如何为 Spring RESTful Web 服务创建 Spring 拦截器

Question

我有一些 Spring RESTful (RestControllers) 没有 web.xml 的网络服务，我正在使用 Spring 启动来启动这些服务。

我想为 Web 服务添加授权层，并希望在实际调用 Web 服务本身之前将所有 http 请求路由到一个前端控制器。（我有一个代码可以在授权层模拟会话行为，根据我从客户端发送的每个 httpRequest 生成的密钥来验证用户）。

是否有关于将所有请求路由到过滤器/前端控制器的标准 Spring 解决方案？

提前致谢，普拉尼思

编辑：添加我的代码

控制器： `

@RestController
public class UserService {
    UserDAO userDAO = new UserDAO();

    @RequestMapping(value="/login", method = RequestMethod.POST)
    @LoginRequired
    public String login(@RequestParam(value="user_name") String userName, @RequestParam(value="password") String password, HttpServletRequest request){
        return userDAO.login(userName, password);
    }
}`

拦截器：

`

public class AuthenticationInterceptor implements HandlerInterceptor  {
    @Override
    public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler)
        throws Exception {
        System.out.println("In Interceptor");
        //return super.preHandle(request, response, handler);
        return true;
    }
    @Override
    public void postHandle( HttpServletRequest request, HttpServletResponse response,
            Object handler, ModelAndView modelAndView) throws Exception {
        System.out.println("---method executed---");
    }
    @Override
    public void afterCompletion(HttpServletRequest request, HttpServletResponse response,
            Object handler, Exception ex) throws Exception {
        System.out.println("---Request Completed---");
    }
}

`

接口。 `

@Target({ElementType.METHOD, ElementType.TYPE})
@Retention(RetentionPolicy.RUNTIME)
public @interface LoginRequired {
}

`

Answer 1

这里是一个拦截器的例子：

public class AuthenticationInterceptor implements HandlerInterceptor  {
    @Override
    public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler)
        throws Exception {
         HandlerMethod handlerMethod = (HandlerMethod) handler;
        LoginRequired loginRequired = handlerMethod.getMethod().getAnnotation(LoginRequired.class);
        if (loginRequired == null) {
            return true;
        }

        String token = httpServletRequest.getParameter("token");

        if (StringUtils.isBlank(token)) {
            throw new MissingParameterException();
        }

        authenticationService.checkToken(token);

        return super.preHandle(httpServletRequest, httpServletResponse, handler);
    }
    @Override
    public void postHandle( HttpServletRequest request, HttpServletResponse response,
            Object handler, ModelAndView modelAndView) throws Exception {
        System.out.println("---method executed---");
    }
    @Override
    public void afterCompletion(HttpServletRequest request, HttpServletResponse response,
            Object handler, Exception ex) throws Exception {
        System.out.println("---Request Completed---");
    }

我们可以创建注释：

 @Target({ElementType.METHOD, ElementType.TYPE})
        @Retention(RetentionPolicy.RUNTIME)
        public @interface LoginRequired {
        }

然后在控制器上，我们有这个注释：

@RequestMapping(value = "/protected/controller")
@LoginRequired
public ResponseEntity<BaseResponse> controller() {
   ...
}

这只是 template/example 给你一个想法。希望对您有所帮助。

Answer 2

你应该添加这个来注册你的拦截器

@Configuration
public class MyConfiguration extends WebMvcConfigurerAdapter {

    @Bean
    AuthenticationInterceptor getAuthenticationInterceptor() {
        return new AuthenticationInterceptor();
    }

    @Override
    public void addInterceptors (InterceptorRegistry registry) {
        registry.addInterceptor(getAuthenticationInterceptor());

    }
}

Answer 3

可以采取以下步骤来实现带有 Spring 的拦截器：

实现拦截器 class 扩展 HandlerInterceptorAdapter class。以下是代码的样子：

public class LoginInterceptor extends HandlerInterceptorAdapter {

    @Override
    public void afterCompletion(HttpServletRequest request, HttpServletResponse response, Object handler, Exception exception)
    throws Exception {
    // TODO Auto-generated method stub

    }

    @Override
    public void postHandle(HttpServletRequest request, HttpServletResponse response, Object handler, ModelAndView modelAndView)
    throws Exception {
    // TODO Auto-generated method stub

    }

    @Override
    public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler) throws Exception {

        HandlerMethod handlerMethod = (HandlerMethod) handler;

        String emailAddress = request.getParameter("emailaddress");
        String password = request.getParameter("password");

        if(StringUtils.isEmpty(emailAddress) || StringUtils.containsWhitespace(emailAddress) ||
        StringUtils.isEmpty(password) || StringUtils.containsWhitespace(password)) {
            throw new Exception("Invalid User Id or Password. Please try again.");
        }

        return true;
    }


}

实施 AppConfig class 或在现有配置之一中添加 addInterceptors class。请注意使用 LoginInterceptor 实例指定的路径模式

@Configuration  
public class AppConfig extends WebMvcConfigurerAdapter  {  

    @Override
    public void addInterceptors(InterceptorRegistry registry) {
       registry.addInterceptor(new LoginInterceptor()).addPathPatterns("/account/login");
    }
}

实现如下控制器方法：

@Controller
@RequestMapping("/account/login")
public class LoginController {

    @RequestMapping(method = RequestMethod.GET)
    public String login() {
        return "login";
    }
}

Answer 4

这种事情有一个默认的解决方案。 spring 安全。你只需要实现类似的东西：

@Configuration
@Order(SecurityProperties.ACCESS_OVERRIDE_ORDER)
class SecurityConfig extends WebSecurityConfigurerAdapter {

    @Autowired
    private UserDetailsService userDetailsService;

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http.authorizeRequests()
                .formLogin()
                .loginPage("/login")
                .failureUrl("/login?error")
                .usernameParameter("email")
                .permitAll()
                .and()
                .logout()
                .logoutUrl("/logout")
                .logoutSuccessUrl("/")
                .permitAll();
    }

    @Override
    public void configure(AuthenticationManagerBuilder auth) throws Exception {
        auth
                .userDetailsService(userDetailsService)
                .passwordEncoder(new BCryptPasswordEncoder());
    }
}

它的依赖是：

<dependency>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-starter-security</artifactId>
</dependency>

Answer 5

Spring 5 之后：实现应该是这样的：我们应该有一个实现

的 class

HandlerInterceptor 

    public class CustomInterceptor implements HandlerInterceptorr{
}

然后我们可以通过实现WebMvcConfigurer的class注册这个拦截器并覆盖方法 addInterceptors

public class ServiceInterceptorAppConfig implements WebMvcConfigurer {
  @Autowired
  CustomInterceptor customInterceptor;

  @Override
  public void addInterceptors(InterceptorRegistry registry) {
    registry.addInterceptor(customInterceptor);
  }
}

Answer 6

如果您正在寻找 spring 启动应用程序的简单答案..

public class HttpInterceptor implements HandlerInterceptor {
    public boolean preHandle(HttpServletRequest request, HttpServletResponse response, Object handler) throws Exception {
        // do something here. 
        return true;
    }
}

和

@Configuration
public class AppConfig implements WebMvcConfigurer {

    @Override
    public void addInterceptors(InterceptorRegistry registry) {
        registry.addInterceptor(new HttpInterceptor());
       // registry.addInterceptor(new HttpInterceptor()).addPathPatterns("/account/login"); you can add specific end point as well.
    }
}

如何为 Spring RESTful Web 服务创建 Spring 拦截器

How to create a Spring Interceptor for Spring RESTful web services

spring-mvc

spring-restcontroller

spring-rest