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java - 如何使用 JAXB 在 Java 中解析这个 XML?

转载 作者:IT老高 更新时间:2023-10-28 20:49:47 25 4
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我有以下 XML,没有 XSD 或模式,我想使用 JAXB 解析为 java 对象,因为我听说它比 SAX 更好。有没有办法通过注释或更好的方法来实现这一点?它是否使我只需要一个 FosterHome 类(class)?我无法找到如何做到这一点,任何帮助都非常感谢。

我最初想开设 FosterHome、Family 和 Child 类(class)。使用 JAXB,是否不再需要 3 个类?由于 ChildID 出现在两个不同的区域,我不确定如何处理。

<?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>

最佳答案

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:

福斯特之家

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;

}

家庭

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;

}

演示

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);
}

}

输入/输出

<?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>

更多信息


更新

Is there I easy way I can iterate/print out all the ChildID in the Family class?

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);
}
}

关于java - 如何使用 JAXB 在 Java 中解析这个 XML?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8186896/

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