在使用Hibernate针对数据库架构验证一组JPA实体时出现以下错误:
Caused by: org.hibernate.HibernateException: Wrong column type in public.postal_code for column country. Found: bpchar, expected: bytea
at org.hibernate.mapping.Table.validateColumns(Table.java:282)
at org.hibernate.cfg.Configuration.validateSchema(Configuration.java:1268)
at org.hibernate.tool.hbm2ddl.SchemaValidator.validate(SchemaValidator.java:155)
at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:460)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1737)
at org.hibernate.ejb.EntityManagerFactoryImpl.<init>(EntityManagerFactoryImpl.java:84)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:904)
... 9 more
底层数据库是PostgreSQL 9.1,因此定义了相关的数据库表:
CREATE TABLE country
(
code_alpha2 character(2) NOT NULL, -- ISO 3166 alpha2 code
code_alpha3 character(3), -- ISO 3166 alpha3 code
CONSTRAINT country_pkey PRIMARY KEY (code_alpha2)
)
WITH (
OIDS=FALSE
);
CREATE TABLE postal_code
(
country character(2) NOT NULL, -- ISO 3166 alpha2 country-code
code character varying(12) NOT NULL, -- Postal code proper
CONSTRAINT postal_code_pk PRIMARY KEY (country, code),
CONSTRAINT country_fk FOREIGN KEY (country)
REFERENCES country (code_alpha2) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
WITH (
OIDS=FALSE
);
实体定义如下:
@Entity
public class Country implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@Column(name="code_alpha2", columnDefinition="bpchar")
private String codeAlpha2;
@Column(name="code_alpha3", columnDefinition="bpchar")
private String codeAlpha3;
public Country() {
}
public String getCodeAlpha2() {
return this.codeAlpha2;
}
public void setCodeAlpha2(String codeAlpha2) {
this.codeAlpha2 = codeAlpha2;
}
public String getCodeAlpha3() {
return this.codeAlpha3;
}
public void setCodeAlpha3(String codeAlpha3) {
this.codeAlpha3 = codeAlpha3;
}
}
@Entity
@IdClass(PostalCodePK.class)
@Table(name="postal_code")
public class PostalCode implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@ManyToOne(fetch=FetchType.EAGER)
@JoinColumn(name="country")
private Country country;
@Id
private String code;
public PostalCode() {
}
public Country getCountry() {
return country;
}
public void setCountry(Country country) {
this.country = country;
}
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
}
最后,定义postal_code主键的类:
@Embeddable
public class PostalCodePK implements Serializable {
//default serial version id, required for serializable classes.
private static final long serialVersionUID = 1L;
@Column(columnDefinition="bpchar")
private Country country;
private String code;
public PostalCodePK() {
}
public Country getCountry()
return this.country;
}
public void setCountry(Country country) {
this.country = country;
}
public String getCode() {
return this.code;
}
public void setCode(String code) {
this.code = code;
}
public boolean equals(Object other) {
if (this == other) {
return true;
}
if (!(other instanceof PostalCodePK)) {
return false;
}
PostalCodePK castOther = (PostalCodePK)other;
return
this.country.equals(castOther.country)
&& this.code.equals(castOther.code);
}
public int hashCode() {
final int prime = 31;
int hash = 17;
hash = hash * prime + this.country.hashCode();
hash = hash * prime + this.code.hashCode();
return hash;
}
}
为什么Hibernate期望在列国家中使用bytea?我怎样才能说服验证者按原样接受架构?
答案 0 :(得分:1)
回答我自己的问题:Hibernate不是将依赖关系视为一个实体,而是将其作为一个对象进行处理,并将其存储在数据库中。解决方案是将PostalCodePK中字段“country”的数据类型更改为String并切换到使用嵌入ID,同时将注释@MapsId添加到PostalCode中的字段“country”:
@EmbeddedId
PostalCodePK id;
@MapsId("country")
@ManyToOne
@JoinColumn(name="country")
private Country country;