我想添加一个包装器,其命名是在运行时确定的,因为它取决于类名(我可以使用@JsonRootName,但我不想这样做,因为我必须在每个子类上使用它,即效率不高)。
我想我应该使用 @JsonSerialize 来覆盖默认的序列化器,但我只想创建包装器;我不想自己序列化对象字段(而且我在一个抽象类中,所以我什至不知道子类的字段!)。我不关心它们,我只关心包装!所以我希望默认序列化器为我处理这些字段,或者类似的东西。
@JsonSerialize(using = CustomSerializer.class)
public abstract class Request {
public static class CustomSerializer extends JsonSerializer<Request > {
@Override
public void serialize(Request request, JsonGenerator jgen, SerializerProvider provider) throws IOException {
// Doing my stuff to determine the wrapper name based on request.class.getSimpleName()
// Then what should I wright to serialize the fields?
// Basically I just want a function to generate the same json that the default serializer would generate!
// I tried the following, but obviously it gives a com.fasterxml.jackson.databind.JsonMappingException: Infinite recursion
jgen.writeObject(value);
// Same error for the function below
provider.defaultSerializeValue(value, jgen);
}
}
最佳答案
要创建包装序列化程序,您需要使用 com.fasterxml.jackson.databind.ser.BeanSerializerModifier
类(class)。您可以使用 com.fasterxml.jackson.databind.module.SimpleModule
注册它。下面的示例展示了如何做到这一点的端到端解决方案:
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.BeanDescription;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.SerializationConfig;
import com.fasterxml.jackson.databind.SerializationFeature;
import com.fasterxml.jackson.databind.SerializerProvider;
import com.fasterxml.jackson.databind.json.JsonMapper;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.databind.ser.BeanSerializerModifier;
import com.fasterxml.jackson.databind.util.NameTransformer;
import java.io.IOException;
import java.util.UUID;
public class JsonPathApp {
public static void main(String[] args) throws Exception {
SimpleModule wrappersModule = new SimpleModule("requestWrapper");
wrappersModule.setSerializerModifier(new BeanSerializerModifier() {
@Override
public JsonSerializer<?> modifySerializer(SerializationConfig config, BeanDescription beanDesc, JsonSerializer<?> serializer) {
if (Request.class.isAssignableFrom(beanDesc.getBeanClass())) {
return new RequestWrapperJsonSerializer(serializer);
}
return serializer;
}
});
ObjectMapper mapper = JsonMapper.builder()
.enable(SerializationFeature.INDENT_OUTPUT)
.addModule(wrappersModule)
.build();
System.out.println(mapper.writeValueAsString(new Request1(1, "POST")));
System.out.println(mapper.writeValueAsString(new Request2(2, UUID.randomUUID())));
}
}
class RequestWrapperJsonSerializer extends JsonSerializer<Request> {
private final JsonSerializer baseSerializer;
public RequestWrapperJsonSerializer(JsonSerializer baseSerializer) {
this.baseSerializer = baseSerializer;
}
@Override
public void serialize(Request value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
gen.writeStartObject();
gen.writeFieldName(value.getClass().getSimpleName() + "Wrapper");
gen.writeStartObject();
baseSerializer.unwrappingSerializer(NameTransformer.NOP).serialize(value, gen, serializers);
gen.writeEndObject();
gen.writeEndObject();
}
}
abstract class Request {
private int id;
//constructor, getters, setters, toString
}
class Request1 extends Request {
private String body;
//constructor, getters, setters, toString
}
class Request2 extends Request {
private UUID uuid;
//constructor, getters, setters, toString
}
上面的代码打印:
{
"Request1Wrapper" : {
"id" : 1,
"body" : "POST"
}
}
{
"Request2Wrapper" : {
"id" : 2,
"uuid" : "dd4cccb5-1cf5-4dd4-8bc7-97cb101e5d7d"
}
}
相反unwrappingSerializer
您可以使用的方法 serialize
方法并删除额外的包装调用。
关于java - @JsonSerialize - 如何在运行时创建包装器并为对象字段使用默认序列化?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59411455/