java-将XML转换为嵌套地图的地图

我有一个带有嵌套元素和重复标签的XML.例如：

<person>
    <name>Rama</name>
    <age>27</age>
    <gender>male</gender>
    <address>
        <doornumber>234</doornumber>
        <street>Kanon</street>
        <city>Hyderabad</city>
    </address>
    <qualification>
        <degree>M.Sc</degree>
        <specialisation>Maths</specialisation>
    </qualification>
    <qualification>
        <degree>B.E.</degree>
        <specialisation>Electrical</specialisation>
    </qualification>
</person>

现在,我需要一个API来将XML转换为Java Maps：

{name="Rama",age="27",gender="male",address={doornumber=234,street="Kanon",city="Hyderabad"},qualification=[{degree="M.Sc",specialisation="Maths"},{degree="B.E.",specialisation="Electrical"}]}

我知道我们可以使用XStream API来实现这一点.在这里,我只是想知道使用XStream是否有缺点,是否还有其他更好的Java API可以实现这一点.有什么建议么？

注意：这应该以一般方式完成,即Java API应该适用于任何XML,而不仅适用于上述XML.

解决方法:

不过已经很晚了.但是我已经为XStream API编写了一个自定义MapEntryConverter,它可以处理XML数据的任何复杂性,无论其深度如何.即使它将支持Duplicate标签名(将存储为ArrayList).

XStream xStream = new XStream(new DomDriver());
xStream.registerConverter(new MapEntryConverter());
xStream.alias("xml", java.util.Map.class);

// from XML, convert back to map
Map<String, List<Object>> map = (Map<String, List<Object>>) xStream.fromXML(xmlData);
/*System.out.println("MAP: \n" + map.entrySet().toString());*/

String xml = xStream.toXML(map);
/*System.out.println("XML: \n"+xml);*/

MapEntryConverter.java

import java.util.AbstractMap;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

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;

public class MapEntryConverter implements Converter{
    @SuppressWarnings("rawtypes")
    public boolean canConvert(Class clazz) {
        return AbstractMap.class.isAssignableFrom(clazz);
    }

    @SuppressWarnings({ "unchecked" })
    public void marshal(Object value, HierarchicalStreamWriter writer, MarshallingContext context) {
        AbstractMap<String, List<?>> map = (AbstractMap<String, List<?>>) value;
        List<Map<String, ?>> list = (List<Map<String, ?>>) map.get("xml");
        for( Map<String, ?> maps: list ) {
            for( Entry<String, ?> entry: maps.entrySet() ) {
                mapToXML(writer, entry);
            }
        }
    }

    @SuppressWarnings("unchecked")
    private void mapToXML(HierarchicalStreamWriter writer, Entry<String, ?> entry) {
        writer.startNode(entry.getKey());
        if( entry.getValue() instanceof String ) {
            writer.setValue(entry.getValue().toString());
        }else if(  entry.getValue() instanceof ArrayList ) {
            List<?> list = (List<?>) entry.getValue();
            for( Object object: list ) {
                Map<String, ?> map = (Map<String, ?>) object;
                for( Entry<String, ?> entryS: map.entrySet() ) {
                    mapToXML(writer, entryS);
                }
            }
        }
        writer.endNode();
    }

    public Object unmarshal(HierarchicalStreamReader reader, UnmarshallingContext context) {
        Map<String, List<Object>> map = new HashMap<String, List<Object>>();
        map = xmlToMap(reader, new HashMap<String, List<Object>>());
        return map;
    }

    private Map<String, List<Object>> xmlToMap(HierarchicalStreamReader reader, Map<String, List<Object>> map) {
        List<Object> list = new ArrayList<Object>();
        while(reader.hasMoreChildren()) {
            reader.moveDown();
            if( reader.hasMoreChildren() ) {
                list.add(xmlToMap(reader, new HashMap<String, List<Object>>()));
            }else {
                Map<String, Object> mapN = new HashMap<String, Object>();
                mapN.put(reader.getNodeName(), reader.getValue());
                list.add(mapN);
            }
            reader.moveUp();
        }
        map.put(reader.getNodeName(), list);
        return map;
    }

}

请注意,Map中的每个值(如果有任何子级)都存储为列表.根据我的说法,解组部分非常整洁.我将在空闲时间完善并简化编组部分.

输入数据应在xml标记中提供,如下所示.

<xml>
    <person>
        <name>Rama</name>
        <age>27</age>
        <gender>male</gender>
        <address>
            <doornumber>234</doornumber>
            <street>Kanon</street>
            <city>Hyderabad</city>
        </address>
        <qualification>
            <degree>M.Sc</degree>
            <specialisation>Maths</specialisation>
        </qualification>
        <qualification>
            <degree>B.E.</degree>
            <specialisation>Electrical</specialisation>
        </qualification>
    </person>
</xml>

将我的输出与您的预期输出进行比较时,会有一点变化.重复的键值不会合并为数组,而是它们是单独的列表.

标签：xstream,jaxb,xml,spring,java
来源： https://codeday.me/bug/20191122/2056380.html