Question

    @POST
    @Path("/{companyid}/location")
    public Response addLocationCommand(LocationCommandDto addLocationCommand, @PathParam("companyid") String companyID) throws Throwable{
        addLocationCommand.setCompanyId(companyID);
        Response response = processApiRequest(addLocationCommand, LocationCommandDtoToAddLocationCommandConverter.class,
                AddLocationCommandProcessor.class);
        return response;

现在我以json格式发送请求

{
    "locationNam1e": "ytjtjtyj",
    "locationType":"new",
    "timeZone":"",
    "mondayWorkingType":"halfday"
}

现在我的LocationCommandDto类包含以下属性。

public class LocationCommandDto implements CommandDTO {

        private String companyId;
        private String locationName;
        private String locationType;
        private String timeZone;
}

现在在json我有属性为“locationNam1e”，这是不正确的，在我的LocationCommandDto中它是locationName，所以如果请求body的参数与我们正在转换的class属性不匹配，如何映射它们并抛出异常？

Answer 1

应在ObjectMapper

中配置此类内容

ObjectMapper objectMapper = getObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, true);

Answer 2

如果你正在使用Spring Boot，你可以在application.properties文件中设置以下属性：

spring.jackson.deserialization.fail-on-unknown-properties=true

参数转换为REST中的Object

2 个答案: