休眠：外键具有错误的列数 [英] Hibernate: Foreign key has the wrong number of columns

问题描述

我定义了两个实体类User和Permission之间的多对多关系。用户具有用户名和县ID的主键组合，并且我的权限表具有常规整数ID。表UserPermission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，所以我不会有机会做正确的事情（™），并在用户上建立一个整数主键。

我已经在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>权限; 
  

Permission.class表示：

 @ManyToMany（cascade = {CascadeType.PERSIST，CascadeType.MERGE}，mappedBy =permissions，targetEntity = User.class）
 private Collection< User>用户; 
  

我想这是要走的路，但是当我启动使用Hibernate 3的Spring上下文，我得到：
$ b $ pre $ 引起：org.hibernate.AnnotationException：引用com.mydomain.data.entities.User的外键来自com.mydomain.data.entities.Permission的列号错误。应该是1


我在注释中做了什么错误？它应该是2，而不是1。

---

更新：

Arthur建议我添加referencedColumnName，但是这给了我一个新的异常：

pre $ 引起：org.hibernate.AnnotationException：referencedColumnNames（用户名，县级）com.mydomain.data.entities.Permission.permissions引用com.mydomain.data.entities.User没有映射到一个单一的属性


在他的请求，这里按照代码：
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 =; 
 $ b @ManyToOne（fetch = FetchType.LAZY）
 @JoinColumn（name =CountyId，nullable = false）
 @ForeignKey（name =FK_CountyID）
县私立县; 
 
 @Column（name =Permission，nullable = true）
 private Integer权限= 1; 
 
 @ManyToMany（cascade = {CascadeType.PERSIST，CascadeType.MERGE}，
 mappedBy =Permissions，
 targetEntity = Item.class）
 private Collection<项>项目; 
 
 @ManyToMany（cascade = {CascadeType.PERSIST，CascadeType.MERGE}，
 mappedBy =Permissions，
 targetEntity = User.class）
 private Collection<使用者名称>用户; 
 $ b $ ** Getters and Setters ** / 
} 
  

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; 
导入org.springframework.security.core.userdetails.UserDetails; 
 $ b $实体
 @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; 
 $ b @ManyToOne（fetch = FetchType.LAZY）
 @JoinColumn（name =CountyId，nullable = false，insertable = false，updatable = false）
 @ForeignKey =FK_CountyID）
私人郡县; 
 
 @Column（name =Name，length = 50，nullable = true）
私有字符串名称; 
 
 @Column（name =Password，length = 30，nullable = true）
 private String password; 
 
 @Column（name =Role，nullable = false）
私人整数角色; 
 $ b @ManyToMany（targetEntity = Permission.class，
 cascade = {CascadeType.PERSIST，CascadeType.MERGE}）
 @JoinTable（name =tblUserPermission，
 joinColumn（name =Username，referencedColumnName =Username），@JoinColumn（name =CountyId，referencedColumnName =CountyId）}，
 inverseJoinColumns = {@JoinColumn（name =PermissionId ，referencedColumnName =PermissionId）}）
 private Collection< Permission>权限; 
 
 @OneToMany（fetch = FetchType.LAZY，mappedBy =county）
 @IndexColumn（name =version）
 private List< Version>版本; 
 $ b $ ** Getters and setters ** / 
} 
  

干杯

Nik

解决方案为了解决referencedColumnName异常

在用户放入

  @ManyToMany（cascade = {CascadeType.PERSIST ，cascadeType.MERGE}）
 private Collection< Permission>权限; 
  

并在权限中

  @ManyToMany（mappedBy =permissions）
 @JoinTable（name =tblUserPermission，
 joinColumns = {@ JoinColumn（name =permissionId，referencedColumnName =permissionId） }，
 inverseJoinColumn = {
 @JoinColumn（name =username，referencedColumnName =username），
 @JoinColumn（name =countyId，referencedColumnName =countyId）}） 
私人收藏<使用者>用户; 
  

UserId class

  public class UserId实现Serializable {
 
私有字符串用户名; 
 
 private Integer countyId; 
 
 // getter's和setter's 
 
 public boolean equals（Object o）{
 
 if（o == null）
 return false ; 
 
 if（！（o instanceof UserId））
 return false; 
 
 UserId id =（UserId）o; （！（getUsername（）。equals（id.getUsername（）））
返回false; 
 $ b如果（！（getCountyId（）。equals（id.getCountyId ）））
返回false; 
 
返回true; 
} 
 
 public int hach​​code（）{
 // hashcode 
} 
 
} 
  

然后在User class放置

  @Entity 
 @Table（name =tblUser）
 @IdClass（UserId.class）
 public class User ... {
 
 @Id 
私有字符串用户名; 
 
 @Id 
私有整数县ID; 
 
 } 
  

问候，

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.

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.

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 says this:

@ManyToMany( cascade = {CascadeType.PERSIST, CascadeType.MERGE}, mappedBy = "permissions", targetEntity = User.class )
private Collection<User> users;

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

What have I done wrong in my annotation? It should be 2, not 1.

---

Update:

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

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 **/
}

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 **/
}

Cheers

Nik

解决方案

In order to solve referencedColumnName exception

In User put

@ManyToMany(cascade={CascadeType.PERSIST, cascadeType.MERGE})
private Collection<Permission> permissions;

And in Permission

@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 class

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
    }

}

Then in User class put

@Entity
@Table(name="tblUser")
@IdClass(UserId.class)
public class User ... {

    @Id
    private String username;

    @Id
    private Integer countyId;

}

regards,

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