创建通用的抽象类JAXB [英] Creating abstract generic jaxb class

问题描述

我有以下简单的JAXB类，它的通用类型E

I have the following simple jaxB class that takes generic type 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;
    }

}

我需要删除提及NameSearchResults @XmlElement（类型= NameSearchResults.class）制造基地实际上通用的，但如果我这样做，我得到的错误。

I need to remove the reference to NameSearchResults @XmlElement(type=NameSearchResults.class) to make the base actually generic, but if I do I get the error.

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.]

这是一个扩展它的类的实例

This is an example of a class that extends it

扩展分类

@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;
    }
}

如何让我的基类实际上是通用的？

How do i make the base class actually generic?

preferably我想我的扩展类格式工作 SearchResponse＆LT; E＆GT;扩展SearchResponseBase＆LT; E＆GT; 并使用它作为一个泛型类型太

preferably i would like my extended class to work in the format SearchResponse<E> extends SearchResponseBase<E> and use it as a generic type too.

如果我这样做保罗建议我可以得到德班：

if i do as paul suggested i can get teh class to:

@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;
    }  
}

有没有办法，我可以推NameSearchResults出这＆LT; e延伸NameSearchResults＆GT;

推荐答案

感谢@PaulBellora的帮助下，基地和扩展类都将成为抽象的巡航能力，然后一个名称执行力度，是这样的：

Thanks to @PaulBellora for the help, the base and extend class will both become abstract then haveing a Name implimentation, like this:

基本

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

    public abstract E getSearchResults();

    public abstract void setSearchResults(E mSearchResults);

}

扩展的基本

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

    public abstract SearchRequest getSearchRequest();

    public abstract void setSearchRequest(SearchRequest value);   
}

命名执行力度

@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;
    }
}

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