Java 选择要序列化的属性

在我的程序中，我需要用XML存储对象。但我不希望所有属性都序列化为xml。我该怎么做

public class Car implements ICar{
//all variables has their own setters and getters
private String manufacturer;
private String plate;
private DateTime dateOfManufacture;
private int mileage;
private int ownerId;
private Owner owner; // will not be serialized to xml
.....
}


//code for serialize to xml
public static String serialize(Object obj)
{
    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    XMLEncoder encoder = new XMLEncoder(baos);
    encoder.writeObject(obj);
    encoder.close();        
    return baos.toString();
}

退房。下面是一个更新的示例

BeanInfo info = Introspector.getBeanInfo(Car.class);
PropertyDescriptor[] propertyDescriptors = info.getPropertyDescriptors();
for (int i = 0; i < propertyDescriptors.length; ++i) {
    PropertyDescriptor pd = propertyDescriptors[i];
    if (pd.getName().equals("dateOfManufacture")) {
        pd.setValue("transient", Boolean.TRUE);
    }
}

BeanInfo info=Introspector.getBeanInfo（Car.class）；
PropertyDescriptor[]propertyDescriptors=info.getPropertyDescriptors（）；
对于（int i=0；i

我选择使用带有注释的JAXB序列化。这是最好、最简单的选择。谢谢大家的帮助

public static String serialize(Object obj) throws JAXBException
{
    StringWriter writer = new StringWriter();
    JAXBContext context = JAXBContext.newInstance(obj.getClass());
    Marshaller m = context.createMarshaller();

    m.marshal(obj, writer);
    return writer.toString();
}

public static Object deserialize(String xml, Object obj) throws JAXBException
{
    StringBuffer xmlStr = new StringBuffer(xml);
    JAXBContext context = JAXBContext.newInstance(obj.getClass());
    Unmarshaller um = context.createUnmarshaller();

    return um.unmarshal(new StreamSource(new StringReader(xmlStr.toString())));
}

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这是否适用于XmlEncoder的序列化，因为这是原始海报所使用的？@zdarsky.peter：看看如何将JavaBean属性标记为瞬态。或者使用JAXB进行序列化，并使用

@xmltransive

注释将属性标记为瞬态。我认为这是正确的，似乎是使用xmlcoder标记道具瞬态的唯一方法。我更新了示例，并添加了指向Oracle文档的链接，解释了使用xmlcoder的瞬态字段。