[英]Jpa mapping of composite key
Hi Below is my tables and its desc. 嗨,下面是我的桌子和它的desc。
Employee:
emp_Id primary_key
emp_Name
emp_Address
Address:
emp_id -- foreign key of employee table
addres_type
type_id
AddressType:
Type_id -- foreign key of Address table
Type
Desc
How can i create JPA entities with this relationship. 如何使用此关系创建JPA实体。
Thanks in advance 提前致谢
you'd better use AbstractPersistable to generate the id : 你最好使用AbstractPersistable来生成id:
so Address and AddressType will have a primary_key also 所以Address和AddressType也会有一个primary_key
1-Employee 1-雇员
@Entity
@Table(name = "Employee ")
public class Employee extends AbstractPersistable<Long>{
// Employee here have a generated id from AbstractPersistable
private static final long serialVersionUID = 1L;
@Column(name = "emp_Name ")
private String emp_Name ;
@ManyToOne(optional = true)
@JoinColumn(name = "emp_Address")
private Address emp_Address;
}
2-Address 2,地址
@Entity
@Table(name = "Address")
public class Address extends AbstractPersistable<Long>{
//Address here have a generated id from AbstractPersistable
private static final long serialVersionUID = 1L;
//you can delete this column because you have already the ManyToOne in Employee
@ManyToOne(optional = true)
@JoinColumn(name = "emp_id")
private Employee emp_id;
//*****************
@Column(name = "addres_type")
private String addres_type;
@ManyToOne(optional = true)
@JoinColumn(name = "Type_id")
private AddressType Type_id ;
}
3-AddressType 3地址类型
@Entity
@Table(name = "AddressType")
public class AddressType extends AbstractPersistable<Long>{
//AddressType here have a generated id from AbstractPersistable
private static final long serialVersionUID = 1L;
@Column(name = "Type")
private String Type;
@Column(name = "Desc")
private String Desc;
}
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