Unmarshalling推土机映射文件提供一个映射库

Question

我试图解开一些推土机映射文件，以便为许多应用程序提供一个映射可用性库。但我不能让JaxB注释正常工作。无论是映射列表我们unmarshalled为空或空

从映射文件，所有我感兴趣的是。

<?xml version="1.0" encoding="UTF-8" standalone="no"?> 
<mappings> 
    <mapping> 
     <class-a>package.MySourceClass</class-a> 
     <class-b>other.package.DestinationClass</class-b> 
    </mapping> 
</mappings> 

我有一个映射类

@XmlRootElement(name="mappings") 
@XmlAccessorType(XmlAccessType.FIELD) 
public class Mappings { 

    @XmlElementWrapper(name="mappings") 
    private List<Mapping> mappingEntries = null; 

//Getters and setters omitted 

和映射类

@XmlRootElement(name="mapping") 
@XmlAccessorType(XmlAccessType.FIELD) 
public class Mapping { 


    @XmlElement(name ="class-a") 
    private String classA; 

    @XmlElement(name = "class-b") 
    private String classB; 

我已经试过了注释的众多组合，我无法弄清楚我做错了什么。

有人可以指出我在正确的方向。

Answer 1

你可以做到以下几点：

映射

package forum11193953; 

import java.util.List; 
import javax.xml.bind.annotation.*; 

@XmlRootElement(name="mappings") // Match the root element "mappings" 
@XmlAccessorType(XmlAccessType.FIELD) 
public class Mappings { 

    @XmlElement(name="mapping") // There will be a "mapping" element for each item. 
    private List<Mapping> mappingEntries = null; 

} 

映射

package forum11193953; 

import javax.xml.bind.annotation.*; 

@XmlAccessorType(XmlAccessType.FIELD) 
public class Mapping { 


    @XmlElement(name ="class-a") 
    private String classA; 

    @XmlElement(name = "class-b") 
    private String classB; 

} 

演示

package forum11193953; 

import java.io.File; 
import javax.xml.bind.*; 

public class Demo { 

    public static void main(String[] args) throws Exception { 
     JAXBContext jc = JAXBContext.newInstance(Mappings.class); 

     Unmarshaller unmarshaller = jc.createUnmarshaller(); 
     File xml= new File("src/forum11193953/input.xml"); 
     Mappings mappings = (Mappings) unmarshaller.unmarshal(xml); 

     Marshaller marshaller = jc.createMarshaller(); 
     marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true); 
     marshaller.marshal(mappings, System.out); 
    } 

} 

的input.xml /输出

<?xml version="1.0" encoding="UTF-8" standalone="yes"?> 
<mappings> 
    <mapping> 
     <class-a>package.MySourceClass</class-a> 
     <class-b>other.package.DestinationClass</class-b> 
    </mapping> 
</mappings> 

Answer 2

尝试JMapper框架：http://code.google.com/p/jmapper-framework/

随着JMapper你有静态代码的性能和更多的动态映射的所有优点。