[英]Simplest way to pass a Serialized object from client to Spring controller and get back serialized object response?
[英]How to get serialized object without their relationships on spring
我有下一個代碼:
我的實體:
@Entity
@NamedEntityGraphs({
@NamedEntityGraph(
name = "client",
attributeNodes = {
@NamedAttributeNode( value = "country" )}),
@NamedEntityGraph(
name = "only_client",
attributeNodes = {})
})
public class Client{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name="ClientId")
private int id;
private String firstName;
private String lastName;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "CountryId")
private Country country;
//Constructor, getters and setters...
}
//Country class...
我的資料庫:
@Repository
public interface ClienteRepository extends JpaRepository<Cliente, Serializable> {
// #1
@Override
@EntityGraph(value = "client",type = EntityGraph.EntityGraphType.FETCH)
List<Cliente> findAll();
// #2
@EntityGraph(value = "only_client")
List<Cliente> findAllByLastNameContaining(String lastName);
}
因此,當我使用第一方法時,效果很好,但是當我嘗試第二方法時,控制台將拋出:
無法編寫JSON:沒有為類org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer找到序列化器,也沒有發現創建BeanSerializer的屬性(為避免異常,請禁用SerializationFeature.FAIL_ON_EMPTY_BEANS); 嵌套的異常是com.tfasterxml.jackson.databind.JsonMappingException:沒有為org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer類找到序列化器,也沒有發現創建BeanSerializer的屬性(為避免異常,請禁用SerializationFeature.FAIL_ON_EMPTY_BEANS)(通過引用鏈) :java.util.ArrayList [0]-> com.spring.demo.entity.Client [“ country”]-> com.spring.demo.entity.Country _ $$ _ jvst9a4_1 [“ handler”])
我了解傑克遜曾嘗試序列化國家/地區bean,但我的目的是僅獲取Client的主要參數,而不獲取其關系。
PD:我已經做了控制台要求我執行的操作:
ObjectMapper objMapper = new ObjectMapper();
objMapper .configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false);
return objMapper .writeValueAsString(listOnlyClient)
這是可行的,但是作為第一種方法,因此不是一種選擇。
感謝前進。
嘗試添加以下內容:
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new Hibernate5Module());
根據您的版本休眠挑Hibernate5Module/4 or 3
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