如何使用JAXB解析Java中的XML？

  • Z时代
  • 2024-01-10
  • 分类：问答

我有以下XML，没有XSD或架构，我想使用JAXB解析为java对象，因为我听说它比SAX更好。有没有一种方法可以通过注释或更好的方法来完成此任务？它是否使我只需要一个FosterHome类？我很难找到如何执行此操作的任何帮助，不胜感激。

我本来是想开设FosterHome，Family和Child班的。使用JAXB，不再需要3个类吗？我不确定如何处理此问题，因为ChildID在两个不同的区域显示。

<?xml version="1.0" encoding="UTF-8"?>
<FosterHome>
    <Orphanage>Happy Days Daycare</Orphanage>
    <Location>Apple Street</Location>
    <Families>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child1</ChildID>
                <ChildID>Child2</ChildID>
            </ChildList>
        </Family>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child3</ChildID>
                <ChildID>Child4</ChildID>
            </ChildList>
        </Family>
    </Families>
    <RemainingChildList>
        <ChildID>Child5</ChildID>
        <ChildID>Child6</ChildID>
    </RemainingChildList>
</FosterHome>

回答：

您可以执行以下操作。通过利用，@XmlElementWrapper您可以减少所需的类数量：

package nov18;
import java.util.List;
import javax.xml.bind.annotation.*;
@XmlRootElement(name="FosterHome")
@XmlAccessorType(XmlAccessType.FIELD)
public class FosterHome {
    @XmlElement(name="Orphanage")
    private String orphanage;
    @XmlElement(name="Location")
    private String location;
    @XmlElementWrapper(name="Families")
    @XmlElement(name="Family")
    private List<Family> families;
    @XmlElementWrapper(name="RemainingChildList")
    @XmlElement(name="ChildID")
    private List<String> remainingChildren;
}

package nov18;
import java.util.List;   
import javax.xml.bind.annotation.*;
@XmlAccessorType(XmlAccessType.FIELD)
public class Family {
    @XmlElement(name="ParentID")
    private String parentID;
    @XmlElementWrapper(name="ChildList")
    @XmlElement(name="ChildID")
    private List<String> childList;
}

package nov18;
import java.io.File;
import javax.xml.bind.*;
public class Demo {
    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(FosterHome.class);
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        FosterHome fosterHome = (FosterHome) unmarshaller.unmarshal(new File("src/nov18/input.xml"));
        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(fosterHome, System.out);
    }
}

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<FosterHome>
    <Orphanage>Happy Days Daycare</Orphanage>
    <Location>Apple Street</Location>
    <Families>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child1</ChildID>
                <ChildID>Child2</ChildID>
            </ChildList>
        </Family>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child3</ChildID>
                <ChildID>Child4</ChildID>
            </ChildList>
        </Family>
    </Families>
    <RemainingChildList>
        <ChildID>Child5</ChildID>
        <ChildID>Child6</ChildID>
    </RemainingChildList>
</FosterHome>

  • http://blog.bdoughan.com/2010/09/jaxb-collection-properties.html

有没有简单的方法可以迭代/打印出Family类中的所有ChildID？

您可以执行以下操作：

    for(Family family : fosterHome.getFamilies()) {
        System.out.println(family.getParentID());
        for(String childID : family.getChildList()) {
            System.out.println("    " + childID);
        }
    }

以上是 如何使用JAXB解析Java中的XML？ 的全部内容， 来源链接： utcz.com/qa/409647.html

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