休眠:外键的列数错误
我在两个实体类 User 和 Permission 之间定义了多对多关系.用户有一个用户名和县 ID 的主键组合，我的权限表有一个常规整数 ID.表 UserPermission 具有三个外键作为主键:username、countyId 和 permissionId.
I have defined a many-to-many relationship between my two entity classes User and Permission. User has a primary key composite of username and countyId, and my Permission table has a regular integer Id. The table UserPermission has the three foreign keys as its primary key: username, countyId and permissionId.
由于这是一个遗留数据库，我将没有机会做正确的事情 (™) 并在 User 上创建一个整数主键.
Since this is a legacy database, I won't have the opportunity to do the Right Thing(™) and make an integer primary key on User.
我在 User.class 中定义了这样的多对多关系:
I've defined the many-to-many relationship like this in User.class:
@ManyToMany(targetEntity=Permission.class, cascade={ CascadeType.PERSIST, CascadeType.MERGE } )
@JoinTable(name="tblUserPermission",
joinColumns = { @JoinColumn(name="username"), @JoinColumn(name="countyId") },
inverseJoinColumns = { @JoinColumn(name="permissionId") })
private Collection<Permission> permissions;
Permission.class 是这样说的:
Permission.class says this:
@ManyToMany( cascade = {CascadeType.PERSIST, CascadeType.MERGE}, mappedBy = "permissions", targetEntity = User.class )
private Collection<User> users;
我认为这是要走的路，但是当我启动使用 Hibernate 3 的 Spring 上下文时，我得到:
I thought this was the way to go, but when I fire up my Spring context that uses Hibernate 3, I get:
Caused by: org.hibernate.AnnotationException: A Foreign key refering com.mydomain.data.entities.User from com.mydomain.data.entities.Permission has the wrong number of column. should be 1
我在注释中做错了什么?它应该是 2，而不是 1.
What have I done wrong in my annotation? It should be 2, not 1.
更新:
Arthur 建议我添加 referencedColumnName，但这给了我一个新的例外:
Arthur suggested I add referencedColumnName, but that gave me a new exception:
Caused by: org.hibernate.AnnotationException: referencedColumnNames(username, countyId) of com.mydomain.data.entities.Permission.permissions referencing com.mydomain.data.entities.User not mapped to a single property
应他的要求，请遵循以下代码:Permission.Class:
On his request, here follow the code: Permission.Class:
package com.mydomain.data.entities;
import java.io.Serializable;
import java.util.Collection;
import javax.persistence.*;
import org.hibernate.annotations.ForeignKey;
@Entity
@Table(name = "tblPermission")
public class Permission extends PublishableEntityImpl implements Serializable, Cloneable {
private static final long serialVersionUID = 7155322069731920447L;
@Id
@Column(name = "PermissionId", length = 8, nullable = false)
private String PermissionId = "";
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name = "CountyId", nullable = false)
@ForeignKey(name="FK_CountyID")
private County county;
@Column(name = "Permission", nullable = true)
private Integer permission = 1;
@ManyToMany( cascade = {CascadeType.PERSIST, CascadeType.MERGE},
mappedBy = "Permissions",
targetEntity = Item.class )
private Collection<Item> items;
@ManyToMany( cascade = {CascadeType.PERSIST, CascadeType.MERGE},
mappedBy = "Permissions",
targetEntity = User.class )
private Collection<User> users;
/** Getters and Setters **/
}
和 User.class
and User.class
package com.mydomain.data.entities;
import java.util.*;
import java.io.Serializable;
import javax.persistence.*;
import org.hibernate.annotations.ForeignKey;
import org.hibernate.annotations.IndexColumn;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.GrantedAuthorityImpl;
import org.springframework.security.core.userdetails.UserDetails;
@Entity
@Table(name = "tblUser")
public class User extends PublishableEntityImpl implements Serializable, Cloneable {
@Id
@Column(name = "CountyId", nullable = false)
private Integer countyId;
@Id
@Column(name = "Username", length = 25, nullable = false)
private String username;
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name = "CountyId", nullable = false, insertable=false, updatable=false)
@ForeignKey(name="FK_CountyID")
private County county;
@Column(name = "Name", length = 50, nullable = true)
private String name;
@Column(name = "Password", length = 30, nullable = true)
private String password;
@Column(name = "Role", nullable = false)
private Integer role;
@ManyToMany(targetEntity=Permission.class,
cascade={ CascadeType.PERSIST, CascadeType.MERGE } )
@JoinTable(name="tblUserPermission",
joinColumns = { @JoinColumn(name="Username", referencedColumnName="Username"), @JoinColumn(name="CountyId", referencedColumnName="CountyId") },
inverseJoinColumns = { @JoinColumn(name="PermissionId", referencedColumnName="PermissionId") })
private Collection<Permission> permissions;
@OneToMany(fetch=FetchType.LAZY, mappedBy="county")
@IndexColumn(name="version")
private List<Version> versions;
/** Getters and setters **/
}
干杯
尼克
推荐答案
为了解决referencedColumnName异常
In order to solve referencedColumnName exception
用户输入
@ManyToMany(cascade={CascadeType.PERSIST, cascadeType.MERGE})
private Collection<Permission> permissions;
并且在许可中
@ManyToMany(mappedBy="permissions")
@JoinTable(name="tblUserPermission",
joinColumns={@JoinColumn(name="permissionId", referencedColumnName="permissionId")},
inverseJoinColumns={
@JoinColumn(name="username", referencedColumnName="username"),
@JoinColumn(name="countyId", referencedColumnName="countyId")})
private Collection<User> users;
UserId 类
public class UserId implements Serializable {
private String username;
private Integer countyId;
// getter's and setter's
public boolean equals(Object o) {
if(o == null)
return false;
if(!(o instanceof UserId))
return false;
UserId id = (UserId) o;
if(!(getUsername().equals(id.getUsername()))
return false;
if(!(getCountyId().equals(id.getCountyId()))
return false;
return true;
}
public int hachcode() {
// hashcode
}
}
然后在用户类中放
@Entity
@Table(name="tblUser")
@IdClass(UserId.class)
public class User ... {
@Id
private String username;
@Id
private Integer countyId;
}
问候，
相关文章