xmljava类_将xml表示为java类

注意&＃xff1a;我是

EclipseLink JAXB (MOXy)领导者,也是

JAXB (JSR-222)专家组的成员.

IS there a way to generate java classes dynamicaly for a structure

like this ?

JAXB实现提供了从XML模式生成Java模型的能力.从Java SE 6开始的JDK中包含的参考实现可在以下位置获得&＃xff1a;

/bin/xjc

可以在此处找到从XML模式生成对象模型的示例&＃xff1a;

A small correction,i don’t have an xsd for the xml

如果您没有XML模式,您可以找到一个实用程序来从XML文档生成XML模式&＃xff1a;

或者从代码开始.

从代码开始

您还可以从代码开始并注释模型以映射到现有的XML结构.

根

package forum11213872;

import java.util.List;

import javax.xml.bind.annotation.*;

&＃64;XmlRootElement(name&＃61;"Root")

&＃64;XmlAccessorType(XmlAccessType.FIELD)

public class Root {

&＃64;XmlElement(name&＃61;"Book")

private List books;

}

书

package forum11213872;

import javax.xml.bind.annotation.*;

&＃64;XmlAccessorType(XmlAccessType.FIELD)

public class Book {

&＃64;XmlAttribute

private String name;

&＃64;XmlAttribute

private int price;

}

演示

package forum11213872;

import java.io.File;

import javax.xml.bind.*;

public class Demo {

public static void main(String[] args) throws Exception {

JAXBContext jc &＃61; JAXBContext.newInstance(Root.class);

Unmarshaller unmarshaller &＃61; jc.createUnmarshaller();

File xml &＃61; new File("src/forum11213872/input.xml");

Root root &＃61; (Root) unmarshaller.unmarshal(xml);

Marshaller marshaller &＃61; jc.createMarshaller();

marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT,true);

marshaller.marshal(root,System.out);

}

}

input.xml中/输出