我试图在Person和Auth表之间创建OnetoOne关系.问题是当生成DB表“Auth”时,我没有在AUTH表中看到应该引用Person的外键.对象是让Auth表使用Person Table的相同主键.
@MappedSuperclass
public abstract class DomainBase {
@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE)
private Long id;
@Version
@Column(name="OPLOCK")
private Integer version;
}
@Entity
@Table(name = "person")
public class Person extends DomainBase {
@OnetoOne(cascade=CascadeType.ALL)
@JoinColumn(name="auth_id")
private Auth auth;
}
@Entity
public class Auth {
@Id
@GeneratedValue(generator="foreign")
@GenericGenerator(name="foreign", strategy = "foreign", parameters={
@Parameter(name="property", value="person")
})
@Column(name="person_id")
private int personId;
---------------------------------
@OnetoOne(cascade = CascadeType.ALL)
@PrimaryKeyJoinColumn
private Person person;
}
CREATE TABLE auth
(
person_id integer NOT NULL,
activate boolean,
activationid character varying(255),
last_login_attempt_date timestamp without time zone,
last_login_attempt_timezone character varying(255),
last_login_date timestamp without time zone,
last_login_timezone character varying(255),
nonlocked boolean,
num_login_attempts integer,
CONSTRAINT auth_pkey PRIMARY KEY (person_id),
CONSTRAINT uk_d68auh3xsosyrjw3vmwseawvt UNIQUE (activationid)
)
WITH (
OIDS=FALSE
);
ALTER TABLE auth
OWNER TO postgres;
解决方法:
似乎问题是你在“person”表和“auth”表之间声明两次@OnetoOne注释,而没有指定它们之间的关系.看一下hibernate documentation,在2.2.5.1点,有一些关于使用一对一关联的例子.
对我来说,最好的方法是在一个表中设置关联,一个声明foreing键列,另一个在另一个对象中使用mappedBy参数.在您的代码中,这将是:
@Entity
@Table(name = "person")
public class Person extends DomainBase {
@OnetoOne(cascade=CascadeType.ALL)
@JoinColumn(name="auth_id")
private Auth auth;
}
@Entity
public class Auth {
@Id
@GeneratedValue(generator="foreign")
@GenericGenerator(name="foreign", strategy = "foreign", parameters={
@Parameter(name="property", value="person")
})
@Column(name="person_id")
private int personId;
@OnetoOne(mappedBy = "auth")
private Person person;
....
}
这是hibernate文档中的第二个示例,在句子“在下面的示例中,关联实体通过显式外键列链接”之后引入.我测试了这段代码,出现了“auth_id”列.
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