我已经定义了我的两个实体类User和Permission之间的多对多关系.用户具有username和countyId的主键组合,我的Permission表具有常规整数Id.表UserPermission具有三个外键作为其主键:username,countyId和permissionId.
由于这是一个遗留数据库,我将无法使用Right Thing(™)并在User上创建一个整数主键.
我在User.class中定义了这样的多对多关系:
@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.class说:
@ManyToMany( cascade = {CascadeType.PERSIST,CascadeType.MERGE},mappedBy = "permissions",targetEntity = User.class )
private Collection
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:
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?应该是2,而不是1.
更新:
Arthur建议我添加referencedColumnName,但这给了我一个新的异常:
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
根据他的要求,请遵循以下代码:
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,mappedBy = "Permissions",targetEntity = Item.class )
private Collection
和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,CascadeType.MERGE } )
@JoinTable(name="tblUserPermission",joinColumns = { @JoinColumn(name="Username",referencedColumnName="Username"),@JoinColumn(name="CountyId",referencedColumnName="CountyId") },inverseJoinColumns = { @JoinColumn(name="PermissionId",referencedColumnName="PermissionId") })
private Collection
干杯
聂
最佳答案
为了解决referencedColumnName异常
原文链接:/spring/432522.html在用户放
@ManyToMany(cascade={CascadeType.PERSIST,cascadeType.MERGE})
private Collection
并在权限
@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
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User类中放
@Entity
@Table(name="tblUser")
@IdClass(UserId.class)
public class User ... {
@Id
private String username;
@Id
private Integer countyId;
}
问候,