Hibernate 一对多，运算符不存在（bytea = integer）

Question

我正在尝试使用一对多关系，但由于这个错误而卡住了：

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) BŁĄD: operator ndoesnt exist: bytea = integer

我不知道为什么会出现此异常。我不在任何地方使用 bytea 运算符。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;
    }

}

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;
    }   
}

我认为这是一个简单的问题，但我不知道如何解决它。

Answer 1

UserRole使用的是field-access，因为@Id是放在一个field上的。 user @ManyToOne 关联被放置在属性上。您也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 table FK 是从未注释的 user 字段加载的，因此使用了 null。

我还将提取属性更改为 LAZY，因为 EAGER 提取是 a bad idea。

另一个小改动与 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 attached/detached 实体。

Hibernate 一对多，运算符不存在（bytea = integer）

Hibernate one-to-many , operator doesn't exist ( bytea = integer )

java

database

hibernate

jpa

hibernate-mapping