Generics 如何让XStream很好地输出Scala列表?我可以写一个自定义转换器吗?
此代码:Generics 如何让XStream很好地输出Scala列表?我可以写一个自定义转换器吗?,generics,list,scala,xstream,Generics,List,Scala,Xstream,此代码: println(new XStream.toXML(List(1,2,3))) 生成以下XML: <scala.coloncolon serialization="custom"> <unserializable-parents/> <scala.coloncolon> <int>1</int> <int>2</int> <int>3</int>
println(new XStream.toXML(List(1,2,3)))
生成以下XML:
<scala.coloncolon serialization="custom">
<unserializable-parents/>
<scala.coloncolon>
<int>1</int>
<int>2</int>
<int>3</int>
<scala.ListSerializeEnd/>
</scala.coloncolon>
</scala.coloncolon>
1.
2.
3.
相反,我希望这样:
<list>
<int>1</int>
<int>2</int>
<int>3</int>
</list>
1.
2.
3.
这与泛型java集合的序列化方式类似。最好的方法是什么
我已经实现了我自己的转换器,但我仍然停留在解组方法上,不清楚如何实例化空列表
class ListConverter( _mapper : Mapper ) extends AbstractCollectionConverter(_mapper) {
/** Helper method to use x.getClass
*
* See: http://scalide.blogspot.com/2009/06/getanyclass-tip.html
*/
def getAnyClass(x: Any) = x.asInstanceOf[AnyRef].getClass
def canConvert( clazz: Class[_]) = {
classOf[::[_]] == clazz
}
def marshal( value: Any, writer: HierarchicalStreamWriter, context: MarshallingContext) = {
val list = value.asInstanceOf[List[_]]
for ( item <- list ) {
writeItem(item, context, writer)
}
}
def unmarshal( reader: HierarchicalStreamReader, context: UnmarshallingContext ) = {
println(context.getRequiredType())
var list : List[_] = createCollection(context.getRequiredType()).asInstanceOf[List[_]]
while (reader.hasMoreChildren()) {
reader.moveDown();
val item = readItem(reader, context, list);
list = item :: list
reader.moveUp();
}
list
}
}
object ListConverter {
def configureXStream( stream: XStream ) = {
stream.alias("list", classOf[::[_]])
stream.registerConverter( new ListConverter(stream.getMapper) )
}
}
class ListConverter(\u-mapper:mapper)扩展了AbstractCollectionConverter(\u-mapper){
/**使用x.getClass的Helper方法
*
*见:http://scalide.blogspot.com/2009/06/getanyclass-tip.html
*/
def getAnyClass(x:Any)=x.asInstanceOf[AnyRef].getClass
def canConvert(clazz:Class[])={
类别[::[]]==clazz
}
def marshal(值:Any,writer:hierarchycalstreamwriter,上下文:MarshallingContext)={
val list=值.asInstanceOf[list[\u]]
对于(item发布问题后的几秒钟,我得到了答案,以下是解组的一个工作实现:
def unmarshal( reader: HierarchicalStreamReader, context: UnmarshallingContext ) = {
var list : List[_] = Nil
while (reader.hasMoreChildren()) {
reader.moveDown();
val item = readItem(reader, context, list);
list = list ::: List(item) // be sure to build the list in the same order
reader.moveUp();
}
list
}
空列表只有一个实例,那就是对象Nil
你能详细说明一下吗?我看不出你的答案与你问的如何实例化空列表的问题有什么关联。我只是说只有一个空列表,那就是Nil
。所以你需要参考它——就像你在e您在答案中显示的代码。