java - 创建抽象的通用jaxb类

我有以下简单的jaxB类，采用通用类型E

@XmlAccessorType(XmlAccessType.FIELD)
@XmlTransient
@XmlRootElement(name = "searchResponseBase")
public abstract class SearchResponseBase<E>{

    @XmlElement(type=NameSearchResults.class)
    protected E searchResults;

    public E getSearchResults()
    {
        return searchResults;
    }

    public void setSearchResults(E mSearchResults)
    {
        this.searchResults = mSearchResults;
    }

}

我需要删除对NameSearchResults @XmlElement(type=NameSearchResults.class)的引用，以使基数实际上是通用的，但是如果出现这种情况，我会收到错误消息。

错误

[com.sun.istack.internal.SAXException2: class au.test.nameSearch.NameSearchResults nor any of its super class is known to this context.
javax.xml.bind.JAXBException: class au.test.nameSearch.NameSearchResults nor any of its super class is known to this context.]

这是扩展它的类的示例

扩展类

@SuppressWarnings("javadoc")
@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(propOrder = {
    "searchRequest",
    "searchResults"
})
@XmlRootElement(name = "searchResponse")
public class SearchResponse extends SearchResponseBase<NameSearchResults> {

    @XmlElement(required = true)
    protected SearchRequest searchRequest;

    public SearchRequest getSearchRequest() {
        return searchRequest;
    }

    public void setSearchRequest(SearchRequest value) {
        this.searchRequest = value;
    }
}

我如何使基类实际上是泛型的？

最好我希望我的扩展类以SearchResponse<E> extends SearchResponseBase<E>格式工作，并将其也用作通用类型。

如果我按照保罗的建议去做，我可以上课：

@XmlRootElement(name = "searchResponse")
public class SearchResponse<E extends NameSearchResults> extends SearchResponseBase<E> {

    @XmlElement(required = true)
    protected SearchRequest searchRequest;

    protected E searchResults;

    public SearchRequest getSearchRequest() {
        return searchRequest;
    }

    public void setSearchRequest(SearchRequest value) {
        this.searchRequest = value;
    }

    @Override
    public E getSearchResults() {
        return searchResults;
    }

    @Override
    public void setSearchResults(E mSearchResults) {
        this.searchResults = mSearchResults;
    }
}

有什么方法可以将NameSearchResults推出此<E extends NameSearchResults>吗？

最佳答案

感谢@PaulBellora的帮助，基类和扩展类都将变为抽象，然后具有名称隐含形式，如下所示：

基础

@XmlRootElement(name = "searchResponseBase")
public abstract class SearchResponseBase<E>{

    public abstract E getSearchResults();

    public abstract void setSearchResults(E mSearchResults);

}

扩展基地

@XmlRootElement(name = "searchResponse")
public abstract class SearchResponse<E> extends SearchResponseBase<E>{

    public abstract SearchRequest getSearchRequest();

    public abstract void setSearchRequest(SearchRequest value);
}

名称含义

@XmlRootElement(name = "nameSearchResponse")
public class NameSearchResponse extends SearchResponse<NameSearchResults>{

    @XmlElement(required = true)
    protected SearchRequest searchRequest;

    protected NameSearchResults searchResults;

    @Override
    public NameSearchResults getSearchResults() {
        return searchResults;
    }

    @Override
    public void setSearchResults(NameSearchResults mSearchResults) {
        this.searchResults = mSearchResults;
    }

    @Override
    public SearchRequest getSearchRequest() {
        return searchRequest;
    }

    @Override
    public void setSearchRequest(SearchRequest value) {
        this.searchRequest = value;
    }
}