Spring 返回 401 而不是 200 状态

Question

作为学习 Spring 的一部分，我编写了一个应用程序，但是当我测试身份验证时，我收到 401 状态而不是 200。我正在寻找错误的原因，在我看来 Authentication authentication = authenticationManager.authenticate(new UsernamePasswordAuthenticationToken(email, password)); 行返回 null .但是，我不知道如何解决这个问题。

@Component
public class AuthenticationServiceUsernamePassword {
    private static final Logger LOGGER = LoggerFactory.getLogger(AuthenticationServiceUsernamePassword.class);
    @Autowired
    @Qualifier("customAuthenticationManager")
    private AuthenticationManager authenticationManager;
    @Autowired
    private TokenManager tokenManager;

    public SignedJWT authenticate(final String email, final String password){
        try {
            Authentication authentication = authenticationManager
                .authenticate(new UsernamePasswordAuthenticationToken(email, password));        
            SecurityContextHolder.getContext()
                .setAuthentication(authentication);

            if (authentication.getPrincipal() != null) {
                return tokenManager.createNewToken((PrincipalUser) authentication.getPrincipal());
            }
        } catch (AuthenticationException authException) {
            LOGGER.debug("Authentication failed for user:\"" + email + ".\" Reason " + authException.getClass());
        }

        return null;
    }
}

控制器

@Controller
public class AuthController {
    @Value("${jwt.result}")
    private String defaultTokenResponse;
    @Autowired
    private AuthenticationServiceUsernamePassword authUserPassword;

    @RequestMapping(value = "/authentication", method = RequestMethod.POST, produces = MediaType.APPLICATION_JSON_VALUE)
    public ResponseEntity<String> authenticate(String email, String password, HttpServletRequest request,
                                           HttpServletResponse response){
        if (email != null && password != null){
            try {
                SignedJWT token = authUserPassword.authenticate(email, password);

                if (token != null){
                    return new ResponseEntity<String>(String.format(defaultTokenResponse, token.serialize()),
                        HttpStatus.OK);
                } else {
                    return new ResponseEntity<String>(HttpStatus.UNAUTHORIZED);
                }
            } catch (BadCredentialsException badCredentials) {
                return new ResponseEntity<String>(HttpStatus.UNAUTHORIZED);
            }
        } else {
            return new ResponseEntity<String>(HttpStatus.UNAUTHORIZED);
        }
    }
}

测试类：

@RunWith(SpringJUnit4ClassRunner.class)
@SpringBootTest(classes = Application.class)
@WebAppConfiguration
public class ConnectControllerTest {
    protected MockMvc mockMvc;
    @Autowired
    private WebApplicationContext context;
    @Autowired
    private Filter springSecurityFilterChain;

    @Before
    public void setup() {
        mockMvc = MockMvcBuilders.webAppContextSetup(context)
            .addFilters(springSecurityFilterChain)
            .defaultRequest(get("/"))
            .build();
    }

    @Test
    public void shouldTestAuthentication() throws Exception {
        String result = mockMvc.perform(post("/authentication")
            .param("email", "user@test.pl").param("password", "password"))
            .andExpect(status().isOk())
            .andExpect(jsonPath("$.token").exists())
            .andReturn().getResponse().getContentAsString();
    }
}

如果有人对此处的其余代码感兴趣，请点击链接：repository

Answer 1

您已向单元测试用例添加参数。不是有效载荷。如果添加有效负载，则必须使用 @RequestBody。在这里你必须使用@RequestParam。在控制器中使用以下代码：

public ResponseEntity<String> authenticate(@RequestParam String email, @RequestParam String password, HttpServletRequest request,HttpServletResponse response){
 ..do stuff
}

这将工作..！

Answer 2

好的。第一要务

Email 和Password 传递正确

问题来了

public SignedJWT authenticate(final String email, final String password){
        try {
            System.out.println("test => "+email+" : "+password);
            Authentication authentication = authenticationManager
                    .authenticate(new UsernamePasswordAuthenticationToken(email, password));
            SecurityContextHolder.getContext().setAuthentication(authentication);

            if (authentication.getPrincipal() != null) {
                return tokenManager.createNewToken((PrincipalUser) authentication.getPrincipal());
            }
        } catch (AuthenticationException authException) {
            authException.printStackTrace();
            LOGGER.debug("Authentication failed for user:\"" + email + ".\" Reason " + authException.getClass());
        }
        System.out.println("return nulll");
        return null;
    }

如果你运行你的测试用例，它会抛出以下错误

org.springframework.security.authentication.BadCredentialsException: Bad credentials
    at org.springframework.security.authentication.dao.DaoAuthenticationProvider.additionalAuthenticationChecks(DaoAuthenticationProvider.java:98)
    at org.springframework.security.authentication.dao.AbstractUserDetailsAuthenticationProvider.authenticate(AbstractUserDetailsAuthenticationProvider.java:166)
    at org.springframework.security.authentication.ProviderManager.authenticate(ProviderManager.java:174)
    at org.springframework.security.authentication.ProviderManager.authenticate(ProviderManager.java:199)
    at org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter$AuthenticationManagerDelegator.authenticate(WebSecurityConfigurerAdapter.java:504)
    at com.github.springjwt.security.jwt.service.AuthenticationServiceUsernamePassword.authenticate(AuthenticationServiceUsernamePassword.java:30)
    at com.github.springjwt.web.api.controller.AuthController.authenticate(AuthController.java:31)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImp

这意味着您的测试用例的 username 和 password 与 UserRepository 类用户详细信息不匹配

在您的UserRepository 班级中 您需要设置正确的散列密码及其设置为 null 的盐值。

当您调用authenticate.authenticate 时，它会在内部获取密码和哈希并将其与传递的值匹配。

如果值不匹配，则会引发错误凭据错误

PS：我是在本地运行你的代码后得出这个结论

Answer 3

您的代码大部分是正确的，但在您的控制器定义中出现了错误：

public ResponseEntity<String> authenticate(String email, String password, HttpServletRequest request,
                                           HttpServletResponse response){

默认情况下，Spring 不知道如何检索电子邮件和密码变量。您需要使用 @RequestBody 注释对这些进行注释，例如：

public ResponseEntity<String> authenticate(@RequestBody String email, @RequestBody String password, HttpServletRequest request,
                                           HttpServletResponse response){

但是，如果您的整个控制器将用作 API，您还可以使用 @RestController 注释您的控制器，这告诉 spring 对每个参数使用 @RequestBody，并且每个方法都应该使用 @ResponseBody 注释，这将告诉 spring应该将返回值转换为 JSON（这对 API 来说很方便）。

参考资料：

Spring’s RequestBody and ResponseBody Annotations