首页 文章

Hibernate:Jointable加入特定的子表

提问于
浏览
0

我有两组继承表:1 . TableA(由TableA1和TableA2继承),2 . TableB(由TableB1和TableB2继承)

TableA的DiscriminatorColumn是'type',DiscriminatorValues分别是TableA1和TableA2的'A1'和'A2' . 类似地,TableB的DiscriminatorColumn是'type',DiscriminatorValues分别是TableB1和TableB2的'B1'和'B2' . 我有一个连接表TableA_B用于连接TableA和TableB . 我知道连接表TableA_B只能将TableA1链接到TableB1而TableA2只链接到TableB2,这根据'type'列进行 . 即,A1始终链接到类型B1,A2始终链接到类型B2 .

现在,我的问题是,当我从TableA1进行查询时,@ JoinTable生成连接TableA1,TableA,TableB,TableB1和TableB2的sql . 这里,从TableB2查询是不需要的 . 有没有办法限制这个?正如我所提到的,唯一的区别因素是“类型”列 .

TableA实体:

@Entity
@Table(name = "TableA")
@Inheritance(strategy=InheritanceType.JOINED)
@DiscriminatorColumn(name="type")
public class TableA {

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(unique = true, nullable = false)
    private Long id;
    private String name;
    private String type;

    @ManyToMany(fetch = FetchType.EAGER)
        @JoinTable(
        name="TableA_B",
        joinColumns={@JoinColumn(name="A_id")},
        inverseJoinColumns={@JoinColumn(name="B_id")}
        )
    private Set<DataMaster> inputData = new HashSet<DataMaster>(0);


//getters & setters ...
}

@Entity
@Table(name = "TableA1")
@PrimaryKeyJoinColumn(name="AId")
@DiscriminatorValue("A1")
public class TableA1 extends TableA{

    @Column(insertable=false, updatable=false)
    private Long AId;
    private String value1;
//getters & setters ...
}

@Entity
@Table(name = "TableA2")
@PrimaryKeyJoinColumn(name="AId")
@DiscriminatorValue("A2")
public class TableA2 extends TableA{

    @Column(insertable=false, updatable=false)
    private Long AId;
    private String value2;
//getters & setters ...
}

表B如下:

@Entity
@Table(name = "TableB")
@Inheritance(strategy=InheritanceType.JOINED)
@DiscriminatorColumn(name="type")
public class TableB {

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(unique = true, nullable = false)
    private Long id;
    private String name;
    private String type;
//getters & setters ...
}

@Entity
@Table(name = "TableB1")
@PrimaryKeyJoinColumn(name="BId")
@DiscriminatorValue("B1")
public class TableB1 extends TableB{
    @Column(insertable=false, updatable=false)
    private Long dataId;
    private String value1;
//getters & setters ...
}

@Entity
@Table(name = "TableB2")
@PrimaryKeyJoinColumn(name="BId")
@DiscriminatorValue("B2")
public class TableB2 extends TableB{
    @Column(insertable=false, updatable=false)
    private Long dataId;
    private String value2;
//getters & setters ...
}

DAO:

public TableA1 find(Long id) {
      Session session = DBSessionManager.getFactory().openSession();
      Transaction tx = null;
      TableA1 tableA1 = null;
      try{
         tx = session.beginTransaction();
         tableA1 = (TableA1)session.get(TableA1.class, id);
         tx.commit();
      }catch (HibernateException e) {
         if (tx!=null) tx.rollback();
      }finally {
         session.close(); 
      }     
      return tableA1;
}

我尝试使用FilterJoinTable:

@FilterJoinTable(name="dataTypeFilter", condition=":dataType = type")

在DAO,

session.enableFilter("dataTypeFilter").setParameter("dataType", "B1");

但是,这也没有给出预期的结果 .

这是我添加过滤器的方式:

@Entity
@Table(name = "TableA")
@Inheritance(strategy=InheritanceType.JOINED)
@DiscriminatorColumn(name="type")
@FilterDef(name="dataTypeFilter", parameters = {
        @ParamDef(name = "dataType", type = "string")
        })
public class TableA {

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(unique = true, nullable = false)
    private Long id;
    private String name;
    private String type;

    @ManyToMany(fetch = FetchType.EAGER)
        @JoinTable(
        name="TableA_B",
        joinColumns={@JoinColumn(name="A_id")},
        inverseJoinColumns={@JoinColumn(name="B_id")}
        )
    @FilterJoinTable(name="dataTypeFilter", condition=":dataType = type")
    private Set<DataMaster> inputData = new HashSet<DataMaster>(0);


//getters & setters ...
}

DAO:

public TableA1 find(Long id) {
          Session session = DBSessionManager.getFactory().openSession();
          Transaction tx = null;
          TableA1 tableA1 = null;
          try{
             tx = session.beginTransaction();
         session.enableFilter("dataTypeFilter").setParameter("dataType", "B1");
             tableA1 = (TableA1)session.get(TableA1.class, id);
             tx.commit();
          }catch (HibernateException e) {
             if (tx!=null) tx.rollback();
          }finally {
             session.close(); 
          }     
          return tableA1;
    }

1 回答

  • 0

    根据您的描述,您应该在TableA1和TableB1之间 Build ManyToMany关联,并在TableA2和TableB2之间 Build 另一个ManyToMany关联 . 但是您将这两个关联映射为TableA和TableB之间的单个关联 . 您应该将这两个关联映射为它们的实际内容:

    private class TableA1 extends TableA {
        @ManyToMany
        private Set<TableB1> tablesB1s;
    }
    
    private class TableA2 extends TableA {
        @ManyToMany
        private Set<TableB2> tablesB2s;
    }
    

    我不会使用相同的连接表来映射两个不同的关联 . 我甚至不确定它是否有可能 .

相关问题