1

我正在尝试解组一些推土机映射文件，以便为许多应用程序提供映射可用性库。但我无法让 JaxB 注释正常工作。我们解组为 null 或空的映射列表

从映射文件中，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

和一个映射类

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


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

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

我已经尝试了许多注释组合，但我无法弄清楚我做错了什么。

有人可以指出我正确的方向。

4

2 回答 2

1

您可以执行以下操作：

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);
    }

}

输入.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>
于 2012-06-25T17:27:16.343 回答
0

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

使用 JMapper，您将拥有动态映射的所有优势以及静态代码的性能等等。

于 2012-09-24T08:12:56.993 回答