要转换的xml
- XML code
<?xml version="1.0" encoding="UTF-8"?><User> <Age>23</Age> <NameList> <name>name1</name> <name>name2</name> <name>name3</name> <name>name4</name> </NameList></User>
对应的 User 类
- Java code
public class User { private int age; private List nameList;}
请问:对于这样的xml转换成bean 用apache 的 simpleXml 该怎样实现呢? 谢谢!
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推荐用dom4j来解析xml,这个不难,更多信息请百度
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- Java code
File file= new File(xmlFileName); DocumentBuilderFactory ch = DocumentBuilderFactory.newInstance(); DocumentBuilder db = ch.newDocumentBuilder() ; Document dom = db.parse(file) ; NodeList nl_sr = dom.getElementsByTagName("source-rightCode");
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其实就是解析xml文件,然后赋值给你的bean对象。
用什么解析方式得看你的具体需求
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- Java code
import com.thoughtworks.xstream.*;import com.google.common.io.*;import com.google.common.base.*;public class Main { public static void main(final String[] args) throws Exception{ XStream xstream = new XStream(); xstream.alias("User",User.class); xstream.aliasField("Age",User.class,"age"); xstream.aliasField("NameList",User.class,"nameList"); xstream.alias("name",String.class); User user = (User)xstream.fromXML(Resources.toString(Resources.getResource(Main.class,"user.xml"),Charsets.UTF_8)); System.out.println(xstream.toXML(user)); }}
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最后一句的输出是
- XML code
<User> <Age>23</Age> <NameList> <name>name1</name> <name>name2</name> <name>name3</name> <name>name4</name> </NameList></User>
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simple-xml 的话
- Java code
@ElementList(name="NameList",entry="name")private List<String> nameList;
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xstream 也有 annotation ,功能好像比api弱多了。
- Java code
@XStreamAlias("Age")private int age;