8

我有一个Map<String, String>.
每个人的第一个想法是将其转换为List<Pair<String,String>>（Pair作为自定义类）。

我试过@XmlAdapter这样的：

public class MapPropertiesAdapter extends XmlAdapter<List<Property>, Map<String,String>> { ... }

但是我使用的 JAXB impl Eclipse MOXy 以ClassCastException“无法将 HashMap 转换为 Collection”而告终。

JAXB 是否支持这种转换？还是我忽略了一些解释为什么不是的文档部分？

PS：我想得到这样的XML：

<properties>
    <property name="protocol"/>
    <property name="marshaller"/>
    <property name="unmarshaller"/>
    <property name="timeout"/>
    ...
</properties>

我明白了，只需要使用一个中间类。在 XMLCompositeObjectMappingNodeValue.marshalSingleValue( XMLCompositeObjectMappingNodeValue.java:161)中的处理 NPE 中也有描述

4

1 回答 1

9

您应该将其调整为具有属性的对象，而不是将其调整Map为 a 。ListList

XmlAdapter (MapPropertiesAdapter)

import java.util.*;
import java.util.Map.Entry;
import javax.xml.bind.annotation.*;
import javax.xml.bind.annotation.adapters.XmlAdapter;

public class MapPropertiesAdapter extends XmlAdapter<MapPropertiesAdapter.AdaptedProperties, Map<String, String>>{

    public static class AdaptedProperties {
        public List<Property> property = new ArrayList<Property>();
    }

    public static class Property {
        @XmlAttribute
        public String name;

        @XmlValue
        public String value;
    }

    @Override
    public Map<String, String> unmarshal(AdaptedProperties adaptedProperties) throws Exception {
        if(null == adaptedProperties) {
            return null;
        }
        Map<String, String> map = new HashMap<String, String>(adaptedProperties.property.size());
        for(Property property : adaptedProperties.property) {
            map.put(property.name, property.value);
        }
        return map;
    }

    @Override
    public AdaptedProperties marshal(Map<String, String> map) throws Exception {
        if(null == map) {
            return null;
        }
        AdaptedProperties adaptedProperties = new AdaptedProperties();
        for(Entry<String,String> entry : map.entrySet()) {
            Property property = new Property();
            property.name = entry.getKey();
            property.value = entry.getValue();
            adaptedProperties.property.add(property);
        }
        return adaptedProperties;
    }

}

领域模型（根）

下面是一个具有Map属性的模型对象。注解@XmlJavaTypeAdapter用于指定.XmlAdapter

import java.util.Map;
import javax.xml.bind.annotation.*;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;

@XmlRootElement
@XmlAccessorType(XmlAccessType.FIELD)
public class Root {

    @XmlJavaTypeAdapter(MapPropertiesAdapter.class)
    private Map<String, String> properties;

}

演示

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

public class Demo {

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

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        File xml = new File("src/forum17024050/input.xml");
        Root root = (Root) unmarshaller.unmarshal(xml);

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

}

输入.xml/输出

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <properties>
        <property name="A">a</property>
        <property name="B">b</property>
    </properties>
</root>
于 2013-06-10T13:31:26.580 回答