2011-10-05 84 views
0

我有兩個分類。第一堂課是TNota。EclipseLink中的CASCADETYPE JPA

@Entity 
@Table(name = "t_nota") 
public class TNota implements Serializable {  
@Id 
@SequenceGenerator(name="seq_t_nota", sequenceName="seq_t_nota", initialValue=37, allocationSize=1) 
@GeneratedValue(generator="seq_t_nota") 
@Basic(optional = false) 
@Column(name = "id_nota", nullable = false) 
private double idNota; 

@Basic(optional = false) 
@Column(name = "nota", nullable = false, length = 255) 
private String nota; 

@JoinColumn(name = "id_tipo_nota", referencedColumnName = "id", nullable = false) 
@ManyToOne(optional = false) 
private NTipoNota nTipoNota; 

public TNota() { 
} 

public TNota(Long idNota) { 
    this.idNota = idNota; 
} 

public double getIdNota() { 
    return idNota; 
} 

public void setIdNota(double idNota) { 
    this.idNota = idNota; 
} 

public String getNota() { 
    return nota; 
} 

public void setNota(String nota) { 
    this.nota = nota; 
} 

public NTipoNota getNTipoNota() { 
    return nTipoNota; 
} 

public void setNTipoNota(NTipoNota nTipoNota) { 
    this.nTipoNota = nTipoNota; 
} 
} 

和其他類NtipoNota ..

@Entity 
@Table(name = "n_tipo_nota") 
public class NTipoNota implements Serializable { 
@Id 
@Basic(optional = false) 
@Column(name = "id", nullable = false) 
private Integer id; 
@Basic(optional = false) 
@Column(name = "nombre", nullable = false, length = 255) 
private String nombre; 
@OneToMany(cascade = CascadeType.ALL, mappedBy = "nTipoNota",fetch=FetchType.LAZY) 
private List<TNota> tNotaList; 

public NTipoNota() { 
} 

public NTipoNota(Integer id) { 
    this.id = id; 
} 

public Integer getId() { 
    return id; 
} 

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

public String getNombre() { 
    return nombre; 
} 

public void setNombre(String nombre) { 
    this.nombre = nombre; 
}  

public List<TNota> getTNotaList() { 
    return tNotaList; 
} 

public void setTNotaList(List<TNota> tNotaList) { 
    this.tNotaList = tNotaList; 
} 
} 

我擁有所有類型的存儲在數據庫中的筆記。我只是想堅持一個新的TNota如下,但我得到一個錯誤,因爲它堅持一個新的NTipoNota與id = 5已經存在於數據庫中。使用的TopLink我從未有過這樣的煩惱:

TNota note = new TNota(); 
note.setNota("Hola mundo"); 
note.setNTipoNota(new NTipoNota(5)); 
manager.persist(note); 

我確定如下:

TNota note = new TNota(); 
note.setNota("Hola mundo"); 
note.setNTipoNota(manager.find(NTipoNota.class, 5); 
manager.persist(note); 

我想不會有所有代碼變化,由於這個問題。有什麼形式可以讓我們在創建對象的新實例時不會堅持對象?

謝謝大家。

回答

1

你的新代碼是正確的,而你以前的代碼是不正確的。您也可以調用合併來解決關係。