使用XStream来映射xml中列表元素到java对象

网上很多讲述使用XStream来映射xml文件到java对象的文章，但是几乎没有一篇文章讲述，如何使用XStream来映射xml中的同级的列表元素到java对象的文章。本文示例中将使用XStream 1.3.1。请去XStream的官方站点上下载相应的包。

在Eclipse中新建一个Java Project，然后引入XStream的包：xstream－1.3.1.jar和xpp3-min-1.1.4c.jar。在这里使用xpp的xml解析技术来解析xml，而不是传统的DOM或者sax来进行xml的解析。

在src下新建sample.xml文件。代码如下：

<?xml version="1.0" encoding="UTF-8"?>
<person id="1" name="sample">
	<wife name="wife_name" age="25"></wife>
	<house name="house1" address="address1"></house> 
	<house name="house1" address="address1"></house> 
	<house name="house1" address="address1"></house> 
	<house name="house1" address="address1"></house> 
</person>

新建House类来映射house元素，House类的源代码如下：

public class House {
	private String name = "";
	private String address = "";
	public String getName() {
		return name;
	}
	public void setName(String name) {
		this.name = name;
	}
	public String getAddress() {
		return address;
	}
	public void setAddress(String address) {
		this.address = address;
	}
	
	
}

新建Wife类来映射wife标签，源代码如下：

public class Wife {
	private String name = "";
	private int age;
	public String getName() {
		return name;
	}
	public void setName(String name) {
		this.name = name;
	}
	public int getAge() {
		return age;
	}
	public void setAge(int age) {
		this.age = age;
	}
	
	
}

新建Person类来映射person元素，源代码如下：

import java.util.List;


public class Person {
	private int id;
	private String name;
	private Wife wife;
	private List<House> houseList;
	
	public int getId() {
		return id;
	}
	public void setId(int id) {
		this.id = id;
	}
	public String getName() {
		return name;
	}
	public void setName(String name) {
		this.name = name;
	}
	public Wife getWife() {
		return wife;
	}
	public void setWife(Wife wife) {
		this.wife = wife;
	}
	public List<House> getHouseList() {
		return houseList;
	}
	public void setHouseList(List<House> houseList) {
		this.houseList = houseList;
	}
}

新建TestMain方法，包含main方法。首先，构建XStream的实例：

	XStream xstream = new XStream();

然后，就是配置xml文件和java对象之间的映射，源代码如下：

		xstream.alias("person", Person.class);
		xstream.aliasAttribute(Person.class, "id", "id");
		xstream.aliasAttribute(Person.class, "name", "name");
		
		xstream.alias("wife", Wife.class);
		xstream.aliasAttribute(Wife.class, "name", "name");
		
		xstream.addImplicitCollection(Person.class, "houseList");
		xstream.alias("house", House.class);
		xstream.aliasAttribute(House.class,"name", "name");
		xstream.aliasAttribute(House.class, "address", "address");

注意代码中的这句“xstream.addImplicitCollection(Person.class, "houseList");”，如果不加，xml配置文件的house元素将作为houselist的子元素。将变成

<houseList>
<house name="house1" address="address1"></house> 
	<house name="house1" address="address1"></house> 
	<house name="house1" address="address1"></house> 
	<house name="house1" address="address1"></house> 
</houseList>

。
最后，使用XStream.fromXML来将对应的xml转化成java对象。代码如下：

Person person = (Person)xstream.fromXML(TestMain.class.getClassLoader()
.getResourceAsStream("sample.xml"));

由于sample.xml文件位于classpath下，所以使用了ClassLoader来加载sample.xml。

测试代码如下：

        System.out.println("[person.name=" + person.getName() + "]");
	System.out.println("[wife.name=" + person.getWife().getName() + ", wife.age=" + person.getWife().getAge() + "]");
	System.out.println("[person.houseList.size=" + person.getHouseList().size() + "]");
	System.out.println("[person.houseList.house1.name=" +                person.getHouseList().get(0).getName() + ",address=" + person.getHouseList().get(0).getAddress() +"]");

测试结果如下：
[person.name=sample]
[wife.name=wife_name, wife.age=0]
[person.houseList.size=4]
person.houseList.house1.name=house1,address=address1]

评论

1 楼 yonge812 2013-08-28  

如果xml既包含命名空间（节点包含前缀）又包含与java属性的映射，如何转换成对象？