Pra*_*nth 4 java xml spring xstream jaxb
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>
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现在我想要一个 API 可以将此 XML 转换为 Java 中的地图:
{name="Rama",age="27",gender="male",address={doornumber=234,street="Kanon",city="Hyderabad"},qualification=[{degree="M.Sc",specialisation="Maths"},{degree="B.E.",specialisation="Electrical"}]}
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我知道我们可以使用 XStream API 来实现这一点。这里我只是想知道使用 XStream 是否有任何缺点以及是否存在其他更好的 Java API 来实现这一点。有什么建议么?
注意:这应该以通用方式完成,即 Java API 应该适用于任何 XML,而不仅仅是上述 XML。
虽然已经很晚了。但我已经为 XStream API 编写了一个自定义 MapEntryConverter,它可以处理任何复杂的 XML 数据,无论它有多深。即使它支持重复标签名称(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);*/
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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;
}
}
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请注意,如果 Map 中有任何子项,则 Map 中的每个值都会存储为列表。我认为解组部分非常简洁。我会在空闲时间完善和简化编组部分。
输入数据应在 xml 标记内提供,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>
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将我的输出与您的预期输出进行比较时,有一点变化。重复的键值不会合并为数组,而是单独的列表。
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