2

我想序列化一个使用Simple扩展 HashMap 的元素。

@Root(name = "settings")
@ElementMap(entry="element", key="id", attribute=true, required=true, empty=true)
public class Settings extends HashMap<String, Object> {

    ...

每当我序列化它时,我都没有收到任何错误,但我也会得到一个如下所示的空文件:

<settings/>

有没有办法在不创建内部对象的情况下做到这一点,然后必须将所有方法委托给它?

4

1 回答 1

1

我想原因是,Simple不能转换 HashMaps。

如果我运行此代码...

    Map<String, Object> map = new HashMap<>();
    map.put("a", "b");
    map.put("c", 3);
    map.put("d", new Date());


    ser.write(map, new File("test2.xml"));

...我得到以下异常:

org.simpleframework.xml.transform.TransformException: Transform of class java.util.HashMap not supported


现在,这是我为使您的课程序列化所做的工作:

我写了一个转换器,用于转换Settings.

Converter班级:

public class SettingsConverter implements Converter<Settings>
{
    private Transformer transformer;


    public SettingsConverter()
    {
        this.transformer = new Transformer(new RegistryMatcher());
    }



    @Override
    public Settings read(InputNode node) throws Exception
    {
        Settings settings = new Settings();
        InputNode child = node.getNext();

        while( child != null )
        {
            final String key = child.getAttribute("key").getValue();
            final Class c = Class.forName(child.getAttribute("class").getValue());

            settings.put(key, transformer.read(child.getAttribute("value").getValue(), c));
            child = node.getNext();
        }

        return settings;
    }


    @Override
    public void write(OutputNode node, Settings value) throws Exception
    {
        for( Map.Entry<String, Object> entry : value.entrySet() )
        {
            OutputNode child = node.getChild("setting");

            child.setAttribute("key", entry.getKey());
            child.setAttribute("class", entry.getValue().getClass().getName());
            child.setAttribute("value", transformer.write(entry.getValue(), entry.getValue().getClass())); 
        }   
    }
}

Settings班级:

@Root()
@Convert(value=SettingsConverter.class)
public class Settings extends HashMap<String, Object>
{
    // ...
}

测试:

final File testFile = new File("test.xml");

Settings settings = new Settings();
settings.put("a", "b");
settings.put("c", 3);
settings.put("d", new Date());

// Serialize - make shure you use an AnnotationStrategy here
Serializer ser = new Persister(new AnnotationStrategy());
ser.write(settings, testFile); 


// Deserialize
Settings in = ser.read(Settings.class, testFile);

System.out.println(settings.equals(in));

文件 test.xml:

<settings>
   <setting key="d" class="java.util.Date" value="2012-08-28 17:15:13.152 MESZ"/>
   <setting key="c" class="java.lang.Integer" value="3"/>
   <setting key="a" class="java.lang.String" value="b"/>
</settings>

我想有更好的方法(如果不是更多)可以做到这一点,但也许这会有所帮助。

还有一点,@Default注解在这里不起作用(异常),可能解决这个问题可能会解决整个问题

于 2012-08-28T15:16:28.893 回答