Question

我有这两个实体：

@Entity
@Table(name = "BA_USER", schema = "MYSCHEMA")
@XmlType(namespace = BusinessOperations.NAMESPACE)
@JsonIgnoreProperties(ignoreUnknown = true)
public class UserWeb implements Serializable, UserDetails, SecurityUserInfo, Auditable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "userGenerator")
    @SequenceGenerator(name="userGenerator", sequenceName="SEQ_USER")
    @Column(name = "USERID", unique = true, nullable = false, precision = 22, scale = 0)
    private long userid;

    @ManyToOne(targetEntity = Country.class, fetch = FetchType.LAZY)
    @JoinColumn(name = "COUNTRYID")
    private Country country;

    @ManyToOne(targetEntity = Menu.class, fetch = FetchType.LAZY)
    @JoinColumn(name = "MENUID")
    private Menu menu;

    //Getters and Setters
}

和

@Entity
@Table(name = "BA_PREMIUMUSER", schema = "MYSCHEMA")
public class PremiumUser implements java.io.Serializable {

    @Id
    @OneToOne(targetEntity = UserWeb.class, fetch = FetchType.LAZY)
    @JoinColumn(name = "USERID")
    private UserWeb userWeb;

    @ManyToOne(targetEntity = Customer.class, fetch = FetchType.LAZY)
    private Customer customer;

    //Getters and Setters
}

然后，在我的DAO中，我有以下方法：

public List<PremiumUser> findAllUsers(Long userid, Long customerId, Menu menu) {

    Criteria criteria = getSession().createCriteria(PremiumUser.class);

    criteria.add(Restrictions.eq("clienteProsegur", customerId));
    criteria.createAlias("userWeb", "userweb", CriteriaSpecification.INNER_JOIN);
    criteria.add(Restrictions.eq("userweb.menu", menu));

    return criteria.list();
}

Hibernate正在创建以下查询：

select this_.USERID as USERID52_0_, this_.CUSTOMER as CUSTOMER2_52_0_ 
from MYSCHEMA.BA_PREMIUMUSER this_ 
where userweb1_.MENUID=2 and this_.CUSTOMERID=3

我希望createAlias方法在查询中创建一个join语句，比如_

select this_.USERID as USERID52_0_, this_.CUSTOMER as CUSTOMER2_52_0_ 
from MYSCHEMA.BA_PREMIUMUSER this_ 
inner join MYSCHEMA.BA_USER userweb1_ on this_.USERID = userweb1.USERID
where userweb1_.MENUID=2 and this_.CUSTOMERID=3

我正在使用此版本的hibernate：Hibernate-core-3.6.0.Final

我想我在这里遗漏了一些东西。

Answer 1

在这种情况下，您应该使用方法链接，这可能是您遇到此问题的原因：

public List<PremiumUser> findAllUsers(Long userid, Long customerId, Menu menu)
  return getSession().createCriteria( PremiumUser.class )
         .add( Restrictions.eq( "clienteProsegur", customerId ) )
         .createAlias( "userWeb", "userweb", CriteriaSpecification.INNER_JOIN )
         .add( Restrictions.eq( "userweb.menu", menu ) )
         .list();
}

Hibernate Criteria.createAlias没有连接表

1 个答案: