0

我遇到了这个问题，这有点接近这个问题，但是当我完成所有步骤后，我仍然有这样一个例外：

org.codehaus.jackson.map.JsonMappingException: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer 也没有发现创建 BeanSerializer 的属性（为避免异常，禁用 SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS））（通过引用链: java.util.ArrayList[0]->com.myPackage.SomeEntity["mainEntity"]->com.myPackage.MainEntity["subentity1"]->com.myPackage.Subentity1_$$_javassist_8["handler"])

这是我的实体的代码：

@JsonAutoDetect
public class MainEntity {

    private Subentity1 subentity1;

    private Subentity2 subentity2;

    @JsonProperty
    public Subentity1 getSubentity1() {
        return subentity1;
    }

    public void setSubentity1(Subentity1 subentity1) {
        this.subentity1 = subentity1;
    }

    @JsonProperty
    public Subentity2 getSubentity2() {
        return subentity2;
    }

    public void setSubentity2(Subentity2 subentity2) {
        this.subentity2 = subentity2;
    }
}



@Entity
@Table(name = "subentity1")
@JsonAutoDetect
public class Subentity1 {

    @Id
    @Column(name = "subentity1_id")
    @GeneratedValue
    private Long id;

    @Column(name = "name", length = 100)
    private String name;

    @JsonIgnore
    @OneToMany(mappedBy = "subentity1")
    private List<Subentity2> subentities2;

     @JsonProperty
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    @JsonProperty
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
    //here I didin't add @JsonProperty, cause it leads to cycling during serialization
    public List<Subentity2> getSubentity2s() {
        return subentity2s;
    }

    public void setSubentity2s(List<Subentity2> subentity2s) {
        this.subentity2s = subentity2s;
    }

}

@Entity
@Table(name = "subentity2")
@JsonAutoDetect
public class Subentity2 {
    @Id
    @Column(name = "subentity2_id")
    @GeneratedValue
    private Long id;

    @Column(name = "name", length = 50)
    private String name;

    @ManyToOne
    @JoinColumn(name = "subentity1_id")
    private Subentity1 subentity1;

    @JsonProperty
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    @JsonProperty
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @JsonProperty
    public Subentity1 getSubentity1() {
        return subentity1;
    }

    public void setSubentity1(Subentity1 subentity1) {
        this.subentity1 = subentity1;
    }

这是我的转换方法的代码：

 private String toJSON(Object model) {
        ObjectMapper mapper = new ObjectMapper();
        String result = "";
        try {
            result = mapper.writeValueAsString(model);
        } catch (JsonGenerationException e) {
            LOG.error(e.getMessage(), e);
        } catch (JsonMappingException e) {
            LOG.error(e.getMessage(), e);
        } catch (IOException e) {
            LOG.error(e.getMessage(), e);
        }
        return result;
    }

我将非常感谢任何帮助、建议或代码:)

UPD

另外，我忘了从我的控制器中添加一段代码：

String result = "";
        List<SomeEntity> entities = someEntityService.getAll();
        Hibernate.initialize(entities);
        for (SomeEntity someEntity : entities) {
            Hibernate.initialize(someEntity.mainEntity());
            Hibernate.initialize(someEntity.mainEntity().subentity1());
            Hibernate.initialize(someEntity.mainEntity().subentity2());
        }
        result = this.toJSON(entities);

我不能忽略任何字段，因为我需要它们

4

3 回答 3

1

基本上，您的一些字段被包装到惰性休眠代理中。在序列化您的对象之前调用Hibernate.initialize(model)，它将加载您的惰性集合和引用。

但我不会混合使用数据库和视图模型，这是一种不好的做法。为您的 RESTful 模型创建一组类，并在序列化之前将数据库实体转换为它们。

于 2012-11-19T15:44:56.780 回答
0

我创建了一个带有平面字段（字符串、布尔值、双精度等）的 Bean，它们在我的类中，并为转换创建了方法

于 2012-11-20T08:24:27.590 回答
0

如果您使用延迟加载，请添加此

@JsonIgnoreProperties({"hibernateLazyInitializer", "handler"})
于 2017-05-03T15:16:26.643 回答