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java - 使用 AuthenticationFailureHandler 在 Spring Security 中自定义身份验证失败响应

转载 作者:塔克拉玛干 更新时间:2023-11-03 04:48:32 31 4
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目前,每当用户身份验证失败时,spring security 都会响应:

{"error": "invalid_grant","error_description": "Bad credentials"}

我想用如下响应代码增强此响应:

{"responsecode": "XYZ","error": "invalid_grant","error_description": "Bad credentials"}

经过一番探索,我似乎需要做的是实现一个 AuthenticationFailureHandler,我已经开始这样做了。但是,每当我提交无效的登录凭据时,似乎永远不会达到 onAuthenticationFailure 方法。我已经逐步完成代码,并将日志记录放在 onAuthenticationFailure 方法中以确认它没有被访问。

我的失败处理程序是:

@Component
public class SSOAuthenticationFailureHandler extends SimpleUrlAuthenticationFailureHandler{

@Override
public void onAuthenticationFailure(HttpServletRequest request, HttpServletResponse response,
AuthenticationException exception) throws IOException, ServletException {
super.onAuthenticationFailure(request, response, exception);
response.addHeader("responsecode", "XYZ");
}
}

我的 WebSecurityConfigurerAdapter 包含:

@Configuration
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {

@Autowired SSOAuthenticationFailureHandler authenticationFailureHandler;

@Override
protected void configure(HttpSecurity http) throws Exception {
http.csrf().disable();
http.formLogin().failureHandler(authenticationFailureHandler);
}

@Autowired
public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
auth.userDetailsService(service).passwordEncoder(passwordEncoder());
auth.authenticationEventPublisher(defaultAuthenticationEventPublisher());
}

@Bean
public DefaultAuthenticationEventPublisher defaultAuthenticationEventPublisher(){
return new DefaultAuthenticationEventPublisher();
}

@Override
@Bean
public AuthenticationManager authenticationManagerBean() throws Exception {
return super.authenticationManagerBean();
}

@Bean
public SSOAuthenticationFailureHandler authenticationHandlerBean() {
return new SSOAuthenticationFailureHandler();
}

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

我的问题是:

  1. 这是实现我想要的结果的正确方法吗? (自定义spring security认证响应)
  2. 如果是这样,我是不是在尝试设置我的身份验证失败处理程序时做错了什么(因为错误的登录似乎没有到达 onAuthenticationFailure 方法?

谢谢!

最佳答案

您可以通过在配置方法中对 HttpSecurity 对象调用 .exceptionHandling() 来为 Spring Security 添加异常处理。如果您只想处理错误的凭证,您可以忽略 .accessDeniedHandler(accessDeniedHandler())。

拒绝访问处理程序处理您已在方法级别保护应用程序的情况,例如使用@PreAuthorized、@PostAuthorized 和@Secured。

您的安全配置示例可能是这样的

SecurityConfig.java
/*
The following two are the classes we're going to create later on.
You can autowire them into your Security Configuration class.
*/
@Autowired
private CustomAuthenticationEntryPoint unauthorizedHandler;

@Autowired
private CustomAccessDeniedHandler accessDeniedHandler;

/*
Adds exception handling to you HttpSecurity config object.
*/
@Override
protected void configure(HttpSecurity http) throws Exception {
http.csrf()
.disable()
.exceptionHandling()
.authencationEntryPoint(unauthorizedHandler) // handles bad credentials
.accessDeniedHandler(accessDeniedHandler); // You're using the autowired members above.


http.formLogin().failureHandler(authenticationFailureHandler);
}

/*
This will be used to create the json we'll send back to the client from
the CustomAuthenticationEntryPoint class.
*/
@Bean
public Jackson2JsonObjectMapper jackson2JsonObjectMapper() {
ObjectMapper mapper = new ObjectMapper();
mapper.configure(JsonParser.Feature.ALLOW_COMMENTS, true);
return new Jackson2JsonObjectMapper(mapper);
}

CustomAuthenticationEntryPoint.java

您可以在其自己的单独文件中创建它。这是入口点处理无效凭据。在方法内部,我们必须创建自己的 JSON 并将其写入 HttpServletResponse 对象。出色地使用我们在安全配置中创建的 Jackson 对象映射器 bean。

 @Component
public class CustomAuthenticationEntryPoint implements AuthenticationEntryPoint, Serializable {

private static final long serialVersionUID = -8970718410437077606L;

@Autowired // the Jackson object mapper bean we created in the config
private Jackson2JsonObjectMapper jackson2JsonObjectMapper;

@Override
public void commence(HttpServletRequest request,
HttpServletResponse response,
AuthenticationException e) throws IOException {

/*
This is a pojo you can create to hold the repsonse code, error, and description.
You can create a POJO to hold whatever information you want to send back.
*/
CustomError error = new CustomError(HttpStatus.FORBIDDEN, error, description);

/*
Here we're going to creat a json strong from the CustomError object we just created.
We set the media type, encoding, and then get the write from the response object and write
our json string to the response.
*/
try {
String json = jackson2JsonObjectMapper.toJson(error);
response.setStatus(HttpServletResponse.SC_UNAUTHORIZED);
response.setContentType(MediaType.APPLICATION_JSON_VALUE);
response.setCharacterEncoding(StandardCharsets.UTF_8.toString());
response.getWriter().write(json);
} catch (Exception e1) {
e1.printStackTrace();
}

}
}

CustomAccessDeniedHandler.java

这会处理授权错误,例如尝试在没有适当的特权。您可以按照我们上面使用错误凭据异常执行的相同方式来实现它。

@Component
public class CustomAccessDeniedHandler implements AccessDeniedHandler {

@Override
public void handle(HttpServletRequest request, HttpServletResponse response,
AccessDeniedException e) throws IOException, ServletException {

// You can create your own repsonse here to handle method level access denied reponses..
// Follow similar method to the bad credentials handler above.
}

}

希望这对您有所帮助。

关于java - 使用 AuthenticationFailureHandler 在 Spring Security 中自定义身份验证失败响应,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42839910/

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