我尝试使用Resource和Simple Object返回对象，但得到错误：
网页浏览器。http。转换器。HttpMessageConversionException：类型定义错误：[simple type，class java.io.BufferedInputStream];嵌套异常为com.fasterxml.Jackson.databind.exc。无效定义异常：未找到java.io.BufferedInputStream类的序列化器，也未发现用于创建BeanSerializer的属性（为避免异常，请禁用SerializationFeature.FAIL_ON_EMPTY_BEANS）（通过引用链：ru.fabit.visor.ui.web.http.model.response. complex.mobilecomplex.DeviceSummaryResponse[“route”]-〉org.springframework.core.io.InputStreamResource[“inputStream”]）
静置方法：

@GetMapping(DEVICE_ID_PATH)
    public ResponseEntity<DeviceSummaryResponse> getDeviceSummary(@PathVariable(name = "id") Long id) {

        log.debug("REST request to get summary for complex with id = {}", id);

        var deviceSummary = mobileComplexService.getSummary(id);

        return ResponseEntity.ok().body(deviceSummaryMapper.toResponse(deviceSummary));
    }

响应实体：

@Data
@AllArgsConstructor
@NoArgsConstructor
@Builder
public class DeviceSummaryResponse {

    @ApiModelProperty(value = "Конфигурация устройства", required = true)
    private ConfigResponse config;

    @ApiModelProperty(value = "GPX файл с маршрутом устройства", required = true)
    private Resource route;
}

如何返回此对象？