JAXB解析XML

来源：互联网发布：德州扑克人工智能编辑：程序博客网时间：2024/05/16 14:09

尝试过dom4j和XStream，二者对于XML转java bean都比较无力。最终JAXB解烦忧。

主要问题出现在构建pojo类上：

*出现在XML里的标签，都应该是pojo中的字段名。

*List都以数组形式表达，数组的字段名变成每一个元素的标签名。

<ALL><PNRNO /><PASSENGERLIST>    <PASSENGER>        <NAME>1</NAME>    </PASSENGER>    <PASSENGER>        <NAME>2</NAME>    </PASSENGER>    <PASSENGER>        <NAME>3</NAME>    </PASSENGER></PASSENGERLIST></ALL>

应表示为

@XmlRootElement(name = "ALL")public class ALL {    @XmlElement    public PassengerList PASSENGERLIST;        @XmlRootElement    public static class PassengerList {        @XmlElement        Passenger[] PASSENGER;    }}

注解使用参考：

http://blog.csdn.net/chen7788/article/details/7384315

http://blog.csdn.net/czplplp_900725/article/details/7888896

JAXB实现XML和java bean的互转：

    /**     * 将Java对象序列化为XML字符串     *     * @param pojo 待序列化的Java对象     * @throws JAXBException     * @return     */public static String toXml(Object pojo) throws JAXBException {JAXBContext jaxbContext = JAXBContext.newInstance(pojo.getClass());Marshaller marshaller = jaxbContext.createMarshaller();marshaller.setProperty(Marshaller.JAXB_ENCODING, "UTF-8"); //编码格式marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true); //是否格式化生成的XMLmarshaller.setProperty(Marshaller.JAXB_FRAGMENT, false); // 是否省略XML头声明信息StringWriter stringWriter = new StringWriter();marshaller.marshal(pojo, stringWriter);return stringWriter.toString();}

    /**     * 将XML字符串反序列化为Java对象     *     * @param xmlStr 待反序列化的XML字符串     * @param pojoClass　需要反序列化的类型     * @throws JAXBException     * @return     */public static <T> Object fromXml(String xmlStr, Class<T> pojoClass) throws JAXBException {JAXBContext jaxbContext = JAXBContext.newInstance(pojoClass);Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();Object object = unmarshaller.unmarshal(new ByteArrayInputStream(xmlStr.getBytes()));return object;}

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