Question

我使用平针织物构建了Rest，我尝试处理异常。我注册了jackson如何使用这个类对json映射的所有操作默认json处理器：

@javax.ws.rs.ApplicationPath("/")
public class MyApplication extends ResourceConfig {
   public MyApplication() {
       packages("controller");
       register(JacksonFeature.class);       
   }
}

我定义了一个异常映射器来捕获一些异常，例如＆＃34; UnauthorizedException＆＃34;和＆＃34; BadRequestException＆＃34;使用以下课程：

@Provider
public class ClientExceptionMapper implements ExceptionMapper<BaseException>
{
    @Override
    public Response toResponse(BaseException ex) 
    {
        System.out.println("qwertweryt");
        if(ex instanceof UnauthorizedException)
        {
            System.out.println("Unauthorized!");
            return (((UnauthorizedException)ex).getResponse());
        }
        else if(ex instanceof BadrequestException)
        {
            System.out.println("Badrequest!");
            return (((BadrequestException)ex).getResponse());
        }
        return null;
    }
}

其中BaseException，UnauthorizedException和BarequestException分别为：

public abstract class BaseException extends WebApplicationException
{
    private Response r;

    public BaseException(Response response)
    {
        r = response;
    }

    public Response getResponse()
    {
        return r;
    }
}

public class UnauthorizedException extends BaseException
{
    public UnauthorizedException(Response response)
    {
        super(response);
    }
}

public class BadrequestException extends BaseException
{
    public BadrequestException(Response response)
    {
        super(response);
    }
}

所以，一切正常，直到我处理jackson异常...实际上我定义了另一个异常映射器，类似于第一个异常映射器来处理jackson生成的所有异常（例如JsonMappingException）但是没有工作。我向您报告我的申请所获得的错误：

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "serverr" (Class controller.server.control.response.GetServerDetailsResponse), not marked as ignorable
at [Source: org.glassfish.jersey.message.internal.EntityInputStream@701d7463; line: 1, column: 14] (through reference chain: controller.server.control.response.GetServerDetailsResponse["serverr"])
at org.codehaus.jackson.map.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:53)
at org.codehaus.jackson.map.deser.StdDeserializationContext.unknownFieldException(StdDeserializationContext.java:267)
at org.codehaus.jackson.map.deser.std.StdDeserializer.reportUnknownProperty(StdDeserializer.java:673)
at org.codehaus.jackson.map.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:659)
at org.codehaus.jackson.map.deser.BeanDeserializer.handleUnknownProperty(BeanDeserializer.java:1365)
at org.codehaus.jackson.map.deser.BeanDeserializer._handleUnknown(BeanDeserializer.java:725)
at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:703)
at org.codehaus.jackson.map.deser.BeanDeserializer.deserialize(BeanDeserializer.java:580)
at org.codehaus.jackson.map.ObjectMapper._readValue(ObjectMapper.java:2704)
at org.codehaus.jackson.map.ObjectMapper.readValue(ObjectMapper.java:1315)
at org.codehaus.jackson.jaxrs.JacksonJsonProvider.readFrom(JacksonJsonProvider.java:419)
at org.glassfish.jersey.message.internal.ReaderInterceptorExecutor$TerminalReaderInterceptor.aroundReadFrom(ReaderInterceptorExecutor.java:181)
at org.glassfish.jersey.message.internal.ReaderInterceptorExecutor.proceed(ReaderInterceptorExecutor.java:134)
at org.glassfish.jersey.server.internal.MappableExceptionWrapperInterceptor.aroundReadFrom(MappableExceptionWrapperInterceptor.java:72)
at org.glassfish.jersey.message.internal.ReaderInterceptorExecutor.proceed(ReaderInterceptorExecutor.java:134)

我的第一个问题是：这两个类可能存在冲突，即同一个项目中没有两个异常映射器？第二个问题是：我该如何解决这个问题？我希望我很清楚，并提前感谢你的帮助！

嗨，我刚读了一篇文章，其中提出的问题与我的问题相同;我向您报告问题的链接：Jersey: Returning 400 error instead of 500 when given invalid request body。首先，我尝试在帖子的第一个答案中编写但与之前结果相同的代码：当我故意在请求正文中插入错误时，我总是获得500错误代码而不是400。我还尝试将 @JsonIgnoreProperties（ignoreUnknown = true）放在处理拦截请求体的类之上，但对我来说并不起作用。

Answer 1

我认为问题在于您没有将ExceptionMapper注册到应用程序运行时加载类提供程序中。尝试使用以下代码将mapper类添加到它：

   @javax.ws.rs.ApplicationPath("/")
    public class MyApplication extends ResourceConfig {
       public MyApplication() {
           packages("controller");
           register(JacksonFeature.class);
           register(ClientExceptionMapper.class);          
       }
    }

希望它可以提供帮助：）

处理杰克逊的例外

1 个答案: