如何将XML转换为java.util.Map，反之亦然 [英] How to convert XML to java.util.Map and vice versa

问题描述

我正在搜索轻量级API（首选单个类）来转换

I'm searching a lightweight API (preferable single class) to convert a

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

到xml，反之亦然，将XML转换回Map。

to xml and, vice versa, convert the XML back to a Map.

示例：

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>

返回：

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

我不想使用 JAXB 或 JSON转换API 。它不需要处理嵌套的地图或属性或其他任何东西，只是这个简单的情况。
有什么建议吗？

I dont want to use JAXB or JSON conversion API. It doesnt have to take care of nested maps or attributes or anything else, just that simple case. Any suggestions?

编辑：我创建了一份工作副本&粘贴样品。感谢 fvu 和 Michal Bernhard 。

Edit: I created a working copy & paste sample. Thanks to fvu and Michal Bernhard.

下载最新的XStream框架，'仅核心'就足够了。

Download latest XStream framework, 'core only' is enough.

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。只是 xstream-xyzjar 就够了。

No converters or anything else is required. Just the xstream-x.y.z.jar is enough.

推荐答案

XStream！

更新：我在评论中要求添加了解组部分..

Updated: I added unmarshal part as requested in comments..

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;
        }

    }

}

这篇关于如何将XML转换为java.util.Map，反之亦然的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！