找不到404的Spring引导控制

Question

我试图找出一种最简单的方法来控制基本Spring Boot RESTful服务的404 Not Found处理程序，例如Spring提供的示例：

https://spring.io/guides/gs/rest-service/

而不是让它返回默认的Json输出：

{
  "timestamp":1432047177086,
  "status":404,
  "error":"Not Found",
  "exception":"org.springframework.web.servlet.NoHandlerFoundException",
  "message":"No handler found for GET /aaa, ..."
}

或者有比抛出并处理NoHandlerFoundException更好的方法吗？

Answer 1

它工作得很好。

当您使用SpringBoot时，它不会显式处理（未找到404)，它使用WebMvc错误响应。如果您的Spring Boot应该处理该异常，那么您应该对Spring Boot进行一些黑客攻击。对于404异常类是NoHandlerFoundException如果要在@RestControllerAdvice类中处理该异常，则必须在应用程序类中添加@EnableWebMVC注释并设置setThrowExceptionIfNoHandlerFound（true）；在DispatcherServlet中。请参阅守则

@SpringBootApplication
@EnableWebMvc
public class Application {  
    @Autowired
    private DispatcherServlet servlet;

    public static void main(String[] args) throws FileNotFoundException, IOException {
        SpringApplication.run(Application.class, args);
    }

    @Bean
    public CommandLineRunner getCommandLineRunner(ApplicationContext context) {
        servlet.setThrowExceptionIfNoHandlerFound(true);
        return args -> {};
    }
}

在此之后，您可以在@RestControllerAdvice类中处理NoHandlerException

@RestControllerAdvice
public class AppException {

    @ExceptionHandler(value={NoHandlerFoundException.class})
    @ResponseStatus(code=HttpStatus.BAD_REQUEST)
    public ApiError badRequest(Exception e, HttpServletRequest request, HttpServletResponse response) {
        e.printStackTrace();
        return new ApiError(400, HttpStatus.BAD_REQUEST.getReasonPhrase());
    }
}   

public class ApiError {
    private int code;
    private String message;
    public ApiError(int code, String message) {
        this.code = code;
        this.message = message;
    }
    public ApiError() {
    }   
    //getter & setter methods...
}