使用oauth2进行休息服务:无法找到令牌的访问令牌

时间:2016-03-28 20:47:27

标签: spring-mvc spring-security spring-security-oauth2 spring-restcontroller

我尝试创建spring rest服务,whis是我自己的oauth2资源服务器的高手。我创建了资源服务器:

@Configuration
@EnableResourceServer
protected static class ResourceServer extends ResourceServerConfigurerAdapter {

    @Autowired
    private TokenStore tokenStore;

    @Override
    public void configure(ResourceServerSecurityConfigurer resources) throws Exception {
        resources.tokenStore(tokenStore).resourceId("mobileapp");
    }

    @Override
    public void configure(HttpSecurity http) throws Exception {
        http.authorizeRequests().antMatchers("/api/shop /**").authenticated().and()
                .authorizeRequests().antMatchers("/auth/**").anonymous();
    }

}

和授权服务器:

@Configuration
@EnableAuthorizationServer
protected static class OAuth2Config extends AuthorizationServerConfigurerAdapter {

    @Autowired
    private AuthenticationManager auth;

    @Autowired
    private DataSource dataSource;

    @Autowired
    private BCryptPasswordEncoder passwordEncoder;

    @Bean
    public JdbcTokenStore tokenStore() {
        return new JdbcTokenStore(dataSource);
    }

    @Bean
    protected AuthorizationCodeServices authorizationCodeServices() {
        return new JdbcAuthorizationCodeServices(dataSource);
    }

    @Override
    public void configure(AuthorizationServerSecurityConfigurer security) throws Exception {
        security.passwordEncoder(passwordEncoder);
    }

    @Override
    public void configure(AuthorizationServerEndpointsConfigurer endpoints) throws Exception {
        endpoints
                .authorizationCodeServices(authorizationCodeServices())
                .authenticationManager(auth)
                .tokenStore(tokenStore())
                .approvalStoreDisabled();
    }

    @Override
    public void configure(ClientDetailsServiceConfigurer clients) throws Exception {
        clients.jdbc(dataSource)
                .passwordEncoder(passwordEncoder);
                .withClient("mobile")
                .authorizedGrantTypes("password", "refresh_token")
                .authorities("ROLE_CLIENT")
                .scopes("read", "write", "trust")
                .autoApprove(true)
               .resourceIds("mobileapp")
                .secret("123456");
    }

当我尝试从服务器接收访问令牌时,使用curl:

  

curl -X POST -vu mobile:123456 http://localhost:8080/oauth/token -H   "接受:application / json" -d   "密码= test123&安培; username=admin@gmail.com& grant_type =密码&安培;范围=读&安培; client_secret = 123456&安培; CLIENT_ID =移动"

我将此错误视为回复消息:

  

{"错误":" SERVER_ERROR"" ERROR_DESCRIPTION":" java.io.NotSerializableException:   org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder"}

在tomcat日志中还有

  

o.s.s.o.p.token.store.JdbcTokenStore - 无法找到访问令牌   令牌

编辑: 密码编码器的Bean定义:

@Bean
public BCryptPasswordEncoder passwordEncoder() {
    BCryptPasswordEncoder bCryptPasswordEncoder = new BCryptPasswordEncoder();
    return bCryptPasswordEncoder;
}

此bean是在类中创建的,其中声明了OAuth2Config和ResourceServer。

我检查了代码,发现了哪个表弹簧使用,表是空的。我的问题是:它应该是自动生成还是我的代码有问题?

提前感谢您的帮助。

3 个答案:

答案 0 :(得分:8)

覆盖JdbcTokenStore类并将此函数替换为。

public OAuth2AccessToken readAccessToken(String tokenValue) {
    OAuth2AccessToken accessToken = null;

    try {
        accessToken = new DefaultOAuth2AccessToken(tokenValue);
    }
    catch (EmptyResultDataAccessException e) {
        if (LOG.isInfoEnabled()) {
            LOG.info("Failed to find access token for token "+tokenValue);
        }
    }
    catch (IllegalArgumentException e) {
        LOG.warn("Failed to deserialize access token for " +tokenValue,e);
        removeAccessToken(tokenValue);
    }

    return accessToken;
}

解决了无法找到访问令牌的问题。在OAuth2Config中使用此类。

答案 1 :(得分:1)

您的模型必须具有未序列化的BCryptPasswordEncoder。在您的用户bmodel中使其成为瞬态。

private transient BCryptPasswordEncoder passwordEncoder;

答案 2 :(得分:0)

就Postgres而言,该解决方案在{strong> ALL BYTEAtoken列中使用authentication

在此架构参考中,列定义为LONGVARBINARYhttps://github.com/spring-projects/spring-security-oauth/blob/master/spring-security-oauth2/src/test/resources/schema.sql

换句话说,如果您使用的是Postgres,请将LONGVARBINARY替换为BYTEA

欢呼