Java 休眠一对多,操作员不';t exist(bytea=integer)
我正在尝试使用一对多关系,但由于以下错误,我一直无法使用:Java 休眠一对多,操作员不';t exist(bytea=integer),java,database,hibernate,jpa,hibernate-mapping,Java,Database,Hibernate,Jpa,Hibernate Mapping,我正在尝试使用一对多关系,但由于以下错误,我一直无法使用: 20:17:54,621 INFO [stdout] (default task-2) Hibernate: select user0_.user_id as user_id1_1_, user0_.nick as nick2_1_, user0_.password as password3_1_ from users user0_ 20:17:54,655 INFO [stdout] (default task-2) Hiber
20:17:54,621 INFO [stdout] (default task-2) Hibernate: select user0_.user_id as user_id1_1_, user0_.nick as nick2_1_, user0_.password as password3_1_ from users user0_
20:17:54,655 INFO [stdout] (default task-2) Hibernate: select userrole0_.users as users3_1_0_, userrole0_.id as id1_0_0_, userrole0_.id as id1_0_1_, userrole0_.role as role2_0_1_, userrole0_.users as users3_0_1_ from user_roles userrole0_ where userrole0_.users=?
20:17:54,657 WARN [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (default task-2) SQL Error: 0, SQLState: 42883
20:17:54,658 ERROR [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (default task-2) BŁĄD: operator ndoesnt exist: bytea = integer
我不知道为什么会有这种例外。我在任何地方都不使用bytea
操作符。下面,我提供了源代码:
User.java
@Entity
@Table(name = "users")
public class User implements Serializable{
private static final long serialVersionUID = 2051614598479375020L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "user_id")
private int id;
@Column(name = "nick")
private String nick;
@Column(name = "password")
private String password;
@OneToMany(fetch = FetchType.EAGER, mappedBy = "users")
private Set<UserRole> userRole = new HashSet<UserRole>();
public User(){}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getNick() {
return nick;
}
public void setNick(String nick) {
this.nick = nick;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public Set<UserRole> getUserRole() {
return userRole;
}
public void setUserRole(Set<UserRole> userRole) {
this.userRole = userRole;
}
}
我假设这是一个简单的问题,但我不知道如何解决。用户角色使用字段访问,因为@Id放在字段上。用户
@ManyToOne
关联被放置在属性上。您还需要将其移动到字段级别:
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "user_id", nullable = false)
private User user;
这导致了你的问题。当Hibernate加载UserRole时,用户表FK是从未注释的User
字段加载的,因此使用了null
我还将fetch属性更改为LAZY,因为急切的抓取是无效的
另一个小变化与用户
标识符类型有关。它应该使用Integer
而不是int
:
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "user_id")
private Integer id;
使用包装器可以区分临时实体和附加/分离的实体。这两个表上的键列是否具有不同的数据类型?请直接在数据库中检查此列的数据类型。
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "user_id")
private Integer id;