Hibernate一对多,运算符不存在(bytea =整数)

Question

我试图使用一对多的关系但我因为这个错误而卡住了:

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) B??D: operator ndoesnt exist: bytea = integer

我不知道为什么我会得到这个例外.我不在bytea任何地方使用运营商.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;
    }

}

和UserRole.java

@Entity
@Table(name = "user_roles")
public class UserRole implements Serializable{

    private static final long serialVersionUID = 6881696293585198874L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    private Integer id;

    private User users;

    @Column(name = "role")
    private String role;


    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "user_id", nullable = false)
    public User getUser() {
        return users;
    }

    public void setUser(User users) {
        this.users = users;
    }

    public String getRole() {
        return role;
    }

    public void setRole(String role) {
        this.role = role;
    }   
}

我认为这是一个容易出问题,但我不知道如何解决它.

Answer 1

在UserRole使用现场访问,因为@Id被放置在一个领域.所述用户 @ManyToOne关联放置在属性.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;

这导致了您的问题.当Hibernate加载UserRole时,User表FK user是从未注释的字段加载的,因此使用了null.

我还将fetch属性更改为LAZY,因为EAGER提取是一个坏主意.

另一个微小变化与User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;

使用包装器可以让您区分瞬态实体和附加/分离实体.