使用 JAX-RS 处理 REST API 中的错误

Question

任务：我不想在我的堆栈跟踪中接收一般的HTTP 500 Internal Server Error 和在客户端同样可怕的堆栈跟踪，我希望看到我的自定义消息和另一个状态码（例如403），开发人员会更清楚发生了什么。并向用户添加一些关于异常的消息。

以下是我的应用程序中更改的几个类：

服务器部分：

AppException.class - 我所有的服务器响应异常（在返回给客户端之前）我想转换成这个异常。有点标准的实体类

public class AppException extends WebApplicationException {

Integer status;

/** application specific error code */
int code;

/** link documenting the exception */
String link;

/** detailed error description for developers */
String developerMessage;

public AppException(int status, int code, String message, String developerMessage, String link) {
    super(message);
    this.status = status;
    this.code = code;
    this.developerMessage = developerMessage;
    this.link = link;
}

public int getStatus() {
    return status;
}

public void setStatus(int status) {
    this.status = status;
}

public int getCode() {
    return code;
}

public void setCode(int code) {
    this.code = code;
}

public String getDeveloperMessage() {
    return developerMessage;
}

public void setDeveloperMessage(String developerMessage) {
    this.developerMessage = developerMessage;
}

public String getLink() {
    return link;
}

public void setLink(String link) {
    this.link = link;
}

public AppException() {
}

public AppException(String message) {
    super("Something went wrong on the server");
}
}

ÀppExceptionMapper.class - 将我的 AppException 映射到 JAX-RS 运行时，而不是标准异常，客户端接收 AppException。

    @Provider
public class AppExceptionMapper implements ExceptionMapper<AppException> {

    @Override
    public Response toResponse(AppException exception) {
        return Response.status(403)
                .entity("toResponse entity").type("text/plain").build();
    }


}

ApplicationService.class- 我的服务类抛出 AppException

 @Path("/applications")
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
public interface ApplicationService {


    @DELETE
    @Path("/deleteById")
    void deleteById(@NotNull Long id) throws AppException;
}

客户部分：

ErrorHandlingFilter.class- 我的 AppException 响应捕获器。在这里，我想根据状态将每个 Response 异常转换为另一个异常。

@Provider
public class ErrorHandlingFilter implements ClientResponseFilter {

    private static ObjectMapper _MAPPER = new ObjectMapper();

    @Override
    public void filter(ClientRequestContext requestContext, ClientResponseContext responseContext) throws IOException {
        if (responseContext.getStatus() != Response.Status.OK.getStatusCode()) {
            if(responseContext.hasEntity()) {
                Error error = _MAPPER.readValue(responseContext.getEntityStream(), Error.class);
                String message = error.getMessage();

                Response.Status status = Response.Status.fromStatusCode(responseContext.getStatus());
                AppException clientException;

                switch (status) {

                case INTERNAL_SERVER_ERROR:
                    clientException = new PermissionException(message);
                    break;


                case NOT_FOUND:
                    clientException = new MyNotFoundException(message);
                    break;

                default:
                    clientException =  new WhatEverException(message);
                }
                    throw clientException;
        }
    }
    }
}

PermissionException.class - 我想要转换 AppException 的异常，如果它带有 500 状态代码。

public class PermissionException extends AppException{

        public PermissionException(String message) {
    super("403 - Forbidden. You dont have enough rights to delete this Application");

}

Integer status;

/** application specific error code */
int code;

/** link documenting the exception */
String link;

/** detailed error description for developers */
String developerMessage;

public PermissionException(int status, int code, String message, String developerMessage, String link) {
    super(message);
    this.status = status;
    this.code = code;
    this.developerMessage = developerMessage;
    this.link = link;
}

public int getStatus() {
    return status;
}

public void setStatus(int status) {
    this.status = status;
}

public int getCode() {
    return code;
}

public void setCode(int code) {
    this.code = code;
}

public String getDeveloperMessage() {
    return developerMessage;
}

public void setDeveloperMessage(String developerMessage) {
    this.developerMessage = developerMessage;
}

public String getLink() {
    return link;
}

public void setLink(String link) {
    this.link = link;
}

public PermissionException() {}


}

ApplicationPresenter.class- 一段 UI 逻辑，我想要处理 ErrorHandlingFilter 抛出的 PermissionException。

@SpringPresenter
public class ApplicationPresenter implements ApplicationView.Observer {

@Resource
    private ApplicationService applicationService;

    @Resource
    private UiEnvironment uiEnvironment;

@Override
    public void deleteSelectedApplication(BeanItemGrid<Application> applicationGrid) {

        try {
applicationService.deleteById(applicationGrid.getSelectedItem().getId());
                    } catch (PermissionException e) {
                        e.printStackTrace();
                        e.getMessage();
                    } catch (AppException e2) {
                    }
}
}

如何解决我的问题？我仍然收到标准500 InternalErrorException.

几乎将整个问题再更新一次！

Answer 1

当您拥有 ExceptionMapper 时，您不会自己捕获异常，而是让框架在 HTTP 请求上调用资源方法时捕获它。

Answer 2

执行错误处理的正确方法是注册ExceptionMapper 实例，这些实例知道在发生特定（或一般）异常时应返回什么响应。

@Provider
public class PermissionExceptionHandler implements ExceptionMapper<PermissionException>{
    @Override
    public Response toResponse(PermissionException ex){
        //You can place whatever logic you need here
        return Response.status(403).entity(yourMessage).build();
    }  
}

更多详情请看我的其他回答：https://stackoverflow.com/a/23858695/2588800

Answer 3

这是Jersey example，但您可以从here 中提取所需信息。最后我只会抛出一个异常并将这个异常映射到任何想要的响应。

假设你有以下资源方法，抛出异常：

@Path("items/{itemid}/")
public Item getItem(@PathParam("itemid") String itemid) {
  Item i = getItems().get(itemid);
  if (i == null) {
    throw new CustomNotFoundException("Item, " + itemid + ", is not found");
  }

  return i;
}

创建你的异常类：

public class CustomNotFoundException extends WebApplicationException {

  /**
  * Create a HTTP 404 (Not Found) exception.
  */
  public CustomNotFoundException() {
    super(Responses.notFound().build());
  }

  /**
  * Create a HTTP 404 (Not Found) exception.
  * @param message the String that is the entity of the 404 response.
  */
  public CustomNotFoundException(String message) {
    super(Response.status(Responses.NOT_FOUND).
    entity(message).type("text/plain").build());
  }
}

现在添加您的异常映射器：

@Provider
public class EntityNotFoundMapper implements ExceptionMapper<CustomNotFoundException> {
  public Response toResponse(CustomNotFoundException  ex) {
    return Response.status(404).
      entity("Ouchhh, this item leads to following error:" + ex.getMessage()).
      type("text/plain").
      build();
  }
}

最后，您必须注册您的异常映射器，以便在您的应用程序中使用它。这是一些伪代码：

register(new EntityNotFoundMapper());
//or
register(EntityNotFoundMapper.class);

Answer 4

我在这里有不同的方法。您可以在主 java 方法中启动码头服务器时尝试此操作

public static void main(String[] args) throws UnknownHostException, JSONException, IOException, Exception {

        MyMain myMain = new MyMain();

        ServletContextHandler context = new ServletContextHandler(ServletContextHandler.SESSIONS);
        context.setContextPath("/");

        Server jettyServer = new Server(5550);
        jettyServer.setHandler(context);
        context.setErrorHandler(new ErrorHandler());
        // default error handler for resources out of "context" scope
        jettyServer.addBean(new ErrorHandler());

        ServletHolder jerseyServlet = context.addServlet(org.glassfish.jersey.servlet.ServletContainer.class, "/*");
        jerseyServlet.setInitOrder(0);

        // Tells the Jersey Servlet which REST service/class to load.
        jerseyServlet.setInitParameter("jersey.config.server.provider.classnames",
                ControllerInn.class.getCanonicalName() );

        try {
            jettyServer.start();            
            jettyServer.join();

        } catch (Exception ex) {
            Logger.getLogger(ControllerInn.class.getName()).log(Level.SEVERE, null, ex);
        } finally {
            jettyServer.destroy();
        }
    }
    /**
     * Dummy error handler that disables any error pages or jetty related messages and returns our
     * ERROR status JSON with plain HTTP status instead. All original error messages (from our code) are preserved
     * as they are not handled by this code.
     */
    static class ErrorHandler extends ErrorPageErrorHandler {
        @Override
        public void handle(String target, Request baseRequest, HttpServletRequest request, HttpServletResponse response) throws IOException {
            response.getWriter()
            .append("{\"message\":\"HTTP ERROR ")
            .append(String.valueOf(response.getStatus()))
            .append("\"}");
        }
    }

所以你可以得到这样的输出

{"message":"HTTP ERROR 500"}

您可以参考here

Answer 5

上面正确建议，理想的做法是让框架为您捕获异常，因为您已经实现了ExceptionMapper。 但是，概述您正在执行的现象的重要一点：如果您需要处理任何未捕获的异常，您需要有一个实现 ExceptionMapper 的 Exception 类，它映射到 Throwable

public class UncaughtExcep implements ExceptionMapper<Throwable>{

   @Override 
   public Response toResponse(Throwable e){

    }
}

假设您的班级 WhatEverException 满足这一点。如果没有，那么实施是一个好习惯