Question

XML Variant 1：

<root>
  <elements>
    <element />
  </elements>
</root>

XML Variant 2：

<root>
  <element />
</root>

bean结构是XML Variant 1中每个元素的类，它们如图所示相互嵌套。

理想的行为是unmarshaller为Variant 2创建与Variant 1相同的bean。这意味着，它应该创建一个Elements类，即使它在结构中不存在。

这是我用于变体1的绑定：

<xml-bindings
xmlns="http://www.eclipse.org/eclipselink/xsds/persistence/oxm"
package-name="demo">
<java-types>
    <java-type name="Root">
        <xml-root-element name="root"/>
        <java-attributes>
            <xml-element java-attribute="elements" xml-path="elements" type="demo.Elements"/>
        </java-attributes>
    </java-type>
    <java-type name="Elements">
        <java-attributes>
            <xml-element java-attribute="element" xml-path="element" type="demo.Element" container-type="java.util.List"/>
        </java-attributes>
    </java-type>
    <java-type name="Element" />
</java-types>

我尝试将xml-path =“elements”改编为xml-path =“。”并认为这可能适用于变体2，但没有成功。什么是实现我想要的最简单方法？

Answer 1

您可以为您的用例使用多个映射文件。

映射文件 - 变体1

<?xml version="1.0"?>
<xml-bindings
    xmlns="http://www.eclipse.org/eclipselink/xsds/persistence/oxm"
    package-name="demo"
    xml-accessor-type="FIELD">
    <java-types>
        <java-type name="Root">
            <xml-root-element/>
        </java-type>
    </java-types>
</xml-bindings>

映射文件 - 变体2

<?xml version="1.0"?>
<xml-bindings
    xmlns="http://www.eclipse.org/eclipselink/xsds/persistence/oxm"
    package-name="demo"
    xml-accessor-type="FIELD">
    <java-types>
        <java-type name="Root">
            <xml-root-element/>
            <java-attributes>
                <xml-element java-attribute="elements" xml-path="."/>
            </java-attributes>
        </java-type>
    </java-types>
</xml-bindings>

<强>演示

在下面的演示代码中，我们将为具有不同元数据的相同域模型创建两个不同的JAXBContext实例。

package demo;

import java.io.File;
import java.util.*;
import javax.xml.bind.*;
import org.eclipse.persistence.jaxb.JAXBContextProperties;

public class Demo {

    public static void main(String[] args) throws Exception {
        // VARIANT #1
        Map<String, Object> properties1 = new HashMap<String, Object>(1);
        properties1.put(JAXBContextProperties.OXM_METADATA_SOURCE, "demo/oxm1.xml");
        JAXBContext jc1 = JAXBContext.newInstance(new Class[] {Root.class}, properties1);
        Unmarshaller unmarshaller1 = jc1.createUnmarshaller();
        File variant1 = new File("src/demo/variant1.xml");
        Root root = (Root) unmarshaller1.unmarshal(variant1);

        // VARIANT #2
        Map<String, Object> properties2 = new HashMap<String, Object>(1);
        properties2.put(JAXBContextProperties.OXM_METADATA_SOURCE, "demo/oxm2.xml");
        JAXBContext jc2 = JAXBContext.newInstance(new Class[] {Root.class}, properties2);
        Marshaller marshaller2 = jc2.createMarshaller();
        marshaller2.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller2.marshal(root, System.out);
     }

}

EclipseLink MOXy：如何将XML结构变体与同一bean模型匹配

1 个答案: