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我使用下面的代码来解组​​并通过 Xpath 查询解组的对象。我能够在解组后获取该对象，但是在通过 XPath 查询时，该值变为 null。

我需要指定任何 NameSpaceResolver 吗？

如果您正在寻找任何进一步的信息，请告诉我。

我的代码：

         JAXBContext jaxbContext = (JAXBContext) JAXBContextFactory.createContext(new Class[] {Transaction.class}, null);
         Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
         StreamSource streamSource= new StreamSource(new StringReader(transactionXML));
         transaction = unmarshaller.unmarshal(streamSource, Transaction.class).getValue();
         String displayValue = jaxbContext.getValueByXPath(transaction, xPath, null, String.class);

我的 XML：

         <Transaction xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"                          xmlns:xsd="http://www.w3.org/2001/XMLSchema" >
         <SendingCustomer firstName="test">

         </SendingCustomer>
         </Transaction>
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1 回答 1

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由于您的示例中没有命名空间，因此您无需担心利用NamespaceResolver. 您没有提供您遇到问题的 XPath，所以我只在下面的示例中选择了一个。

JAVA模型

交易

import javax.xml.bind.annotation.*;

@XmlRootElement(name="Transaction")
public class Transaction {

    @XmlElement(name="SendingCustomer")
    private Customer sendingCustomer;

}

顾客

import javax.xml.bind.annotation.XmlAttribute;

public class Customer {

    @XmlAttribute
    private String firstName;

    @XmlAttribute
    private String lastNameDecrypted;

    @XmlAttribute(name="OnWUTrustList")
    private boolean onWUTrustList;

    @XmlAttribute(name="WUTrustListType")
    private String wuTrustListType;

}

演示代码

输入.xml

<Transaction xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <SendingCustomer firstName="test" lastNameDecrypted="SMITH"
        OnWUTrustList="false" WUTrustListType="NONE">

    </SendingCustomer>
</Transaction>

演示

import javax.xml.bind.Unmarshaller;
import javax.xml.transform.stream.StreamSource;
import org.eclipse.persistence.jaxb.JAXBContext;
import org.eclipse.persistence.jaxb.JAXBContextFactory;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jaxbContext = (JAXBContext) JAXBContextFactory.createContext(new Class[] {Transaction.class}, null);
        Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
        StreamSource streamSource= new StreamSource("src/forum17687460/input.xml");
        Transaction transaction = unmarshaller.unmarshal(streamSource, Transaction.class).getValue();
        String displayValue = jaxbContext.getValueByXPath(transaction, "SendingCustomer/@firstName", null, String.class);
        System.out.println(displayValue);
    }

}

输出

test
于 2013-07-17T13:16:06.917 回答