Question

我使用JAXB来解析XML。如何通过JAXB xml解析获取特定元素（即子节点），而不将该元素解析为节点。

<?xml version="1.0" encoding="UTF-8"?>
        <Header>
            <From>
            <Credential
                        domain="NetworkId"><Identity>ANXXXNNN</Identity>
            </Credential>
            </From>

            <To>
            <Credential
                        domain="NetworkId"><Identity>ANNNXXXXXT</Identity>
            </Credential>
            </To>

            <To>
            <Credential
                        domain="NetworkId"><Identity>BNNXXXT</Identity>
            </Credential>
            </To>
        </Header>

我已经完成了这样的解组，它运行正常。对于性能，我不希望元素作为节点。还有其他方法吗？

Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setNamespaceAware(true);
DocumentBuilder db = dbf.newDocumentBuilder();
    Document doc;
    doc = db.parse(file);
    NodeList node = (NodeList)doc.getElementsByTagName("TO");

   JAXBElement<ToType> element =  jaxbUnmarshaller.unmarshal(node.item(0),ToType.class);

对象模型就像

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "ToType", propOrder = {
    "credential"
})
public class ToType {

    @XmlElement(name = "Credential", required = true)
    protected CredentialType credential;


    public CredentialType getCredential() {
        return credential;
    }

    public void setCredential(CredentialType value) {
        this.credential = value;
    }

}

Answer 1

可以使用StAX (JSR-173)解析器（包含在从Java SE 6开始的JDK / JRE中）。然后，您可以将XMLStreamReader推进到子节点并从那里解组。

import javax.xml.stream.*;
import javax.xml.transform.stream.StreamSource;
import javax.xml.bind.*;

public class Demo {

    public static void main(String[] args) throws Exception {
        XMLInputFactory xif = XMLInputFactory.newFactory();
        StreamSource xml = new StreamSource("src/forum14358769/input.xml");
        XMLStreamReader xsr = xif.createXMLStreamReader(xml);

        // Advance to the "To" element.
        while(xsr.hasNext()) {
            if(xsr.isStartElement() && "To".equals(xsr.getLocalName())) {
                break;
            }
            xsr.next();
         }

        // Unmarshal from the XMLStreamReader that has been advanced
        JAXBContext jc = JAXBContext.newInstance(ToType.class);
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        ToType toType = unmarshaller.unmarshal(xsr, ToType.class).getValue();
    }

}

了解更多信息

http://blog.bdoughan.com/2012/08/handle-middle-of-xml-document-with-jaxb.html

Answer 2

您必须执行以下代码才能执行JAXB Unmarshalling：

unmarshaler.unmarshal(source);

这将返回您使用xjc命令生成的类的实例。

然后，您可以使用bean方法访问任何属性。

如何通过JAXB xml解析获取特定元素？

2 个答案: