Question

我有两个实体：目的地（密钥id1，密钥id2dept和非密钥名称1）和服务（密钥id3和非密钥名称2）。

当我使用目标 - ＆gt; Service

在Destination中使用OneToMany生成架构时

@OneToMany(cascade=ALL, fetch=FetchType.EAGER)          
private Collection<Destination> d;

我收到一个只有两个键的Join表：Joinedtable（键id3，键id1）我期待Joinedtable（密钥id3，密钥id1，密钥id2）

使用JOIN的SQL知识，我理解如果基于id1 = id3加入，那么我应该收到 Joinedtable（id1，id2dept）。但即使不接受，我哪里出错？

@Entity
@Table(name = "BHU_DST")
public class TMP_Destb implements Serializable {
private long id1; 
private long id2dept; 
private char name;
private static final long serialVersionUID = 1L;        
//getter and setter
}

@Entity
@Table(name = "BHU_SVC")
public class TMP_Servb implements Serializable {
private long id2; 
private char name2;
private static final long serialVersionUID = 1L;    
@OneToMany(cascade=ALL, fetch=FetchType.EAGER)          
private Collection<TMP_Destb> d;
//getter and setters
}

public class TMP_DestinationPK  implements Serializable {      
private long id1;         
private long id2dept;
private static final long serialVersionUID = 1L;
//getters setters and hashcodes
}

 <?xml version="1.0" encoding="UTF-8"?>
<entity-mappings version="1.0" xmlns="http://java.sun.com/xml/ns/persistence/orm"      
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"          
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence/orm http://java.sun.com/xml/ns/persistence/orm_1_0.xsd">
<entity class="mypkg.jpa1.TMP_Destb" access="FIELD">
    <id-class class="TMP_DestinationPK"/>
    <attributes>
        <id name="id1">
        </id>
        <id name="id2dept">
        </id>
    </attributes>
</entity>
<entity class="mypkg.jpa1.TMP_Servb" access="FIELD">
    <attributes>
        <id name="id2">
        </id>
    </attributes>
</entity>

JPA - @ Onetomany和复合键

0 个答案: