如何将 XML 转换为具有二进制数据内容的 Json

Question

我尝试从 xml 转换为 json。它适用于普通的文本数据内容。当 xml 标签中的数据是二进制时，我们无法将 xml 解组为 java 对象。您能帮忙分享一下我们如何将 xml 转换为 json 的二进制文件吗？

Java 代码：

public <T> String xmlToJson(Class<T> clazz, String xmlString, boolean includeRoot) throws CustomException {
    StringWriter writer = new StringWriter();
    String jsonString = "{}";

    try {

        XMLInputFactory xif = XMLInputFactory.newInstance();
        xif.setProperty(XMLInputFactory.IS_SUPPORTING_EXTERNAL_ENTITIES, false);
        xif.setProperty(XMLInputFactory.SUPPORT_DTD, true);
        XMLStreamReader xsr = xif.createXMLStreamReader(new StringReader(xmlString));
        JAXBContext jc = JAXBContext.newInstance(clazz);
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.setProperty("eclipselink.media-type", "application/json");
        marshaller.setProperty("eclipselink.json.include-root", includeRoot);
        marshaller.setProperty(MarshallerProperties.JSON_ATTRIBUTE_PREFIX, "@");
        JAXBElement<T> addressFromXML = unmarshaller.unmarshal(xsr, clazz); //ERROR at this line for binary data
        marshaller.marshal(addressFromXML, writer);
        jsonString = writer.toString();
    } catch (Exception e) {
        logger.error(e.getMessage());
        throw new CustomException(e.getMessage());
    }
    return jsonString;
}

这是错误：

Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Handler dispatch failed; nested exception is java.lang.NoClassDefFoundError: javax/mail/internet/MimeMultipart] with root cause
java.lang.NoClassDefFoundError: javax/mail/internet/MimeMultipart
    at org.eclipse.persistence.internal.oxm.XMLBinaryDataHelper.initializeDataTypes(XMLBinaryDataHelper.java:74)
    at org.eclipse.persistence.internal.oxm.XMLBinaryDataHelper.<init>(XMLBinaryDataHelper.java:54)
    at org.eclipse.persistence.internal.oxm.XMLBinaryDataHelper.getXMLBinaryDataHelper(XMLBinaryDataHelper.java:60)
    at org.eclipse.persistence.internal.oxm.XMLInlineBinaryHandler.endElement(XMLInlineBinaryHandler.java:126)
    at org.eclipse.persistence.internal.oxm.record.deferred.EndElementEvent.processEvent(EndElementEvent.java:37)
    at org.eclipse.persistence.internal.oxm.record.deferred.DeferredContentHandler.executeEvents(DeferredContentHandler.java:64)
    at org.eclipse.persistence.internal.oxm.record.deferred.BinaryMappingContentHandler.executeEvents(BinaryMappingContentHandler.java:75)
    at org.eclipse.persistence.internal.oxm.record.deferred.BinaryMappingContentHandler.processSimpleElement(BinaryMappingContentHandler.java:67)
    at org.eclipse.persistence.internal.oxm.record.deferred.DeferredContentHandler.endElement(DeferredContentHandler.java:122)
    at org.eclipse.persistence.internal.oxm.record.XMLStreamReaderReader.parseEvent(XMLStreamReaderReader.java:150)
    at org.eclipse.persistence.internal.oxm.record.XMLStreamReaderReader.parse(XMLStreamReaderReader.java:100)
    at org.eclipse.persistence.internal.oxm.record.XMLStreamReaderReader.parse(XMLStreamReaderReader.java:87)
    at org.eclipse.persistence.internal.oxm.record.SAXUnmarshaller.unmarshal(SAXUnmarshaller.java:1016)
    at org.eclipse.persistence.internal.oxm.XMLUnmarshaller.unmarshal(XMLUnmarshaller.java:657)
    at org.eclipse.persistence.jaxb.JAXBUnmarshaller.unmarshal(JAXBUnmarshaller.java:460)

Answer 1

java.lang.NoClassDefFoundError: javax/mail/internet/MimeMultipart

您的类路径中缺少 java mail-api。

如果您使用的是 maven，您应该能够通过添加流动依赖项来解决这个问题：

    <dependency>
        <groupId>javax.mail</groupId>
        <artifactId>javax.mail-api</artifactId>
    </dependency>
    <dependency>
        <groupId>com.sun.mail</groupId>
        <artifactId>javax.mail</artifactId>
    </dependency>

更多关于Java Mail API的信息