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使用 Hibernate 针对数据库模式验证一组 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，相关的数据库表是这样定义的：

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 在列国期待一个 bytea？以及如何说服验证器按原样接受架构？

4

1 回答 1

1

回答我自己的问题：Hibernate 不将依赖项视为实体，而是将其视为要序列化并存储在数据库中的对象。解决方案是将 PostalCodePK 中“country”字段的数据类型更改为 String，并切换到使用嵌入式 ID，同时将注释 @MapsId 添加到 PostalCode 中的“country”字段：

    @EmbeddedId
    PostalCodePK id;

    @MapsId("country")
    @ManyToOne
    @JoinColumn(name="country")
    private Country country;
于 2013-08-05T10:07:06.853 回答