java-多类型列表注释转换：JAXB到SimpleXML

我正在尝试将一些JAXB xjc.exe生成的类转换为Simple XML类.我不确定如何注释动态元素.例如,在模式中,我有：

<!-- Message Set Request/Response Pairs and contained requests  -->
<xsd:element name="QBXMLMsgsRq">
    <xsd:complexType>
        <xsd:choice minOccurs="0" maxOccurs="unbounded">
            <xsd:element name="HostQueryRq" type="HostQueryRqType"/>
            <xsd:element name="CompanyQueryRq" type="CompanyQueryRqType"/>
            <xsd:element name="CompanyActivityQueryRq" type="CompanyActivityQueryRqType"/>
            <!-- many more of these choices -->
        </xsd:choice>
        <xsd:attribute name="oldMessageSetID" type="STRTYPE"/>
        <!-- some other attributes -->
    </xsd:complexType>
</xsd:element>

通过xjc.exe运行时,它会为@XmlElement生成以下注释

@XmlElements({
    @XmlElement(name = "HostQueryRq", type = HostQueryRqType.class),
    @XmlElement(name = "CompanyQueryRq", type = CompanyQueryRqType.class),
    @XmlElement(name = "CompanyActivityQueryRq", type = CompanyActivityQueryRqType.class),
    //+ et al
})
protected List<Object> hostQueryRqOrCompanyQueryRqOrCompanyActivityQueryRq;

那么如何将这个JAXB结构转换为带有SimpleXML注释的类结构？

解决方法:

答案是使用ElementListUnion标识列表类型的可用选项.检查“在单个列表中收集各种类型”下的here.例：

@Root
public class Example {

   @ElementListUnion({
      @ElementList(entry="int", type=Integer.class, inline=true),
      @ElementList(entry="date", type=Date.class, inline=true),
      @ElementList(entry="text", type=String.class, inline=true)
   })
   private List<Object> list;
}

标签：jaxb,simple-framework,xsd,xml,java
来源： https://codeday.me/bug/20191031/1976679.html