关系Hibernate

Question

孩子：@EDIT感谢@MithatKonuk

@Entity
@Table(name="child")
public class Child {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name="ID")
    private int id;

    @Column(name="FirstName")
    private String firstName;

    @Column(name="SecondName")
    private String secondName;

    @Column(name="PESEL")
    private String PESEL;

    @Column(name="Sex")
    private String sex;

  @JsonBackReference
  @ManyToOne(cascade = CascadeType.ALL)
  @JoinColumn(name="familyid",referencedColumnName = "id")
  private Family family;

和家人：@EDIT感谢@米塔科努克

@Entity
@Table(name="family")
public class Family {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "ID")
    private int id;



    @OneToOne(cascade = CascadeType.ALL)
    private Father father;

    @JsonManagedReference
    @OneToMany(mappedBy = "family",fetch=FetchType.EAGER,cascade = CascadeType.ALL)
    private List<Child> childList;

我想得到作为家庭一部分的孩子的名单。我尝试进行查询：@编辑感谢@MithatKonuk

@Transactional
public List<Child> findAllChild(int id) {
    // TODO Auto-generated method stub
    HibernateUtil.initManager();
    HibernateUtil.getEntityManager().clear();
    Query query = HibernateUtil.getEntityManager().createQuery("SELECT child FROM Child child FETCH JOIN child.family u WHERE u.id = :userId ");
    query.setParameter("userId",id);
    List<Child> result = query.getResultList();
    HibernateUtil.shutdown();
    return result;
}

当我尝试运行它时，总是会出现错误：

Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: expecting "all", found 'JOIN' near line 1, column 64 [SELECT child FROM com.example.Family3.domain.Child child FETCH JOIN child.family u WHERE u.id = :userId ]
    at org.hibernate.hql.internal.ast.QuerySyntaxException.convert(QuerySyntaxException.java:74) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.hql.internal.ast.ErrorTracker.throwQueryException(ErrorTracker.java:93) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.parse(QueryTranslatorImpl.java:296) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:188) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:143) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:119) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:80) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:153) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:595) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:704) ~[hibernate-core-5.3.5.Final.jar:5.3.5.Final]
    ... 9 common frames omitted

我需要你的帮助。

@编辑好的，所以我将此代码更新为完整的实体。你还需要更多吗？我很抱歉，也许这很容易，但是我正在学习Hibernate，我不知道这个cos出了什么问题。谢谢你的回答。

Answer 1

JOIN应该在FETCH之前，试试这个查询：

"SELECT child FROM Child child JOIN FETCH child.family WHERE child.family.id = :userId"

Answer 2

首先您需要更改查询

@Transactional
public List<Child> findAllChild(int id) {
    // TODO Auto-generated method stub
    HibernateUtil.initManager();
    HibernateUtil.getEntityManager().clear();
    Query query = HibernateUtil.getEntityManager().createQuery("SELECT child FROM Child child JOIN FETCH child.family  WHERE child.family.id = :userId ");
    query.setParameter("userId",id);
    List<Child> result = query.getResultList();
    HibernateUtil.shutdown();
    return result;
}

第二个是请在每个类中添加@Entity和@Table（“孩子”）和@Table（“家庭”），如下所示

@Entity
@Table("child table name")
public class Child{}

@Entity
@Table("family table name")
public class Family{}