杰克逊没有消耗JSON根元素

本文介绍了杰克逊没有消耗JSON根元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在使用JAX-RS + Jersey来使用网络服务请求，杰克逊正在翻译JSON数据：

I'm using JAX-RS + Jersey to consume the web-service request and Jackson to translate JSON data:

@Path("/") public class JAXRSRestController { @Path("/jsonRequest") @POST @Consumes(MediaType.APPLICATION_JSON) public Response submitJsonRequest(SampleObject sampleObject, @Context HttpHeaders headers) { Ack ack = new Ack(); ack.setUniqueId(sampleObject.getId()); ack.setType(sampleObject.getName()); return Response.ok().entity(ack).build(); } }

如果请求的格式如下，不会消耗：

Here if the request is in the below format, it would not be consumed:

{ "sampleObject": { "id": "12345", "name": "somename" } }

但如果请求采用以下格式，则会被使用：

But if the request is in the below format, it will be consumed:

{ "id": "12345", "name": "somename" }

如何让控制器同时使用Json根元素？

SampleObject类：

SampleObject class:

import org.codehaus.jackson.map.annotate.JsonRootName; @XmlRootElement(name = "sampleObject") @JsonRootName(value = "sampleObject") @XmlAccessorType(XmlAccessType.FIELD) @XmlType(name = "SampleObject", propOrder = { "id", "name" }) public class SampleObject { protected String id; protected String name; public SampleObject(){} public SampleObject(String id, String name) { this.id = id; this.name = name; } public String getId() { return id; } public void setId(String id) { this.id = id; } public String getName() { return name; } public void setName(String name) { this.name = name; } }

web.xml：

<?xml version="1.0" encoding= "UTF-8"?> <web-app xmlns:xsi="www.w3/2001/XMLSchema-instance" > <display-name>Wed Application</display-name> <servlet> <servlet-name>Jersey RESTFul WebSerivce</servlet-name> <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class> <init-param> <param-name>com.sun.jersey.config.property.packages</param-name> <param-value>com.jaxrs.rest</param-value> </init-param> <init-param> <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name> <param-value>true</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>Jersey RESTFul WebSerivce</servlet-name> <url-pattern>/*</url-pattern> </servlet-mapping> </web-app>

推荐答案

我能想到两种方法。如果这在您的应用程序中很常见，我建议您在 ObjectMapper 上启用解包。如果这是一次性的情况，包装器对象不是一个糟糕的选择。

There are two approaches I can think of. If this is a common occurrence in your application, I would recommend enabling unwrapping on your ObjectMapper. If this is a one-off situation, a wrapper object is not a bad option.

@JsonRootName 仅适用于在 ObjectMapper上启用展开。您可以使用反序列化功能完成此操作。请注意，这将解包所有请求：

@JsonRootName will only apply if unwrapping is enabled on the ObjectMapper. You can accomplish this with a deserialization feature. Note that this will unwrap all requests:

public CustomObjectMapper() { super(); enable(DeserializationFeature.UNWRAP_ROOT_VALUE); }

如果您还没有自定义 ObjectMapper 已注册，然后您需要添加提供商以向Jersey注册您的自定义配置。 这个答案解释了如何做到这一点。

If you do not already have a custom ObjectMapper registered then you will need to add a provider to register your custom config with Jersey. This answer explains how do accomplish that.

如果您不想全局展开，可以创建一个简单的包装器对象并省略 @JsonRootName 注释：

If you do not want to unwrap globally, you can create a simple wrapper object and omit the @JsonRootName annotation:

public class SampleObjectWrapper { public SampleObject sampleObject; }

然后更新资源方法签名以接受包装器：

Then update your resource method signature to accept the wrapper:

public Response submitJsonRequest(SampleObjectWrapper sampleObjectWrapper, @Context HttpHeaders headers)