休眠：使用複雜的類作爲屬性

Question

我努力做到以下幾點：

public class Distance { 

private Course courseA, courseB; 
private int minDistance; 
double cost; 


private Long id; 


public Distance() { 
    super(); 
} 

public Distance(Course courseA, Course courseB, int minDistance, double cost) { 
    super(); 
    this.courseA = courseA; 
    this.courseB = courseB; 
    this.minDistance = minDistance; 
    this.cost = cost; 
} 


@Override 
public String toString() { 
    return "Distance [courseA=" + courseA + ", courseB=" + courseB 
      + ", MinDistance=" + minDistance + ", Cost=" + cost + "]"; 
} 

public Course getCourseA() { 
    return courseA; 
} 

public void setCourseA(Course courseA) { 
    this.courseA = courseA; 
} 

public Course getCourseB() { 
    return courseB; 
} 

public void setCourseB(Course courseB) { 
    this.courseB = courseB; 
} 

public int getMinDistance() { 
    return minDistance; 
} 

public void setMinDistance(int minDistance) { 
    this.minDistance = minDistance; 
} 

public double getCost() { 
    return cost; 
} 

public void setCost(double cost) { 
    this.cost = cost; 
} 

@Id 
public Long getId() { 
    return id; 
} 

public void setId(Long id) { 
    this.id = id; 
} 

} 

凡課程是另一個類我創建：

public class Course { 


private Long id; 

private String name; 

private Calendar date; 

public Course() { 
    super(); 
} 
public Course(Long id,String name, Calendar date) { 
    super(); 
    this.id = id; 
    this.name = name; 
    this.date = date; 
} 

@Override 
public String toString() { 
    return "Course [id=" + id + ", name=" + name + ", examDate=" + date 
      + "]"; 
} 

@Id 
public Long getId() { 
    return id; 
} 
public void setId(Long id) { 
    this.id = id; 
} 
public String getName() { 
    return name; 
} 
public void setName(String name) { 
    this.name = name; 
} 
public Calendar getDate() { 
    return date; 
} 
public void setDate(Calendar date) { 
    this.date = date; 
} 
/* (non-Javadoc) 
* @see java.lang.Object#hashCode() 
*/ 
@Override 
public int hashCode() { 
    final int prime = 31; 
    int result = 1; 
    result = prime * result + ((id == null) ? 0 : id.hashCode()); 
    return result; 
} 
/* (non-Javadoc) 
* @see java.lang.Object#equals(java.lang.Object) 
*/ 
@Override 
public boolean equals(Object obj) { 
    if (this == obj) { 
     return true; 
    } 
    if (obj == null) { 
     return false; 
    } 
    if (!(obj instanceof Course)) { 
     return false; 
    } 
    Course other = (Course) obj; 
    if (id == null) { 
     if (other.id != null) { 
      return false; 
     } 
    } else if (!id.equals(other.id)) { 
     return false; 
    } 
    return true; 
} 

}

我試圖定義courseA和B作爲「distance.hbm.xml」中距離的屬性，但只是對我大喊一個例外：org.hibernate.MappingException: Could not determine type for: database.datatypes.Course at table:distances...
我曾嘗試聲明courseA a nd B作爲組件，它「成功」，但當我調用session.load(Distance.class,1L)時，它返回了正確的對象，但兩個過程都是空指針。

我該如何定義它？

另外，我怎麼可以這樣做，但是從圖書館綜合類（比如東西出來java.util中的）

謝謝！

更新： 我找到了方法，我可以在距離課程中繞過它，但是有一些重要的事情要處理：課程必須有一個日期對象。我寧願使用java.util.Calendar，但如果這是有問題的，有任何其他方式有一個日期，我可以使用？

再次感謝！

Answer 1

可以實現它的方式如下：

@Entity 
@Table(name="distance") 
public class Distance { 

private Course courseA, courseB; 

@Embedded 
public getCourseA(){ 
return this.courseA; 
} 

@Embedded 
public getCourseB(){ 
return this.courseB; 
} 

} 

現在嵌入類：

@Embeddable 
public class Address implements Serializable{ 

@Transient 
public Long getId() { 
    return id; 
} 

@Column 
public String getName() { 
    return name; 
} 

} 

@Embaddable類不是實體，所以它不應該有任何的主鍵。這就是爲什麼你應該但@Transient在編號屬性