mean exposing yourself to lots of problems. Local accounts in Windows 8.1 however, can use a blank password. On top of traditional passwords, any user account can create and use a 4-digit PIN or a picture password. These concepts were introduced by Microsoft to speed up the sign in process for the Windows 8.x operating system. However, they do not replace the use of a traditional password and can be used only in conjunction with a traditional user account password. Another type of password that you encounter in Windows operating systems is the Homegroup password. In a typical home network, users can use the Homegroup to easily share resources. A Homegroup can be joined by a Windows device only by using the Homegroup password. If you would like to learn more about the Homegroup and how to use it for network sharing, don’t hesitate to read our Windows Networking series. What to Keep in Mind When Creating Passwords, PINs and Picture Passwords When creating passwords, a PIN, or a picture password for your user account, we would like you keep in mind the following recommendations: Do not use blank passwords, even on the desktop computers in your home. You never know who may gain unwanted access to them. Also, malware can run more easily as administrator because you do not have a password. Trading your security for convenience when logging in is never a good idea. When creating a password, make it at least eight characters long. Make sure that it includes a random mix of upper and lowercase letters, numbers, and symbols. Ideally, it should not be related in any way to your name, username, or company name. Make sure that your passwords do not include complete words from any dictionary. Dictionaries are the first thing crackers use to hack passwords. Do not use the same password for more than one account. All of your passwords should be unique and you should use a system like LastPass, KeePass, Roboform or something similar to keep track of them. When creating a PIN use four different digits to make things slightly harder to crack. When creating a picture password, pick a photo that has at least 10 “points of interests”. Points of interests are areas that serve as a landmark for your gestures. Use a random mixture of gesture types and sequence and make sure that you do not repeat the same gesture twice. Be aware that smudges on the screen could potentially reveal your gestures to others. The Security of Your Password vs. the PIN and the Picture Password Any kind of password can be cracked with enough effort and the appropriate tools. There is no such thing as a completely secure password. However, passwords created using only a few security principles are much harder to crack than others. If you respect the recommendations shared in the previous section of this lesson, you will end up having reasonably secure passwords. Out of all the log in methods in Windows 8.x, the PIN is the easiest to brute force because PINs are restricted to four digits and there are only 10,000 possible unique combinations available. The picture password is more secure than the PIN because it provides many more opportunities for creating unique combinations of gestures. Microsoft have compared the two login options from a security perspective in this post: Signing in with a picture password. In order to discourage brute force attacks against picture passwords and PINs, Windows defaults to your traditional text password after five failed attempts. The PIN and the picture password function only as alternative login methods to Windows 8.x. Therefore, if someone cracks them, he or she doesn’t have access to your user account password. However, that person can use all the apps installed on your Windows 8.x device, access your files, data, and so on. How to Create a PIN in Windows 8.x If you log in to a Windows 8.x device with a user account that has a non-blank password, then you can create a 4-digit PIN for it, to use it as a complementary login method. In order to create one, you need to go to “PC Settings”. If you don’t know how, then press Windows + C on your keyboard or flick from the right edge of the screen, on a touch-enabled device, then press “Settings”. The Settings charm is now open. Click or tap the link that says “Change PC settings”, on the bottom of the charm. In PC settings, go to Accounts and then to “Sign-in options”. Here you will find all the necessary options for changing your existing password, creating a PIN, or a picture password. To create a PIN, press the “Add” button in the PIN section. The “Create a PIN” wizard is started and you are asked to enter the password of your user account. Type it and press “OK”. Now you are asked to enter a 4-digit pin in the “Enter PIN” and “Confirm PIN” fields. The PIN has been created and you can now use it to log in to Windows. How to Create a Picture Password in Windows 8.x If you log in to a Windows 8.x device with a user account that has a non-blank password, then you can also create a picture password and use it as a complementary login method. In order to create one, you need to go to “PC settings”. In PC Settings, go to Accounts and then to “Sign-in options”. Here you will find all the necessary options for changing your existing password, creating a PIN, or a picture password. To create a picture password, press the “Add” button in the “Picture password” section. The “Create a picture password” wizard is started and you are asked to enter the password of your user account. You are shown a guide on how the picture password works. Take a few seconds to watch it and learn the gestures that can be used for your picture password. You will learn that you can create a combination of circles, straight lines, and taps. When ready, press “Choose picture”. Browse your Windows 8.x device and select the picture you want to use for your password and press “Open”. Now you can drag the picture to position it the way you want. When you like how the picture is positioned, press “Use this picture” on the left. If you are not happy with the picture, press “Choose new picture” and select a new one, as shown during the previous step. After you have confirmed that you want to use this picture, you are asked to set up your gestures for the picture password. Draw three gestures on the picture, any combination you wish. Please remember that you can use only three gestures: circles, straight lines, and taps. Once you have drawn those three gestures, you are asked to confirm. Draw the same gestures one more time. If everything goes well, you are informed that you have created your picture password and that you can use it the next time you sign in to Windows. If you don’t confirm the gestures correctly, you will be asked to try again, until you draw the same gestures twice. To close the picture password wizard, press “Finish”. Where Does Windows Store Your Passwords? Are They Safe? All the passwords that you enter in Windows and save for future use are stored in the Credential Manager. This tool is a vault with the usernames and passwords that you use to log on to your computer, to other computers on the network, to apps from the Windows Store, or to websites using Internet Explorer. By storing these credentials, Windows can automatically log you the next time you access the same app, network share, or website. Everything that is stored in the Credential Manager is encrypted for your protection.

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  • Spring security problem, Error creating bean with name 'org.springframework.web.servlet.mvc.annotati

    - by benaissa
    Hello; I'm developping a web application with spring mvc, i started by developping the web application after i'm trying to add spring security; but i have this message, and i don't find a solution, thanks 16-04-2010 12:10:22:296 6062 ERROR org.springframework.web.servlet.DispatcherServlet - Context initialization failed org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping': Initialization of bean failed; nested exception is java.lang.NoClassDefFoundError: org/springframework/beans/factory/generic/GenericBeanFactoryAccessor at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:527) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:456) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:286) at org.springframework.web.servlet.DispatcherServlet.createDefaultStrategy(DispatcherServlet.java:770) at org.springframework.web.servlet.DispatcherServlet.getDefaultStrategies(DispatcherServlet.java:737) at org.springframework.web.servlet.DispatcherServlet.initHandlerMappings(DispatcherServlet.java:518) at org.springframework.web.servlet.DispatcherServlet.initStrategies(DispatcherServlet.java:410) at org.springframework.web.servlet.DispatcherServlet.onRefresh(DispatcherServlet.java:398) at org.springframework.web.servlet.FrameworkServlet.onApplicationEvent(FrameworkServlet.java:474) at org.springframework.context.event.GenericApplicationListenerAdapter.onApplicationEvent(GenericApplicationListenerAdapter.java:51) at org.springframework.context.event.SourceFilteringListener.onApplicationEventInternal(SourceFilteringListener.java:97) at org.springframework.context.event.SourceFilteringListener.onApplicationEvent(SourceFilteringListener.java:68) at org.springframework.context.event.SimpleApplicationEventMulticaster.multicastEvent(SimpleApplicationEventMulticaster.java:97) at org.springframework.context.support.AbstractApplicationContext.publishEvent(AbstractApplicationContext.java:301) at org.springframework.context.support.AbstractApplicationContext.finishRefresh(AbstractApplicationContext.java:888) at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:426) at org.springframework.web.servlet.FrameworkServlet.createWebApplicationContext(FrameworkServlet.java:402) at org.springframework.web.servlet.FrameworkServlet.initWebApplicationContext(FrameworkServlet.java:316) at org.springframework.web.servlet.FrameworkServlet.initServletBean(FrameworkServlet.java:282) at org.springframework.web.servlet.HttpServletBean.init(HttpServletBean.java:126) at javax.servlet.GenericServlet.init(GenericServlet.java:212) at org.apache.catalina.core.StandardWrapper.loadServlet(StandardWrapper.java:1173) at org.apache.catalina.core.StandardWrapper.allocate(StandardWrapper.java:809) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:129) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191) at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127) at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102) at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109) at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298) at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852) at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588) at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489) at java.lang.Thread.run(Thread.java:619) Caused by: java.lang.NoClassDefFoundError: org/springframework/beans/factory/generic/GenericBeanFactoryAccessor at org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping.determineUrlsForHandler(DefaultAnnotationHandlerMapping.java:113) at org.springframework.web.servlet.handler.AbstractDetectingUrlHandlerMapping.detectHandlers(AbstractDetectingUrlHandlerMapping.java:79) at org.springframework.web.servlet.handler.AbstractDetectingUrlHandlerMapping.initApplicationContext(AbstractDetectingUrlHandlerMapping.java:57) at org.springframework.context.support.ApplicationObjectSupport.initApplicationContext(ApplicationObjectSupport.java:119) at org.springframework.web.context.support.WebApplicationObjectSupport.initApplicationContext(WebApplicationObjectSupport.java:69) at org.springframework.context.support.ApplicationObjectSupport.setApplicationContext(ApplicationObjectSupport.java:73) at org.springframework.context.support.ApplicationContextAwareProcessor.invokeAwareInterfaces(ApplicationContextAwareProcessor.java:99) at org.springframework.context.support.ApplicationContextAwareProcessor.postProcessBeforeInitialization(ApplicationContextAwareProcessor.java:82) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.applyBeanPostProcessorsBeforeInitialization(AbstractAutowireCapableBeanFactory.java:394) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1405) at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:519) ... 32 more Caused by: java.lang.ClassNotFoundException: org.springframework.beans.factory.generic.GenericBeanFactoryAccessor at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1516) at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1361) at java.lang.ClassLoader.loadClassInternal(ClassLoader.java:320) ... 43 more

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