所有的答案都帮助我达成了最终的解决方案，该解决方案可以用于JPA，而不是专门用于Eclipselink或Hibernate。
import com.fasterxml.jackson.core.type.TypeReference;import com.fasterxml.jackson.databind.ObjectMapper;import java.io.IOException;import javax.json.Json;import javax.json.JsonObject;import javax.persistence.Converter;import org.postgresql.util.PGobject;@Converter(autoApply = true)public class JsonConverter implements javax.persistence.AttributeConverter<JsonObject, Object> { private static final long serialVersionUID = 1L; private static ObjectMapper mapper = new ObjectMapper(); @Override public Object convertToDatabaseColumn(JsonObject objectValue) { try { PGobject out = new PGobject(); out.setType("json"); out.setValue(objectValue.toString()); return out; } catch (Exception e) { throw new IllegalArgumentException("Unable to serialize to json field ", e); } } @Override public JsonObject convertToEntityAttribute(Object dataValue) { try { if (dataValue instanceof PGobject && ((PGobject) dataValue).getType().equals("json")) { return mapper.reader(new TypeReference<JsonObject>() { }).readValue(((PGobject) dataValue).getValue()); } return Json.createObjectBuilder().build(); } catch (IOException e) { throw new IllegalArgumentException("Unable to deserialize to json field ", e); } }}
