@Enumerated(EnumType.STRING)生成int列而不是varchar

Question

我将spring jpa与hibernate和ms Sqlserver结合使用。我有一个实体，它有一个带有@Enumerated(EnumType.STRING)注释的枚举属性。生成数据库架构时，用枚举值填充的列具有int类型而不是char类型。

设置数据库属性的Bean方法：

private Properties getHibernateProperties() {
    Properties properties = new Properties();
    properties.setProperty("hibernate.dialect","org.hibernate.dialect.SQLServer2012Dialect");

    properties.setProperty("hibernate.hbm2ddl.auto","create-drop");       
    properties.setProperty("ejb.naming_strategy","layer.controller.configuration.PhysicalNamingStrategyImpl");
    properties.setProperty("hibernate.search.default.directory_provider","filesystem");


    return properties;
}

@Bean
public LocalContainerEntityManagerFactoryBean entityManagerFactory() {

    HibernateJpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
    vendorAdapter.setShowSql(Boolean.TRUE);        
    LocalContainerEntityManagerFactoryBean factory = new LocalContainerEntityManagerFactoryBean();
    factory.setJpaVendorAdapter(vendorAdapter);
    factory.setDataSource(dataSource());
    factory.setPackagesToScan("layer.domain","layer.application.security.authentication.entities");
    factory.setJpaProperties(getHibernateProperties());
    factory.afterPropertiesSet();

    return factory;
}

枚举和实体：

public enum Elenco {
    A,
    B,
    C,
    D;
}

@Entity
@Table(name="elenco")
public class ElencoClass extends BaseEntity<Long> {

    @Column(name="nomeECognome")
    private String nome;


    @Enumerated(EnumType.STRING)
    @Column(columnDefinition = "varchar")
    private Elenco ele;


    @Id
    @Column(name="id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Override
    protected Long getId() {
        return id;
    }

    @Override
    protected void setId(Long id) {

        super.id=id;
    }

    public String getNome() {
        return nome;
    }

    public void setNome(String nome) {
        this.nome = nome;
    }

    public Elenco getEle() {
        return ele;
    }

    public void setEle(Elenco ele) {
        this.ele = ele;
    }
}

这是spring生成的表：

 CREATE TABLE [dbo].[elenco](
    [id] [bigint] IDENTITY(1,1) NOT NULL,
    [ele] [int] NULL,
    [nome] [varchar](255) NULL,

Answer 1

如果您可以在getter方法上编写@Enumerated(EnumType.STRING)，那么它将会工作。

Answer 2

你试过这个吗？您仍然需要在get方法中使用@Column注释来指定列名，而不是列定义。

@Enumerated(EnumType.STRING)
@Column(name="ele")
public Elenco getEle() {...}