java - 解码简单 xml

Question

我有简单的架构:

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified"
           xmlns:xs="http://www.w3.org/2001/XMLSchema">
    <xs:element name="error"  type="xs:string">
    </xs:element>
</xs:schema>

我使用 JAXB 从 XML 架构生成了 Java 代码。我只有一门课:

@XmlRegistry
public class ObjectFactory {

    private final static QName _Error_QNAME = new QName("", "error");

    /**
     * Create a new ObjectFactory that can be used to create new instances of schema derived classes for package: error
     * 
     */
    public ObjectFactory() {
    }

    /**
     * Create an instance of {@link JAXBElement }{@code <}{@link String }{@code >}}
     * 
     */
    @XmlElementDecl(namespace = "", name = "error")
    public JAXBElement<String> createError(String value) {
        return new JAXBElement<String>(_Error_QNAME, String.class, null, value);
    }

}

我通常使用这段代码来解析XML:

 JAXBContext context = JAXBContext.newInstance(RootGenerateClass.class);
 Unmarshaller unmarshaller = context.createUnmarshaller();
 RootGenerateClass response = (RootGenerateClass) unmarshaller.unmarshal(streamWrapper.getStream());

在这种情况下我应该做什么(我没有任何 rootGenerateClass)？我试试这个:

JAXBContext context = JAXBContext.newInstance(String.class);
Unmarshaller unmarshaller = context.createUnmarshaller();
String response = (String) unmarshaller.unmarshal(streamWrapper.getStream());

当然不行((

Answer 1

假设您的 ObjectFactory 位于 com.example 包中，您应该能够执行此操作

JAXBContext context = JAXBContext.newInstance("com.example");
Unmarshaller unmarshaller = context.createUnmarshaller();
JAXBElement<String> responseElt = (JAXBElement<String>) unmarshaller.unmarshal(streamWrapper.getStream());
String response = responseElt.getValue();

当您为 JAXBContext.newInstance 提供包名称时，它将在该包中查找 ObjectFactory 类。