Question

我正在尝试为JAX-RS编写自定义ExceptionMapper类。我想以某种方式从JAX-RS javax.ws.rs.core.Response对象中读取原始请求的URI。

当我在调试模式下使用IntelliJ IDEA检查响应对象时，我可以在以下路径中看到此信息：响应＆gt;上下文＆gt; solveUri在哪里 context的类型为org.glassfish.jersy.client.ClientResponse。这个类有一个resolvedUri变量，它保存了我需要的信息。

我能以某种方式获得此信息吗？我如何编写getRequestUri(response)方法？

public class MyExceptionMapper implements ExceptionMapper<WebApplicationException> {

    @Override
    public Response toResponse(WebApplicationException error) {

        Response response = error.getResponse();
        ErrorResponse errorResponse = ErrorResponseBuilder
                .builder()
                .httpStatus(getDefaultStatusCodeIfNull(response))
                .errorMessage(getCustomErrorMessage(response))
                .requestedUri(getRequestedUri(response)) <--------- HOW TO READ IT?
                .build();

        return Response
                .status(errorResponse.getHttpStatus())
                .type(ExtendedMediaType.APPLICATION_JSON)
                .entity(errorResponse)
                .build();
    }
}

Answer 1

使用

@Context private HttpServletRequest servletRequest;

并使用HttpServletRequest.getRequestURI()

public class MyExceptionMapper implements 
     ExceptionMapper<WebApplicationException> {

    @Context
    private HttpServletRequest servletRequest;

    @Override
    public Response toResponse(WebApplicationException error) {

        Response response = error.getResponse();
        ErrorResponse errorResponse = ErrorResponseBuilder
                .builder()
                .httpStatus(getDefaultStatusCodeIfNull(response))
                .errorMessage(getCustomErrorMessage(response))
                .requestedUri(servletRequest.getRequestURI())
                .build();

        return Response
               .status(errorResponse.getHttpStatus())
               .type(ExtendedMediaType.APPLICATION_JSON)
               .entity(errorResponse)
               .build();
    }
}

如何从JAX-RS javax.ws.rs.core.Response获取原始请求的URI

1 个答案: