如何在JAX-RS上返回没有父标记的JSON?

问题描述 投票:1回答:1

我正在尝试为Liferay创建一个REST API模块,但是当我尝试从我的webservices生成JSON响应时,我遇到了问题。

我想生成一个这样的简单JSON:

{
    "status": "ok",
    "message": "News not found for ID: 5"
}

但相反,这就是我得到的:

{
    "parentResponse": {
        "status": "ok",
        "message": "News not found for ID: 5"
    }
}

这是我的POJO课程:

@XmlRootElement
public class ParentResponse {

    public String status, message;
    public Object item;

    public ParentResponse() {

    }

    public ParentResponse(String status, String message, Object item) {
        this.status = status;
        this.message = message;
        this.item = item;
    }   
}

我的webservice返回json:

// return a single news based on supplied ID
    @GET
    @Path("{id}")
    @Produces("application/json")
    public Response getNewsById(@PathParam("id") String id) {
        ResponseBuilder builder;
        try {
            News news = findById(new Long(id));
            if (news != null) {
                builder = Response.ok(news);
            } 
            else { // This is my POJO class returned as a JSON
                ParentResponse parentResponse = new ParentResponse("ok", "News not found for ID: " + id, null);

                builder = Response.status(Response.Status.NOT_FOUND).entity(parentResponse);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
        return builder.build();
    }

那么,如何在没有root标签的情况下获取JSON?我尝试在@XmlRootElement注释旁边添加(name =“”),但这不起作用。

java json jax-rs liferay
1个回答
0
投票

我建议你使用jackson提供程序来编组/解组json输入/输出(就像我在你的评论中所做的那样):

compile "com.fasterxml.jackson.jaxrs:jackson-jaxrs-json-provider:2.9.0" 
compile "com.fasterxml.jackson.core:jackson-annotations:2.9.0"

您的模型应如下所示:

@JsonInclude(JsonInclude.Include.NON_EMPTY) 
public class ParentResponse { 

  @JsonProperty("message") 
  public String message; 

  @JsonProperty("status") 
  public String status; 

//... 

}

如果你使用Liferay 7,我建议通过创建一个调用你的服务端点(Response getNewsById(@PathParam("id") String id))的服务来利用OSGi。当您的OSGi服务启动时,它应该设置jackson提供程序并设置Web服务端点URL ...以及其他任何内容来初始化您的服务。


@Component(immediate = false)
public class WonderfulOsgiService{

    private MyWebService myWebService;

    public Response getNews( String id) {
        return myWebService.getNewsById(id);
    }


    @Activate
    @Modified
    private void activate(Map<String, Object> config) {
        // Creates jackson provider for marshalling/unmarshalling json
        ObjectMapper myMapper = new ObjectMapper();
        myMapper.enable(DeserializationFeature.READ_UNKNOWN_ENUM_VALUES_USING_DEFAULT_VALUE);
        JacksonJsonProvider provider = new JacksonJsonProvider(myMapper);
        provider.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
        List<Object> providers = new ArrayList<>();
        providers.add(provider);


        //web service endpoint address
        String address = "http://my-web-service-address.com/url/"
        //MyWebService is the class containing getNewsById(id)
        myWebService = JAXRSClientFactory.create(address, MyWebService.class, providers);   

    }
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