当尝试将具有双向关联的JPA对象转换为JSON时,我不断得到

org.codehaus.jackson.map.JsonMappingException: Infinite recursion (StackOverflowError)

我所找到的是这个帖子,基本上是建议避免双向关联。有人有解决这个春季bug的方法吗?

------ edit 2010-07-24 16:26:22 -------

代码片段:

业务对象1:

@Entity
@Table(name = "ta_trainee", uniqueConstraints = {@UniqueConstraint(columnNames = {"id"})})
public class Trainee extends BusinessObject {

    @Id
    @GeneratedValue(strategy = GenerationType.TABLE)
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(name = "name", nullable = true)
    private String name;

    @Column(name = "surname", nullable = true)
    private String surname;

    @OneToMany(mappedBy = "trainee", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @Column(nullable = true)
    private Set<BodyStat> bodyStats;

    @OneToMany(mappedBy = "trainee", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @Column(nullable = true)
    private Set<Training> trainings;

    @OneToMany(mappedBy = "trainee", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @Column(nullable = true)
    private Set<ExerciseType> exerciseTypes;

    public Trainee() {
        super();
    }

    //... getters/setters ...
}

业务对象2:

import javax.persistence.*;
import java.util.Date;

@Entity
@Table(name = "ta_bodystat", uniqueConstraints = {@UniqueConstraint(columnNames = {"id"})})
public class BodyStat extends BusinessObject {

    @Id
    @GeneratedValue(strategy = GenerationType.TABLE)
    @Column(name = "id", nullable = false)
    private Integer id;

    @Column(name = "height", nullable = true)
    private Float height;

    @Column(name = "measuretime", nullable = false)
    @Temporal(TemporalType.TIMESTAMP)
    private Date measureTime;

    @ManyToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name="trainee_fk")
    private Trainee trainee;
}

控制器:

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.ResponseBody;

import javax.servlet.http.HttpServletResponse;
import javax.validation.ConstraintViolation;
import java.util.*;
import java.util.concurrent.ConcurrentHashMap;

@Controller
@RequestMapping(value = "/trainees")
public class TraineesController {

    final Logger logger = LoggerFactory.getLogger(TraineesController.class);

    private Map<Long, Trainee> trainees = new ConcurrentHashMap<Long, Trainee>();

    @Autowired
    private ITraineeDAO traineeDAO;
     
    /**
     * Return json repres. of all trainees
     */
    @RequestMapping(value = "/getAllTrainees", method = RequestMethod.GET)
    @ResponseBody        
    public Collection getAllTrainees() {
        Collection allTrainees = this.traineeDAO.getAll();

        this.logger.debug("A total of " + allTrainees.size() + "  trainees was read from db");

        return allTrainees;
    }    
}

学员DAO的jpa实现:

@Repository
@Transactional
public class TraineeDAO implements ITraineeDAO {

    @PersistenceContext
    private EntityManager em;

    @Transactional
    public Trainee save(Trainee trainee) {
        em.persist(trainee);
        return trainee;
    }

    @Transactional(readOnly = true)
    public Collection getAll() {
        return (Collection) em.createQuery("SELECT t FROM Trainee t").getResultList();
    }
}

persistence . xml

<persistence xmlns="http://java.sun.com/xml/ns/persistence"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd"
             version="1.0">
    <persistence-unit name="RDBMS" transaction-type="RESOURCE_LOCAL">
        <exclude-unlisted-classes>false</exclude-unlisted-classes>
        <properties>
            <property name="hibernate.hbm2ddl.auto" value="validate"/>
            <property name="hibernate.archive.autodetection" value="class"/>
            <property name="dialect" value="org.hibernate.dialect.MySQL5InnoDBDialect"/>
            <!-- <property name="dialect" value="org.hibernate.dialect.HSQLDialect"/>         -->
        </properties>
    </persistence-unit>
</persistence>

当前回答

重点是将@JsonIgnore放在setter方法中,如下所示。对我来说。

Township.java

@Access(AccessType.PROPERTY)
@OneToMany(fetch = FetchType.LAZY)
@JoinColumn(name="townshipId", nullable=false ,insertable=false, updatable=false)
public List<Village> getVillages() {
    return villages;
}

@JsonIgnore
@Access(AccessType.PROPERTY)
public void setVillages(List<Village> villages) {
    this.villages = villages;
}

Village.java

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "townshipId", insertable=false, updatable=false)
Township township;

@Column(name = "townshipId", nullable=false)
Long townshipId;

其他回答

您可以使用@JsonIgnore来打破循环(引用)。

您需要导入org.codehaus.jackson. annotation. jsonignore(遗留版本)或com.fasterxml.jackson.annotation.JsonIgnore(当前版本)。

我也遇到了同样的问题。我使用@JsonIdentityInfo的ObjectIdGenerators.PropertyGenerator.class生成器类型。

这就是我的解决方案:

@Entity
@Table(name = "ta_trainee", uniqueConstraints = {@UniqueConstraint(columnNames = {"id"})})
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
public class Trainee extends BusinessObject {
...

我有这个问题,但我不想在我的实体中使用注释,所以我通过为我的类创建一个构造函数来解决,这个构造函数必须没有对引用这个实体的实体的引用。假设这种情况。

public class A{
   private int id;
   private String code;
   private String name;
   private List<B> bs;
}

public class B{
   private int id;
   private String code;
   private String name;
   private A a;
}

如果你试图用@ResponseBody发送类B或A到视图,可能会导致一个无限循环。您可以在类中编写构造函数,并像这样使用entityManager创建查询。

"select new A(id, code, name) from A"

这是带有构造函数的类。

public class A{
   private int id;
   private String code;
   private String name;
   private List<B> bs;

   public A(){
   }

   public A(int id, String code, String name){
      this.id = id;
      this.code = code;
      this.name = name;
   }

}

然而,这个解决方案有一些限制,正如你可以看到的,在构造函数中我没有引用List bs,这是因为Hibernate不允许它,至少在3.6.10版本中。最后,所以当我需要在一个视图中显示两个实体时,我做以下工作。

public A getAById(int id); //THE A id

public List<B> getBsByAId(int idA); //the A id.

这个解决方案的另一个问题是,如果您添加或删除一个属性,您必须更新构造函数和所有查询。

@JsonIgnoreProperties是答案。

使用如下代码::

@OneToMany(mappedBy = "course",fetch=FetchType.EAGER)
@JsonIgnoreProperties("course")
private Set<Student> students;

非常重要:如果你正在使用LOMBOK,使用shure排除集合的属性,如Set, List等…

是这样的:

@EqualsAndHashCode(exclude = {"attributeOfTypeList", "attributeOfTypeSet"})