java – Spring Security OAuth2 InvalidGrantException

2023年1月20日 876次阅读

我已经使用
spring-boot框架成功创建了一个Web服务.现在我想用OAuth2(使用spring)保护我的网络服务,并对此提出一些问题：

根据我的研究,spring提供了某种default-url来请求访问令牌(baseURL / oauth / token).我使用postman测试了URL,并返回了一个有效的访问令牌(使用client_credentials grant类型),但没有刷新令牌.但是,此方法不适用于grant_type = password,并导致以下错误响应：

{“error”：“invalid_grant”,“error_description”：“凭据错误”}

我的spring应用程序记录InvalidGrantException.

我用来测试grant_type = password的curl如下：

curl -v -X POST -H "Content-Type: application/json" -H "Authorization: Basic base64encodedclientidandsecret" 'http://localhost:8888/oauth/token?grant_type=password&username=user&password=1234'

我没有使用邮递员测试,因为它不支持grant_type =密码.

如何使用grant_type = password让spring返回accessToken和refreshToken？

我的配置有什么问题吗？

我的spring应用程序(配置)如下所示：

@Configuration
@ComponentScan
@EnableAutoConfiguration(exclude = { MongoAutoConfiguration.class, MongoDataAutoConfiguration.class })
@SpringBootApplication
public class CsWebServerApplication {

    public static final String RESOURCE_ID = "myresource";

    public static final String CLIENT_ID = "myapplication";
    public static final String CLIENT_SECRET = "application_secret";

    public static void main(String[] args) {

        SpringApplication.run(MyWebServerApplication.class, args);
    }

    @Configuration
    @EnableAuthorizationServer
    protected static class OAuth2Config extends AuthorizationServerConfigurerAdapter {

        @Inject
        private AuthenticationManager authenticationManager;

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

        @Override
        public void configure(ClientDetailsServiceConfigurer clients) throws Exception {

            clients.inMemory().withClient(CLIENT_ID)
            .authorizedGrantTypes("client_credentials", "password", "refresh_token")
            .authorities("ROLE_CLIENT", "ROLE_TRUSTED_CLIENT")
            .scopes("read", "write", "trust")
            .secret(CLIENT_SECRET);
        }

        @Override
        public void configure(AuthorizationServerSecurityConfigurer oauthServer) throws Exception {
            super.configure(oauthServer);
        }
    }

    @Configuration
    @EnableResourceServer
    protected static class ResourceConfig extends ResourceServerConfigurerAdapter {

        @Override 
        public void configure(HttpSecurity http) throws Exception {

            http.requestMatchers().antMatchers("/*", "/admin/beans").and().authorizeRequests().anyRequest()
                .access("#oauth2.hasScope('read')"); 
        }

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

    @Configuration
    @EnableWebSecurity
    protected static class WebConfigurer extends WebSecurityConfigurerAdapter {

        @Override
        protected void configure(HttpSecurity http) throws Exception {
            super.configure(http);
        }

        @Override
        public void configure(WebSecurity webSecurity) throws Exception {
            webSecurity.ignoring()
                    // All of Spring Security will ignore the requests
                    .antMatchers("/accessibleservices/**")
        }
    }

}

最佳答案
RFC 6749(OAuth 2.0授权框架)中的“
4.3.2. Access Token Request”如下所述.

The client makes a request to the token endpoint by adding the
following parameters using the “application/x-www-form-urlencoded”
format per Appendix B with a character encoding of UTF-8 in the HTTP
request entity-body:

所以,-H“Content-Type：application / json”是错误的.另外,你的curl命令行是错误的.使用-d选项指定POST的表单参数.

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