Spring CSRF + AngularJs

我已经尝试过许多关于该主题的方法，但没有人适合我。

我在Spring MVC 4.1.7，Spring Security 3.2.3和MySQL + tomcat7上都有基本的CRUD。

问题是，当我尝试使用AngularJS发布表单时，我一直被错误403（访问被拒绝）阻止。

我发现我需要发送带有POST请求的CSRF_TOKEN，但是@R_976_6618@办！

我尝试了许多不同的方法，但没有人起作用。

我的文件

Controller.js

$scope.novo = function novo() {
  if($scope.id){
   alert("update - " + $scope.id);
  }
  else{
      var Obj = {
              descricao : 'Test',saldo_inicial : 0.00,saldo : 33.45,aberto : false,usuario_id : null,ativo : true
      };
      $http.post(urlBase + 'caixas/aDicionar',Obj).success(function(data) {
          $scope.caixas = data;    
      }).error(function(data) {alert(data)});
  }
 };

Spring-security.xml

<beans:beans xmlns="http://www.springframework.org/scheR_868_11845@a/security"
xmlns:beans="http://www.springframework.org/scheR_868_11845@a/beans" xmlns:xsi="http://www.w3.org/2001/XMLscheR_868_11845@a-instance"
xsi:scheR_868_11845@a@R_674_5352@n="http://www.springframework.org/scheR_868_11845@a/beans
http://www.springframework.org/scheR_868_11845@a/beans/spring-beans.xsd
http://www.springframework.org/scheR_868_11845@a/security
http://www.springframework.org/scheR_868_11845@a/security/spring-security.xsd">

<!-- enable use-expressions -->
<http auto-config="true" use-expressions="true">
    <intercept-url pattern="/seguro**"
        access="hasAnyRole('ROLE_USER','ROLE_ADMIN')" />
    <intercept-url pattern="/seguro/financeiro**"
        access="hasAnyRole('ROLE_FINANCEIRO','ROLE_ADMIN')" />

    <!-- access denied page -->
    <access-denied-handler error-page="/negado" />
    <form-login login-page="/home/" default-target-url="/seguro/"
        authentication-failure-url="/home?error" username-parameter="inputEmail"
        password-parameter="inputpassword" />
    <logout logout-success-url="/home?logout" />
    <!-- enable csrf protection -->
    <csrf />
</http>

<!-- SELEct users and user_roles from database -->
<authentication-manager>
    <authentication-provider>
        <password-encoder hash="md5" />
        <jdbc-user-service data-source-ref="datasource"
            users-by-username-query="SELECT login,senha,ativo
               FROM usuarios 
              WHERE login = ?"
            authorities-by-username-query="SELECT u.login,r.role
               FROM usuarios_roles r,usuarios u
              WHERE u.id = r.usuario_id
                AND u.login = ?" />
    </authentication-provider>
</authentication-manager>

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLscheR_868_11845@a-instance"
xsi:scheR_868_11845@a@R_674_5352@n="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

<display-name>Barattie ~ Soluções Integradas</display-name>

<!-- The definition of the Root Spring Container shared by all Servlets 
    and Filters -->
<context-param>
    <param-name>contextConfig@R_674_5352@n</param-name>
    <param-value>
        /WEB-INF/spring/spring-security.xml
        /WEB-INF/spring/spring-database.xml
        /WEB-INF/spring/spring-hibernate.xml
    </param-value>
</context-param>
<context-param>
    <param-name>com.sun.faces.writeStateAtFormEnd</param-name>
    <param-value>false</param-value>
</context-param>

<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<!-- Processes application requests -->
<servlet>
    <servlet-name>appServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfig@R_674_5352@n</param-name>
        <param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>appServlet</servlet-name>
    <url-pattern>/home</url-pattern>
    <url-pattern>/</url-pattern>
</servlet-mapping>

<!-- Spring Security -->
<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>

<error-page>
    <exception-type>java.lang.Throwable</exception-type>
    <@R_674_5352@n>/erro</@R_674_5352@n>
</error-page>

更新

我试图在客户端xsrf上添加重命名，但我一直拒绝访问。

var app = angular.module('myApp',[]).config(function($httpProvider) {
$httpProvider.defaults.xsrfCookiename = '_csrf';
$httpProvider.defaults.xsrfHeaderName = 'X-CSRF-Token';
});

更新2

我试图实现这样的过滤器。

package sys.barattie.util;

import java.io.IOException;

import javax.servlet.FilterChain;
import javax.servlet.ServletException;
import javax.servlet.http.Cookie;
import javax.servlet.http.httpServletrequest;
import javax.servlet.http.httpServletResponse;

import org.springframework.security.web.csrf.CsrfToken;
import org.springframework.web.filter.oncePerrequestFilter;

public class CsrfFilter extends OncePerrequestFilter {

@Override
protected void doFilterInternal(httpServletrequest request,httpServletResponse response,FilterChain filterChain)
        throws ServletException,IOException {
    CsrfToken csrf = (CsrfToken) request.getAttribute(CsrfToken.class.getName());
    if (csrf != null) {
        Cookie cookie = new Cookie("XSRF-TOKEN",csrf.getToken());
        cookie.setPath("/");
        response.addCookie(cookiE);
    }
    filterChain.doFilter(request,responsE);
}
}

并更改了我的弹簧安全性。

     <csrf token-repository-ref="csrfTokenRepository" />
<beans:bean id="csrfTokenRepository" class="org.springframework.security.web.csrf.httpSessionCsrfTokenRepository">
    <beans:property name="headerName" value="X-XSRF-TOKEN" />
</beans:bean>

Web.xml

<filter>
    <filter-name>csrfFilter</filter-name>
    <filter-class>sys.barattie.util.CsrfFilter</filter-class>
</filter>

但这就像服务器未运行筛选器一样，我向其中添加了System.out.println，但在调试中看不到消息