JSP文件
<form:input type="text" class="form-control" path="mName" /> <input type="hidden" name="${_csrf.parameterName}" value="${_csrf.token}" /> <div class="form-action"> <input type="submit" value="Save" class="btn btn-primary" /> </div> </form:form>
我也插入了隐藏标签.现在我该怎么做来验证这个令牌.我在那里迷路了.
在控制器类中,我获取从表单到对象的值,并调用Web ervise来保存数据
@RequestMapping(method = RequestMethod.POST) public String processForm(@modelattribute(value = "userForm") @Valid UserForm userForm,BindingResult result,ModelMap model) { //call the web service }
解决方法
1. Generate new CSRF token and add it to user once on login and store user in http session.
这是在默认的ESAPI实现中完成的,它存储为存储在会话中的User对象的成员变量.
/this code is in the DefaultUser implementation of ESAPI /** This user's CSRF token. */ private String csrfToken = resetCSRFToken(); ... public String resetCSRFToken() { csrfToken = ESAPI.randomizer().getRandomString(8,DefaultEncoder.CHAR_ALPHANUMERICS); return csrfToken; }
2. On any forms or urls that should be protected,add the token as a parameter / hidden field.
对于任何需要CSRF保护的URL,应调用下面的addCSRFToken方法.或者,如果您要创建表单,或者使用另一种呈现URL的技术(例如c:url),请确保添加名称为“ctoken”的参数或隐藏字段以及DefaultHTTPUtilities.getCSRFToken()的值.
//from HTTPUtilitiles interface final static String CSRF_TOKEN_NAME = "ctoken"; //this code is from the DefaultHTTPUtilities implementation in ESAPI public String addCSRFToken(String href) { User user = ESAPI.authenticator().getCurrentUser(); if (user.isAnonymous()) { return href; } // if there are already parameters append with &,otherwise append with ? String token = CSRF_TOKEN_NAME + "=" + user.getCSRFToken(); return href.indexOf( '?') != -1 ? href + "&" + token : href + "?" + token; } ... public String getCSRFToken() { User user = ESAPI.authenticator().getCurrentUser(); if (user == null) return null; return user.getCSRFToken(); }
3. On the server side for those protected actions,check that the submitted token matches the token from the user object in the session.
确保从servlet或spring操作或jsf控制器或用于处理请求的任何服务器端机制调用此方法.这应该在您需要验证CSRF保护的任何请求上调用.请注意,当令牌不匹配时,它被视为可能的伪造请求.
//this code is from the DefaultHTTPUtilities implementation in ESAPI public void verifyCSRFToken(HttpServletRequest request) throws IntrusionException { User user = ESAPI.authenticator().getCurrentUser(); // check if user authenticated with this request - no CSRF protection required if( request.getAttribute(user.getCSRFToken()) != null ) { return; } String token = request.getParameter(CSRF_TOKEN_NAME); if ( !user.getCSRFToken().equals( token ) ) { throw new IntrusionException("Authentication Failed","Possibly forged HTTP request without proper CSRF token detected"); } }
4. On logout and session timeout,the user object is removed from the session and the session destroyed.
在此步骤中,将调用注销.发生这种情况时,请注意会话无效并且当前用户对象被重置为匿名用户,从而删除对当前用户的引用以及相应的csrf令牌.
//this code is in the DefaultUser implementation of ESAPI public void logout() { ESAPI.httpUtilities().killCookie( ESAPI.currentRequest(),ESAPI.currentResponse(),HTTPUtilities.REMEMBER_TOKEN_COOKIE_NAME ); HttpSession session = ESAPI.currentRequest().getSession(false); if (session != null) { removeSession(session); session.invalidate(); } ESAPI.httpUtilities().killCookie(ESAPI.currentRequest(),"JSESSIONID"); loggedIn = false; logger.info(Logger.SECURITY_SUCCESS,"logout successful" ); ESAPI.authenticator().setCurrentUser(User.ANONYMOUS); }
希望这可以帮助你.
Shishir