相关疑难解决方法(0)

我正在使用Jersey为服务器组件创建REST Web服务.

我想在列表中序列化的JAXB注释对象如下所示:

@XmlRootElement(name = "distribution")
@XmlType(name = "tDistribution", propOrder = {
    "id", "name"
})
public class XMLDistribution {
    private String id;
    private String name;
    // no-args constructor, getters, setters, etc
}

我有一个REST资源来检索一个看起来像这样的发行版:

@Path("/distribution/{id: [1-9][0-9]*}")
public class RESTDistribution {
    @GET
    @Produces("application/json")
    public XMLDistribution retrieve(@PathParam("id") String id) {
        return retrieveDistribution(Long.parseLong(id));
    }
    // business logic (retrieveDistribution(long))
}

我还有一个REST资源来检索所有发行版的列表,如下所示:

@Path("/distributions")
public class RESTDistributions {
    @GET
    @Produces("application/json")
    public List<XMLDistribution> retrieveAll() {
        return retrieveDistributions();
    }
    // business logic (retrieveDistributions())
}

我使用ContextResolver来自定义JAXB序列化,当前配置如下:

@Provider
@Produces("application/json")
public class …

我正在开发一个使用Jersey将对象转换为JSON的项目.我希望能够写出嵌套列表,如下所示:

{"data":[["one", "two", "three"], ["a", "b", "c"]]}

我想要转换的对象首先将数据表示为<LinkedList <LinkedList <String >>>,我认为Jersey会做正确的事情.以上输出为空值列表:

{"data":[null, null]}

在阅读了需要包装的嵌套对象之后,我尝试了以下方法:

@XmlRootElement(name = "foo")
@XmlType(propOrder = {"data"})
public class Foo
{
    private Collection<FooData> data = new LinkedList<FooData>();

    @XmlElement(name = "data")
    public Collection<FooData> getData()
    {
        return data;
    }

    public void addData(Collection data)
    {
        FooData d = new FooData();
        for(Object o: data)
        {
            d.getData().add(o == null ? (String)o : o.toString());
        }
        this.data.add(d);
    }

    @XmlRootElement(name = "FooData")
    public static class FooData
    {
        private Collection<String> data = new LinkedList<String>();

        @XmlElement
        public …