如何将XML转换为java.util.Map，反之亦然

  • Z时代
  • 2024-01-10
  • 分类：问答

我正在搜索轻量级API（最好是单个类）以转换

Map<String,String> map = new HashMap<String,String();

到xml，反之亦然，将XML转换回Map。

例：

Map<String,String> map = new HashMap<String,String();
map.put("name","chris");
map.put("island","faranga");
MagicAPI.toXML(map,"root");

结果：

<root>
  <name>chris</chris>
  <island>faranga</island>
</root>

然后回来：

Map<String,String> map = MagicAPI.fromXML("...");

我不想使用JAXB或JSON转换API。只需简单的情况，它就不必处理嵌套的地图或属性或其他任何内容。有什么建议么？

编辑：我创建了一个工作副本并粘贴示例。感谢fvu和Michal Bernhard。

下载最新的XStream框架，“仅核心”就足够了。

Map<String,Object> map = new HashMap<String,Object>();
map.put("name","chris");
map.put("island","faranga");
// convert to XML
XStream xStream = new XStream(new DomDriver());
xStream.alias("map", java.util.Map.class);
String xml = xStream.toXML(map);
// from XML, convert back to map
Map<String,Object> map2 = (Map<String,Object>) xStream.fromXML(xml);

不需要转换器或其他任何东西。仅xstream-xyzjar就足够了。

回答：

XStream！

更新：我按照评论的要求添加了编组部分。

import com.thoughtworks.xstream.XStream;
import com.thoughtworks.xstream.converters.Converter;
import com.thoughtworks.xstream.converters.MarshallingContext;
import com.thoughtworks.xstream.converters.UnmarshallingContext;
import com.thoughtworks.xstream.io.HierarchicalStreamReader;
import com.thoughtworks.xstream.io.HierarchicalStreamWriter;
import java.util.AbstractMap;
import java.util.HashMap;
import java.util.Map;
public class Test {
    public static void main(String[] args) {
        Map<String,String> map = new HashMap<String,String>();
        map.put("name","chris");
        map.put("island","faranga");
        XStream magicApi = new XStream();
        magicApi.registerConverter(new MapEntryConverter());
        magicApi.alias("root", Map.class);
        String xml = magicApi.toXML(map);
        System.out.println("Result of tweaked XStream toXml()");
        System.out.println(xml);
        Map<String, String> extractedMap = (Map<String, String>) magicApi.fromXML(xml);
        assert extractedMap.get("name").equals("chris");
        assert extractedMap.get("island").equals("faranga");
    }
    public static class MapEntryConverter implements Converter {
        public boolean canConvert(Class clazz) {
            return AbstractMap.class.isAssignableFrom(clazz);
        }
        public void marshal(Object value, HierarchicalStreamWriter writer, MarshallingContext context) {
            AbstractMap map = (AbstractMap) value;
            for (Object obj : map.entrySet()) {
                Map.Entry entry = (Map.Entry) obj;
                writer.startNode(entry.getKey().toString());
                Object val = entry.getValue();
                if ( null != val ) {
                    writer.setValue(val.toString());
                }
                writer.endNode();
            }
        }
        public Object unmarshal(HierarchicalStreamReader reader, UnmarshallingContext context) {
            Map<String, String> map = new HashMap<String, String>();
            while(reader.hasMoreChildren()) {
                reader.moveDown();
                String key = reader.getNodeName(); // nodeName aka element's name
                String value = reader.getValue();
                map.put(key, value);
                reader.moveUp();
            }
            return map;
        }
    }
}

以上是 如何将XML转换为java.util.Map，反之亦然 的全部内容， 来源链接： utcz.com/qa/426895.html

回到顶部