问题描述
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,