Question

我正在设置多对多关系，我的联接表JoinTable具有带有两个键的复合Pk，问题是当我尝试保持实体休眠状态时，我总是说他无法转换java.lang.Long设置为所需的类型（User或Weeks）

我尝试将属性更改为Java原语（分别为long和int），但是随后休眠无法将此值设置为目标类型（对象）。

我已经将这种关系映射为：

User.java

@Column(name="id", nullable=false)  
@Id 
@GeneratedValue(generator="USER_ID_GENERATOR")  
@org.hibernate.annotations.GenericGenerator(name="USER_ID_GENERATOR", strategy="native")    
private long id;

@OneToMany(mappedBy="user", targetEntity=com.ti.project.Model.JoinTable.class)  @org.hibernate.annotations.Cascade({org.hibernate.annotations.CascadeType.SAVE_UPDATE, org.hibernate.annotations.CascadeType.LOCK}) 
@org.hibernate.annotations.LazyCollection(org.hibernate.annotations.LazyCollectionOption.TRUE)  
private java.util.Set jointable = new java.util.HashSet();

Weeks.java

@Column(name="week_id", nullable=false) 
@Id 
@GeneratedValue(generator="WEEK_ID_GENERATOR")  
@org.hibernate.annotations.GenericGenerator(name="WEEK_ID_GENERATOR", strategy="native")    
private int week_id;

@OneToMany(mappedBy="week",targetEntity=com.ti.project.Model.JoinTable.class)   
@org.hibernate.annotations.Cascade({org.hibernate.annotations.CascadeType.SAVE_UPDATE, org.hibernate.annotations.CascadeType.LOCK}) 
@org.hibernate.annotations.LazyCollection(org.hibernate.annotations.LazyCollectionOption.TRUE)  
private java.util.Set jointable = new java.util.HashSet();

JoinTable.java

@PrimaryKeyJoinColumn       
@ManyToOne(targetEntity=com.ti.project.Model.User.class, fetch=FetchType.LAZY)  
@org.hibernate.annotations.Cascade({org.hibernate.annotations.CascadeType.LOCK})    
@JoinColumns(value={ @JoinColumn(name="user_id", referencedColumnName="id", nullable=false) }, foreignKey=@ForeignKey(name="FKuser_su465470"))  
private com.ti.project.Model.User user;

@Column(name="user_id", nullable=false, insertable=false, updatable=false)  
@Id     
@org.hibernate.annotations.GenericGenerator(name="USERID_GENERATOR", strategy="foreign", parameters=@org.hibernate.annotations.Parameter(name="property", value="user"))    
private long userId;

@PrimaryKeyJoinColumn   
@ManyToOne(targetEntity=com.ti.project.Model.Weeks.class, fetch=FetchType.LAZY) 
@org.hibernate.annotations.Cascade({org.hibernate.annotations.CascadeType.LOCK})    
@JoinColumns(value={ @JoinColumn(name="week_id", referencedColumnName="week_id", nullable=false) }, foreignKey=@ForeignKey(name="FKweek_su338979")) 
private com.ti.project.Model.Weeks week;

@Column(name="week_id", nullable=false, insertable=false, updatable=false)  
@Id 
@GeneratedValue(generator="WEEKID_GENERATOR")   
@org.hibernate.annotations.GenericGenerator(name="WEEKID_GENERATOR", strategy="foreign", parameters=@org.hibernate.annotations.Parameter(name="property", value="week"))    
private int weekId;

JoinTablePK.java

@ManyToOne(targetEntity=com.ti.project.Model.User.class, fetch=FetchType.LAZY)  
@org.hibernate.annotations.Cascade({org.hibernate.annotations.CascadeType.LOCK})    
@JoinColumns(value={ @JoinColumn(name="user_id", referencedColumnName="id", nullable=false) }, foreignKey=@ForeignKey(name="FKuser_su465470"))  
private com.ti.project.Model.User user;

@ManyToOne(targetEntity=com.ti.project.Model.Weeks.class, fetch=FetchType.LAZY) 
@org.hibernate.annotations.Cascade({org.hibernate.annotations.CascadeType.LOCK})    
@JoinColumns(value={ @JoinColumn(name="week_id", referencedColumnName="week_id", nullable=false) }, foreignKey=@ForeignKey(name="FKweek_su338979")) 
private com.ti.project.Model.Weeks week;

这是生成的异常：

Failed to convert property value of type 'java.lang.Integer' to required type 'com.ti.project.Model.Weeks' for property 'week'; nested exception is java.lang.IllegalStateException: Cannot convert value of type 'java.lang.Integer' to required type 'com.ti.project.Model.Weeks' for property 'week': no matching editors or conversion strategy found

我希望保留这个JoinTable实体，但是这种例外使我发疯了，知道吗？

Answer 1

您的映射都是错误的。您具有@OneToMany和@ManyToOne映射。为什么使用中间表？您只需要@ManyToMany关系中的中间表。删除JoinTable.class，因为您的代码中未引用它。

另一个问题是您的targetClass。查看如何映射@OneToMany和@ManyToOne关系。一定要容易。

https://www.baeldung.com/hibernate-one-to-many

祝你好运！

无法将类型'java.lang.Integer'的属性值转换为所需的类型

1 个答案: