JPA：額外的屬性映射多對多關係

Question

我有很多一對多

employeeId skillSetId numberOfYears 
10   101   2 

我是新來的JPA，無法定義實體Employee和SkillSet表之間的關係，附加列numberOfYears每個關係與關係。我應該爲Employee_SkillSet表定義一個新的實體類嗎？或者我可以在Employee和SkillSet課程中定義多對多關係？我在哪裏指定numberOfYears列？

編輯：似乎重複，但我明確要求使用@IdClass，其中一個實體是@MappedSuperclass，因此必須同時定義ID實例和被引用實體對象。

Answer 1

由於您需要爲元組（Employee，SkillSet）添加一個字段，因此您必須創建另一個實體。

@Entity 
public class Employee implements Serializable { 
    private @Id Long id; 

    @OneToMany(mappedBy="employee") 
    private List<EmployeeSkillSet> skillSets; 
} 

@Entity 
public class SkillSet implements Serializable { 
    private @Id Long id; 
} 

@Entity 
public class EmployeeSkillSet implements Serializable { 
    private @Id Long id; 
    private @ManyToOne Employee employee; 
    private @ManyToOne SkillSet skillSet; 
    private @Basic int numberOfYears; 
} 

當然你也可以選擇使用@IdClass，使（「僱員」，「技能」）的EmployeeSkillSet像這樣的主鍵：

@Entity @IdClass(EmployeeSkillSet.Key.class) 
public class EmployeeSkillSet implements Serializable { 

    private @Id @ManyToOne Employee employee; 
    private @Id @ManyToOne SkillSet skillSet; 
    private @Basic int numberOfYears; 

    public static class Key implements Serializable { 
     private Long employee; // plus getter+setter 
     private Long skillSet; // plus getter+setter 
     // plus hashCode, equals 
    } 
} 

Answer 2

雖然可以使用@ManyToMany註釋爲了模擬多對多關係，更好地考慮性能原因定義了兩對多關係。

您將需要4件文物，有

• 僱員實體
• 技能實體
• EmployeeSkillSet關聯實體（在這裏你可以指定numberOfYears列）
• EmployeeSkillSetPK（用於EmployeeSkillSet關聯實體的主鍵）

代碼會是這樣的

僱員

package <your_package>; 
import javax.persistence.Entity; 
import javax.persistence.GeneratedValue; 
import javax.persistence.GenerationType; 
import javax.persistence.Id; 
import javax.persistence.Table; 

/** 
* 
*/ 
@Entity 
@Table(name="EMPLOYEE") 
public class Employee implements Serializable { 

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

    // Rest of columns and getter and setters for all 
} 

技能

package <your_package>; 
import javax.persistence.Entity; 
import javax.persistence.GeneratedValue; 
import javax.persistence.GenerationType; 
import javax.persistence.Id; 
import javax.persistence.Table; 

/** 
* 
*/ 
@Entity 
@Table(name="SKILLSET") 
public class SkillSet implements Serializable { 

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

    // Rest of columns and getter and setters for all 
} 

EmployeeSkillSetPK

/** 
* 
*/ 
package <your_package>; 

import java.io.Serializable; 

import javax.persistence.Embeddable; 

/** 
* 
*/ 
@Embeddable 
public class EmployeeSkillSetPK implements Serializable { 

    @ManyToOne 
    private Employee emp; 

    @ManyToOne 
    private SkillSet sk; 
    /** 
    * @return the employee 
    */ 
    public Employee getEmployee() { 
     return emp; 
    } 
    /** 
    * @param employee the employee to set 
    */ 
    public void setEmployee(Employee employee) { 
     this.employee = employee; 
    } 
    /** 
    * @return the sk 
    */ 
    public SkillSet getSkillSet() { 
     return sk; 
    } 
    /** 
    * @param sk the sk to set 
    */ 
    public void setSkillSet(SkillSet sk) { 
     this.sk = sk; 
    } 

    public boolean equals(Object o) { 
     if (this == o) { 
      return true; 
     } 

     if (o == null || getClass() != o.getClass()) { 
      return false; 
     } 

     EmployeeSkillSetPK that = (EmployeeSkillSetPK) o; 

     if (employee != null ? !employee.equals(that.employee) : that.employee != null) { 
      return false; 
     } 
     if (sk != null ? !sk.equals(that.sk) : that.sk != null) { 
      return false; 
     } 

     return true; 
    } 

    public int hashCode() { 
     int result; 
     result = (employee != null ? employee.hashCode() : 0); 
     result = 31 * result + (sk != null ? sk.hashCode() : 0); 
     return result; 
    } 
} 

EmployeeSkillSet

這是JPA 1.0，我現在不能測試的代碼，但它應該工作。

請注意，我自己寫了表名和列名。根據需要調整它。