雖然已經很晚了。但是我已經爲XStream API編寫了一個自定義MapEntryConverter,它可以處理任何複雜的XML數據,無論它有多深。即使它會支持重複的標籤名稱(將存儲爲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;
}
}
請注意,在每個地圖值存儲爲一個列表中,如果有任何孩子。根據我的理解部分非常整齊。我會在空閒時間拋光並簡化編組部分。
輸入數據應該在下面的xml標籤中提供。
<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>
將我的輸出與您的預期輸出進行比較時發生了一些變化。重複鍵值不會合併爲一個數組,而是它們是單獨的列表。
嘿,如果API爲你工作,使用它,也可能最終會主要是基於意見。 –