attribute由于ObjectNode对序列化一无所知，因此无法标记给定的属性。您可以对POJO类执行此操作，并且com.fasterxml.jackson.dataformat.xml.ser.ToXmlGenerator仅在@JacksonXmlProperty(isAttribute = true)给定属性使用批注时才进行处理。我建议为POJO需要属性的元素创建元素，并使用JacksonXML批注或实现JsonSerializable接口。它可能如下所示：
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializable;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.jsontype.TypeSerializer;
import com.fasterxml.jackson.databind.node.ObjectNode;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import com.fasterxml.jackson.dataformat.xml.ser.ToXmlGenerator;
import java.io.IOException;
import java.util.LinkedHashMap;
import java.util.Map;
public class XmlMapperApp {
public static void main(String[] args) throws Exception {
Map<String, Object> map = new LinkedHashMap<>();
map.put("puppy", Boolean.TRUE);
map.put("Apple", 2);
map.put("Jet", "Li");
Examples examples = new Examples();
examples.setOverlyComplicated("yes");
examples.setMap(map);
XmlMapper mapper = new XmlMapper();
mapper.enable(SerializationFeature.INDENT_OUTPUT);
ObjectNode rootNode = mapper.createObjectNode();
rootNode.putPOJO("Examples", examples);
ObjectNode currentNode = rootNode.putObject("Single");
currentNode.put("One", 1);
mapper.writeValue(System.out, rootNode);
}
}
class Examples implements JsonSerializable {
@Override
public void serialize(JsonGenerator gen, SerializerProvider serializers) throws IOException {
ToXmlGenerator toXmlGenerator = (ToXmlGenerator) gen;
toXmlGenerator.writeStartObject();
writeAttributes(toXmlGenerator);
writeMap(toXmlGenerator);
toXmlGenerator.writeEndObject();
}
private void writeAttributes(ToXmlGenerator gen) throws IOException {
if (overlyComplicated != null) {
gen.setNextIsAttribute(true);
gen.writeFieldName("overlyComplicated");
gen.writeString(overlyComplicated);
gen.setNextIsAttribute(false);
}
}
private void writeMap(ToXmlGenerator toXmlGenerator) throws IOException {
for (Map.Entry<String, Object> entry : map.entrySet()) {
toXmlGenerator.writeObjectField(entry.getKey(), entry.getValue());
}
}
@Override
public void serializeWithType(JsonGenerator gen, SerializerProvider serializers, TypeSerializer typeSer) throws IOException {
serialize(gen, serializers);
}
private String overlyComplicated;
private Map<String, Object> map;
// getters, setters, toString
}
上面的代码打印：
<ObjectNode>
<Examples overlyComplicated="yes">
<puppy>true</puppy>
<Apple>2</Apple>
<Jet>Li</Jet>
</Examples>
<Single>
<One>1</One>
</Single>
</ObjectNode>
如果要使用相同的对象ExamplePOJO进行JSON序列化，则需要在serialize方法中处理它或创建另一个对象ObjectNode而不是Examlples对象。
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