创建抽象的通用 jaxb 类 [英] Creating abstract generic jaxb class

问题描述

我有以下简单的 jaxB 类，它采用泛型 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;
    }

}

我需要删除对 NameSearchResults @XmlElement(type=NameSearchResults.class) 的引用以使基础真正通用，但如果我这样做，我会收到错误.

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例子

This is an example of a class that extends it

扩展班

@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;
    }
}

如何使基类真正通用?

最好我希望我的扩展类以 SearchResponse<E> 格式工作.扩展 SearchResponseBase 并将其用作泛型类型.

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.

如果我按照保罗的建议去做，我可以去上课:

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;
    }  
}

有什么方法可以将 NameSearchResults 从这个 <E extends NameSearchResults> 中推出?

is there a way i can push the NameSearchResults out of this <E extends NameSearchResults>?

推荐答案

感谢@PaulBellora 的帮助，基类和扩展类都将变为抽象类，然后具有名称实现，如下所示:

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

基础

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

    public abstract E getSearchResults();

    public abstract void setSearchResults(E mSearchResults);

}

扩展基础

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

    public abstract SearchRequest getSearchRequest();

    public abstract void setSearchRequest(SearchRequest value);   
}

名称实现

@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;
    }
}

这篇关于创建抽象的通用 jaxb 类的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！